Higher Mathematics for Students of

Chemistry and Physics

Joseph William Mellor 1869-1938

Chapter VII. How to Solve Differential Equations

Preface to 2009 electronic edition

THE PRIMARY content of this e-book is chapter VII of the 1902 editionof Higher Mathematics for Students of Chemistry and Physics (with spe-cial reference to practical work) by Joseph William Mellor, D.Sc of NewZealand Univerity (published by Longmans, Green, and Company of Lon-don). Chapter VII of this text is an 80 page missive entitled, “How toSolve Differential Equations.” The text is primarily concerned with teach-ing practical methods for arriving at solutions, and contains almost nothingon the theory of differential equations – that is, there are very few theo-rems and proofs. It covers a wide variety of solvable ordinary differentialequations with over a hundred examples with solutions. The later sectionsalso provide an introductory overview of partial differential equations.

Included also is a supplement I have prepared with many of the exam-ples in the main text worked out step-by-step. Some of them I have workedin more than one way. Students using this material are encouraged to at-tempt each example in the main text on their own before referring to thesupplement for my working of that example. In addition the supplementcontains my comments on the main text, as well as descriptions of methodsof solution not covered in the main text. No student should expect to learnthe material here well without working the examples for himself or herself.

Prerequisite skills: Students using this material should be competentin finding derivatives, including applying the product, quotient, and chainrules, in methods for finding integrals, and in the manipulation of poly-nomials, roots, logs, exponentials, and trigonometric functions. Studentsshould also have had at least some exposure to the differential calculus offunctions of more than one variable and partial derivatives.

The text makes scattered use of complex numbers, as does the supple-ment. So students should be familiar with the algebra of complex numbers.They should also be familiar with Euler’s formula relating exponentials ofa complex variable to the sine and cosine functions:

eιx = cos x+ ι sin x hence cosx =eιx + e−ιx

2and sin x =

eιx − e−ιx

2ι.

i

Note the usage of the symbol, ι, for√−1. This notation is used through-

out the text. Most modern texts use the symbol, i, in lieu of ι for this pur-pose. I have continued Mellor’s notation, using ι in the supplement in orderto remain consistent. Mellor uses the adjective, “imaginary,” to describeany non-real complex number. This is at odds with modern nomenclature,in which “imaginary” refers only to real multiples of ι. In the supplement, Iuse the phrase, “purely imaginary,” to indicate such multiples and to makethe distinction from Mellor’s looser usage.

More on notation: Throughout the text and the supplement, the no-tation, log (x), is always to be understood as logarithm to the base, e, oth-erwise known as the natural log. Most modern texts use the notation, ln (x),for natural log. The website, http://jeff560.tripod.com/functions.html, in-dicates that the ln (x) notation was not introduced until 1893, so at thetime of publication of this text, that notation was still new and not yetwidely adopted. The log (x) notation, according to the same source, hadbeen in use since 1748. At any rate, the text uses the older notation, andI have stuck with it in the supplement.

The text consistently uses the following hierarchy for grouping of terms:[{()}]. Again I have stuck with that style in the supplement.

The original text used an unfamiliar notation for limits. Since its nota-tion is no longer in common use, I have converted it to the modern notation,e.g.,

limh→0

sin (h)

h= 1.

Errors in the original text: In preparing the electronic version ofthe text from the original printed material (and especially when I preparedthe supplement), I discovered many errors in the text. Some of these weretype-setting errors, but others were the result of carelessness on the partof the author. Where mistakes were obvious, I have simply corrected themwithout comment. Where there were mistakes of carelessness in the exam-ples and their solutions, I have, in many cases, left the mistake unaltered,but I have tagged it in the right-hand margin with a left-facing arrow.Where you encounter the arrow, you will find my reasons for believing thatthe text is in error in the supplement, together with what I believe thetext should have been. Serious students will attempt to find the error ontheir own before referring to the supplement. In many other places I havealso inserted comments in the right-hand margin, many of which refer thestudent to the supplement for further clarification.

The page numbering of the original printed text is preserved in thiselectronic version. The supplement uses its own page numbering, startingwith S-1, which follows page 359 of the main text. Following that arevarious excerpts from other chapters of the main text that are referred toin chapter VII.

The supplement includes three final sections that are not specificallyreferenced from example problems in the main text. These are Resonantsystems revisited starting on page S-61, which describes how resonant

ii

systems respond to sinusoidal input, Differential equation solution toKeplerian orbits starting on page S-66, which develops the shape of agravitational orbit in a two body problem, and Introduction to Fourier’sheat equation and its solution starting on page S-67, which developsmethods for dealing with a one-dimensional heat equation. These have beenadded because they are important applications of the methods taught inthis tutorial and may be of interest to some students.

The text by J.W. Mellor passes into public domain on thefirst day of the year 2009. My comments and supplement are offeredwithout charge to all under Creative Commons 3.0 license. Seehttp://creativecommons.org/licenses/by/3.0/ for details. I am often receiv-ing emails from students living in Africa, the Middle East, or South CentralAsia, who are eager to learn mathematics but for whom the cost of mod-ern textbooks is burdensome. One of the reasons I have gone to the effortto prepare an electronic edition of this material is for them. If you are amath educator working in those parts of the world and teaching differentialequations, please consider using this material as part of your curriculum soas to allow your students affordable access to the knowledge it provides.

LATEX sources for this material will be made available, free of charge,to anybody who can demonstrate that he or she will put them to gooduse. Translation of the material into languages other than English willalways be considered to be a good use. Email me for details.

The probability that I have transcribed all of this material without erroris vanishingly close to zero. If you discover anything in this document thatappears to you to be in error, please notify me by email so that I maycorrect it.

Karl Hahn, December 2008

iii

Mellor’s Preface to the 1902 edition

IT is almost impossible to follow the later developments of physical orgeneral chemistry without a working knowledge of higher mathematics. Ihave found that the regular textbooks of mathematics rather perplex thanassist the chemical student who seeks a short road to this knowledge, forit is not easy to discover the relation which the pure abstractions of formalmathematics bear to the problems which every day confront the student ofNature’s laws, and realize the complementary character of mathematicaland physical processes.

During the last five years I have taken note of the chief difficultiesmet with in the application of the mathematicians x and y to physicalchemistry, and, as these notes have grown, I have sought to make clearhow experimental results lend themselves to mathematical treatment. Ihave found by trial that it is possible to interest chemical students and togive them a working knowledge of mathematics by manipulating the resultsof physical or chemical observations.

I should have hesitated to proceed beyond this experimental stage if Ihad not found at The Owens College a set of students eagerly pursuingwork in different branches of physical chemistry, and most of them lookingfor help in the discussion of their results. When I told my plan to theProfessor of Chemistry he encouraged me to write this book. It has beenmy aim to carry out his suggestion, so I quote his letter as giving the spiritof the book, which I only wish I could have carried out to the letter.

“THE OWENS COLLEGE,MANCHESTER.”

“MY DEAR MELLOR,“If you will convert your ideas into words and write a book explaining the inwardness

of mathematical operations as applied to chemical results, I believe you will confer abenefit on many students of chemistry. We chemists, as a tribe, fight shy of any symbolsbut our own. I know very well you have the power of winning new results in chemistryand discussing them mathematically. Can you lead us up the high hill by gentle slopes?Talk to us chemically to beguile the way? Dose us, if need be, ‘with learning put lightly,like powder in jam’? If you feel you have it in you to lead the way we will try to follow,and perhaps some of the youngest of us may succeed. Wouldn’t this be a triumph worthworking for? Try.

“Yours very truly,“H.B. DIXON”

THE OWENS COLLEGE,MANCHESTER, May, 1901

iv

CHAPTER VII.

HOW TO SOLVE DIFFERENTIAL EQUATIONS.

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282§ 117. Solution by Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283§ 118. What is a Differential Equation? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286§ 119. Exact Differential Equations of First Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289§ 120. How to find Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .292§ 121. The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295§ 122. Linear Differential Equations of First Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .296§ 123. Equations of First Order and Higher Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298§ 124. Clairault’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300§ 125. Singular Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301§ 126. Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304§ 127. Symbols of Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .304§ 128. The Linear Equation of nth Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305§ 129. Linear Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307§ 130. How to find Particular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310§ 131. The Linear Equation with Variable Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 315§ 132. The Exact Linear Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317§ 133. The Integration of Equations with Missing Terms . . . . . . . . . . . . . . . . . . . . . . . . . 319§ 134. Equations of Motion, Chiefly Oscillatory Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 322§ 135. The Velocity of Simultaneous and Dependent Chemical Reactions . . . . . . . . .330§ 136. Simultaneous Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336§ 137. Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339§ 138. What is the Solution of a Partial Differential Equation? . . . . . . . . . . . . . . . . . . 341§ 139. The Solution of Partial Diff. Equations of First Order . . . . . . . . . . . . . . . . . . . . 344§ 140. Partial Differential Equations of nth Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346§ 141. Linear Partial Diff. Equations with Constant Coefficients . . . . . . . . . . . . . . . . . 347§ 142. The Particular Integral of Linear Partial Diff. Equations . . . . . . . . . . . . . . . . . . 351§ 143. The Linear Partial Equation with Variable Coefficients . . . . . . . . . . . . . . . . . . . 354§ 144. The Integration of Differential Equations in Series . . . . . . . . . . . . . . . . . . . . . . . . 355§ 145. Harmonic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357

Supplemental material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . begins after page 359The Principle of Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S-24The Method of Laplace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .S-31Resonant Systems Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S-61Solution to Keplerian Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .S-66Introduction to Fourier’s Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S-67

Selected sections referenced in chapter VII . . . . . . . . . . . . . . begins after page S-69§ 20. Leibniz’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49§ 22. Euler’s Theorem of Homogeneous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56§ 23. Successive Partial Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57§ 24. Exact Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57§ 25. Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .58§ 67. Envelopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .142§ 72. How to Find a Value for the Integration Constant . . . . . . . . . . . . . . . . . . . . . . . . . . 162

v

Public domain rationale: J.W. Mellor was a British subject and his works originally published byBritish publishers. Hence under British copyright law’s principle of “life plus 70 years,” the works ofJ.W. Mellor passed into the public domain on 1-January-2009.

CHAPTER VII.HOW TO SOLVE DIFFERENTIAL EQUATIONS.

THIS chapter may be looked upon as a sequel to that on the integral cal-culus, but of a more advanced character. The “methods of integration”already described will be found ample for most physico-chemical processes,but chemists are proving every day that more powerful methods will soonhave to be brought in. As an illustration, I may refer to the set of dif-ferential equations which Geitel encountered in his study of the velocity ofhydrolysis of the triglycerides by acetic acid (Journal fur praktische Chemie[2], 55, 429, 1897).

I have previously pointed out that in the effort to find the relationsbetween phenomena, the attempt is made to prove that if a limited numberof hypotheses are prevised, the observed facts are a necessary consequenceof these assumptions. The modus operandi is as follows:

1. To “anticipate Nature” by means of a “working hypothesis,” whichis possibly nothing more than a “convenient, fiction”.

“From the practical point of view,” says Professor Bucker (Presidential Address tothe B. A. meeting at Glasgow, September, 1901), “it is a matter of secondary importancewhether our theories and assumptions are correct, if only they guide us to results inaccord with facts. . . . By their aid we can foresee the results of combinations of causeswhich would otherwise elude us.”

2. Thence to deduce an equation representing the momentary rate ofchange of the two variables under investigation.

3. Then to integrate the equation so obtained in order to reproduce the“working hypothesis” in a mathematical form suitable for experimentalverification (see §§ 18, 69, 88, 89, and elsewhere).

So far as we are concerned this is the ultimate object of our integration.By the process of integration we are said to solve the equation.

282

For the sake of convenience, any equation containing differentials or differ-ential coefficients will, after this, be called a, differential equation.

§ 117. The Solution of a Differential Equation by the Separa-tion of the Variables. The different equations hitherto considered haverequired but little preliminary arrangement before integration. For exam-ple, when preparing the equations representing the velocity of a chemicalreaction of the general type:

dx

dt= kf (t) , (1)

we have invariably collected all the x’s to one side, the t’s, to the other,before proceeding to the integration.

This separation of the variables is nearly always attempted before re-sorting to other artifices for the solution of the differential equation, becausethe integration is then comparatively simple. The following examples willserve to emphasise these remarks:

EXAMPLES. – (1) Integrate the equation, y dx + x dy = 0. Rearrange the terms sothat

dx

x+dy

y= 0; or,

∫

dx

x+

∫

dy

y= C,

by multiplying through with 1/xy. Ansr. log x + log y = C. Two or more apparentlydifferent answers may be the same. Thus, the solution of the preceding equation mayalso be written,

log xy = log eC , i.e., xy = eC ; or log xy = logC′, i.e., xy = C′.

C and logC′ are, of course, the arbitrary constants of integration.(2) The equation for the rectilinear motion of a particle under the influence of an

attractive force from a fixed point is

vdv

dx+

µ

x2= 0.

Solve. Ansr. 12v

2 = µx + C.

(3) Solve(

1 + x2)

dy =√y · dx. Ansr. 2√y − tan−1 x = C.

(4) Solve y − x · dydx = a(

y + dydx

)

. Ansr. y = C (a+ x)1−a

.

(5) In consequence of imperfect insulation, the charge on an electrified body isdissipated at a rate proportional to the magnitude R of the charge. Hence show that ifa is a constant depending on the nature of the body, and R, represents the magnitudeof the charge when t (time) = 0,

E = E0e−at.

Hint. Compound interest law. Integrate by the separation of the variables. Interpretyour result.

(6) Abegg’s formula for the relation between the dielectric constant (D) of a fluidand temperature θ, is

−dDdθ

=D

190.

283

Hence show that D = Ceθ/190, where C is a constant whose value is to be determinedfrom the conditions of the experiment. Put the answer into words.

(7) What curves have a slope −x/y to the x-axis? Ansr. The rectangular hyperbolasxy = C. Hint. Set up the proper differential equation and solve. ←−

(8) The relation between small changes of pressure and volume of a gas underadiabatic conditions, is γp dx+ v dp = 0. Hence show that pvγ = constant.

(9) A lecturer discussing the physical properties of substances at very low tem-peratures, remarked “it appears that the specific heat of a substance decreases withdecreasing temperatures at a rate proportional to the specific heat of the substance it-self ”. Set up the differential equation to represent this “law” and put your result in aform suitable for experimental verification.

(10) Helmholtz equation for the strength of an electric current (C) at the time t, is

C =E

R− L

R

dC

dt,

where E represents the electromotive force in a circuit of resistance R and self-inductionL. If E, R, L, are constants, show that RC = E(1 − e−Rt/L) provided C = 0, whent = 0.

A substitution will often enable an equation to be treated by this simplemethod of solution.

EXAMPLE, – Solve (x − y2)dx + 2xy dy = 0. Ansr. xey2/x = C. Hint, put y2 = v,

divide by x2 ∴ dx/x+ d(v/x) = 0, etc.

If the equation is homogeneous in x and y, that is to say, if thesum of the exponents of the variables in each term is of the same degree,a preliminary substitution of x = ty, or y = tx, according to convenience,will always enable variables to be separated. The rule for the substitutionis to treat the differential coefficient which involves the smallest number ofterms.

EXAMPLES, – (1) x+ y dydx − 2y = 0. Substitute y = tx,

∴

∫

t dt

(1− t)2+

∫

dx

x= C; ∴

1

1− t + log x = C.

Ansr. (y − x)ex/(x−y) = C.(2) If (y − x)dy + ydx = 0, y = Ce−x/y.(3) If x2dy − y2dx− xy dx = 0, x = ex/y + C. ←−(4) (x2 + y2)dx = 2xy dy, x2 − y2 = Cx.

Non-homogeneous equations in x and y can be converted into thehomogeneous form by a suitable substitution.

The most general type of a non-homogeneous equation of the first degreeis,

(ax+ by + c) dx+ (a′x+ b′y + c′) dy = 0. (2)

284

To convert this into an homogeneous equation, assume that

x = v + h and y = w + k,

and substitute in the given equation (2). Thus, we obtain

{av + bw + (ah + bk + c)} dv + {a′v + b′w + (a′h + b′k + c′)} dw = 0. (3)

Find h and k so thatah + bk + c = 0; a′h+ b′k + c′ = 0.

∴ h =b′c− bc′

a′b− ab′; k =

ac′ − a′c

a′b− ab′.

(4)

Substitute these values of h and k in (3). The resulting equation

(av + bw) dv + (a′v + b′w) dw = 0 (5)

is homogeneous and, therefore, may be solved as just indicated.

EXAMPLES, – (1) Solve (3y − 7x− 7)dx + (7y − 3x− 3)dy = 0. Ansr.

(y − x− 1)2 (y + x+ 1)5 = C. Hints. From (2), a = −7, b = 3, c = −7, a′ = −3,b′ = 7, c′ = −3. From (4), h = −1, k = 0. Hence from (3),

3w dv − 7v dv + 7w dw − 3v dv = 0.

To solve this homogeneous equation, substitute w = vt, as above, and separate thevariables.

∴ 7dv

v=

3− 7t

t2 − 1dt; ∴ 7

∫

dv

v+

∫

2dt

t− 1+

∫

5dt

t+ 1= C.

∴ 7 log v + 2 log (t− 1) + 5 log (t+ 1) = C; or v7 (t− 1)2(t+ 5)

5= C.

But x = v + h ∴ v = x+ 1; y = w + k ∴ y = w; ∴ t = w/v = y/ (x+ 1), etc.(2) If (2y − x− 1) dy + (2x− y + 1)dx = 0; x2 − xy + y2 + x− y = C.

If in (3),a : b = a′ : b′ = 1 : m (say),

h and k are indeterminate, since (2) then becomes,

(ax+ by + c) dx+ {m (ax+ by) + c′} dy = 0.

The denominators in equations (4) also vanish. In this case put

z = ax+ by

and eliminate y, thus, we obtain,

a+ bz + c

mz + c+

dz

dx= 0, (6)

an equation which allows the variables to be separated.

EXAMPLES, – (1) Solve (2x+ 3y − 5) dy + (2x+ 3y − 1)dx = 0.

Ansr. x+ y − 4 log (2x+ 3y + 7) = C.

(2) Solve (3y + 2x+ 4) dx− (4x+ 6y − 1) dy = 0.

Ansr. 9 log {(21y + 14x+ 22) /7} − 21 (2y − x) = C. ←−When the variables cannot be separated in a satisfactory manner, spe-

cial artifices must be adopted. We shall find it the simplest plan to adoptthe routine method of referring each artifice to the particular class of equa-tion which it is best calculated to solve.

285

These special devices are sometimes far neater and quicker processes ofsolution than the method just described.

We shall follow the conventional x and y rather more closely than inthe earlier part of this work. The reader will know, by this time, that hisx and y’s, his p and v’s and his s and t’s are not to be kept in “water-tightcompartments.” It is perhaps necessary to make a few general remarks onthe nomenclature.

§ 118. What is a Differential Equation? We have seen that thestraight line,

y = mx+ b, (1)

fulfils two special conditions:1. It cuts one of the coordinate axes at a distance b from the origin.2. It makes an angle tan a = m, with the x-axis.By differentiation.

dy

dx= m. (2)

This equation has nothing at all to say about the constant b. That conditionhas been eliminated. Equation (2), therefore, represents a straight linefulfilling one condition, namely, that it makes an angle tan−1m with thex-axis. Now substitute (2) in (1), the resulting equation,

y =dy

dxx+ b, (3)

in virtue of the constant b, satisfies only one definite condition, (3), there-fore, is the equation of any straight line passing through b. Nothing is saidabout the magnitude of the angle tan−1m.

Differentiate (2). The resulting equation,

d 2y

dx2= 0, (4)

represents any straight line whatever. The special conditions imposed bythe constants m and b in (1), have been entirely eliminated. Equation (4) isthe most general equation of a straight line possible, for it may be appliedto any straight line that can be drawn in a plane.

Let us now find a physical meaning for the differential equation.

286

In § 7, we have found that the third differential coefficient, d 3s/dt3

represents “the rate of change of acceleration from moment to moment”.Suppose that the acceleration d 2s/dt2, of a moving body does not changeor vary in any way. It is apparent that the rate of change of a constantor uniform acceleration must be zero. In mathematical language, this iswritten,

d 3s

dt3= 0. (5)

Now integrate this equation once. We obtain

d 2s

dt2= constant, say, g. (6)

Equation (6) tells us not only that the acceleration is constant, but it fixesthat value to the definite magnitude g ft. per second.

But acceleration measures the rate of change of velocity. Integrate (6),we get,

ds

dt= gt+ C1. (7)

From § 72, we have learnt how to find the meaning of C1. Put t = 0, then § 72 isincludedin thisdocu-mentfollowingthe sup-plement.

dx/dt = C1. This means that when we begin to reckon the velocity, thebody may have been moving with a definite velocity C1. Let C1 = v0 ft.per second. Of course, if the body started from a position of rest, C1 = 0.

Now integrate (7) and find the value of C2 in the result,

s = 12gt2 + v0t+ C2, (8)

by putting t = 0. It is thus apparent that C2 represents the space whichthe body had traversed when we began to study its motion. Let C2 = s0ft. The resulting equation

s = 12gt2 + v0t + s0, (9)

tells us three different things about the moving body at the instant webegan to take its motion into consideration.

l. It had traversed a distance of s0 ft. To use a sporting phrase, if thebody is starting from “scratch,” s0 = 0.

2. The body was moving with a velocity of v0 ft. per second.3. The velocity was increasing at the uniform rate of g ft. per second.Equation (7) tells us the two latter facts about the moving body; equa-

tion (6) only tells us the third fact; equation (5) tells us nothing more thanthat the acceleration is constant. (5), therefore, is true of the motion ofany body moving with a uniform acceleration.

EXAMPLE. – If a body falls in the air, experiment shows that the retarding effect ofthe resisting air is proportional to the square of the velocity of the moving body. Insteadof g, therefore, we must write g − βv2, where β is the

287

variation constant of page 487. For the sake of simplicity, put β = g/a2 and show that

v = aegt/a − e−gt/a

egt/a + e−gt/a; s =

a2

glog

egt/a + e−gt/a

2=a2

glog cosh

gt

a,

since v = 0 when t = 0, and s = 0 when t = 0.

Similar reasoning holds good from whatever sources we may draw ourillustrations. We are, therefore, able to say that a differential equation,freed from constants, is the most general way of expressing a, natural law.

Any equation can be freed from its constants by combining it with thevarious equations obtained by differentiation of the given equation as manytimes as there are constants. The operation is called elimination.

EXAMPLES.–(1) Eliminate the arbitrary constants a and b, from

y = ax+ bx2.

Differentiate twice and combine the results with the original equation. result,

x2d 2y

dx2− 2x

dy

dx+ 2y = 0,

is quite free from the arbitrary restrictions imposed in virtue of the presence of theconstants a and b in the original equation.

(2) Eliminate m from y2 = 4mx. Ansr. y2 = 2x dy/dx. ←−(3) Eliminate α and β from y = α cosx+ β sinx. Ansr. d 2y/dx2 + y = 0.(4) Eliminate α and β from y = αeax + βebx.

Ansr. d 2y/dx2 − (a+ b)dy/dz + aby = 0.

(5) Eliminate k from dx/dt = k(a− x) of § 69. What does the resulting equationmean?

We always assume that every differential equation has been obtainedby the elimination of constants from a given equation called the primitive.In practical work we are not so much concerned with the building up ofa differential equation by the elimination of constants from the primitive,as with the reverse operation of finding the primitive from which the dif-ferential equation has been derived. In other words, we have to find somerelation between the variables which will satisfy the differential equation.Given an expression involving x, y, dy/dx, d 2y/dx2,. . . , to find an equa-tion containing only x, y and constants which can be reconverted into theoriginal equation by the elimination of the constants.

This relation between the variables and constants which satisfies thegiven differential equation is called a general solution, or a completesolution, or a complete integral of the differential

288

equation. A solution obtained by giving particular values to the arbitraryconstants of the complete solution is a particular solution.

Thus y = mx is a complete solution of y = x dy/dx; p = x tan 45◦, is aparticular solution.

A differential equation is ordinary or partial, according as there isone or more than one independent variables present. Ordinary differentia1equations will be treated first.

Equations like (2) and (8) above, are said to be of the first order. For asimilar. reason (4) and (6) are of the second order, (5) of the third order.The order of a differential equation, therefore, is fixed by that of thehighest differential coefficient it contains. The degree of a differentialequation is the highest power of the highest order of differential coeflicientit contains. Thus,

d 2y

dx2+ k

(

dy

dx

)3

+ µx4 = 0,

is of the second order and third degree.It is not difficult to show that the complete integral of a differential

equation, of the nth order, contains n and only n arbitrary constants.We shall first consider equations of the first order.

§ 119. Exact Differential Equations of the First Order. The rea-son many differential equations are so difficult to solve is due to the factthat they have been formed by the elimination of constants as well as bythe elision of some common factor from the primitive. Such an equation,therefore, does not actually represent the complete or total differential ofthe original equation or primitive. The equation is then said to be inexact.On the other hand, an exact differential equation is one that has beenobtained by the differentiation of a function of x and y and performing noother operation involving x and y.

Easy tests were described in §§ 24, 25, to determine whether any givendifferential equation is exact or inexact. It was pointed out that the differ-ential equation,

M dx+N dy = 0, (1)

is the direct result of the differentiation of any function u, provided,

∂M

∂y=

∂N

∂x. (2)

289

This last result was called the criterion of integrability, because, if an equa-tion satisfies the test, the integration can be readily performed by a directprocess. This is not meant to imply that only such equations can be inte-grated as satisfy the test, for many equations which do not satisfy the testcan be solved in other ways.

EXAMPLES.–(1) Apply the test to the equations,

y dx+ x dy = 0 and y dx− x dy = 0.

In the former, M = y, N = x;

∴ ∂M/∂y = 1, ∂N/∂x = 1; ∴ ∂M/∂y = ∂N/∂x.

The test is, therefore, satisfied and the equation is exact. In the other equation, M = y,N = −x, ∴ ∂M/∂y = 1, ∂N/∂x = −1.This does not satisfy the test. In consequence, the equation cannot be solved by themethod for exact differential equations.

(2) Is the equation, (x+ 2y)x dx +(

x2 − y2)

dy = 0 exact? M = x (x+ 2y),

N =(

x2 − y2)

, ∴ ∂M/∂dy = 2x, ∂N/∂x = 2x. The condition is satisfied, the equationis exact.

(3) Show that(

a2y + x2)

dx+(

b3 + a2x)

dy = 0, is exact.(4) Show that (sin y + y cosx) dx+ (sinx+ x cos y) dy = 0, is exact.

To integrate an equation which satisfies the criterion of integrability, wemust remember that M is the differential coefficient of u with respect to x,y being constant, and N is the differential coefficient of N with respect toy, x being constant. Hence we may integrate M dx on the supposition thaty is constant and then treat N dy as if x were a constant. The completesolution of the whole equation is obtained by equating the sum of thesetwo integrals to an undetermined constant. The complete integral is

u = C. (3)

EXAMPLES.–(1) Integrate x (x+ 2y)dx +(

x2 − y2)

dy = 0, from the preceding setof examples. Since the equation is exact,

M = x (x+ 2y) ; N = x2 − y2;

∴

∫

M dx =

∫

x (x+ 2y)dx = 13x

3 + x2y = Y,

where Y is the integration constant which may, or may not, contain y, because y hashere been regarded as a constant.Now the result of differentiating

13x

3 + x2y = Y,

should be the original equation. On trial,

x2dx+ 2xy dx+ x2dy = dY.

On comparison with the original equation, it is apparent that

dY = y2dy; ∴ Y = 13y

3 + C.

Substitute this in the preceding result. The complete solution is, therefore,13x

3 + x2y − 13y

3 = C.

290

To summarise: The method detailed in the example just given may be put into amore practical shape.

To integrate an exact differential equation, first find∫

M dx on the assumption thaty is constant and substitute the result in

∫

M dx+

∫ (

N − ∂

∂y

∫

M dx

)

dy = C. (4)

With the old example, therefore, having found∫

M dx, we may write down at once

13x

3 + x2y +

∫ {

x2 − y2 − ∂

∂y

(

1

3x3 + x2y

)}

dy = C.

∴ 13x

3 + x2y +

∫

(

x2 − y2 − x2)

dy = C.

And the old result follows directly. If we had wished we could have used∫

N dy +

∫ (

M − ∂

∂x

∫

N dy

)

dx = C,

in place of (4).In practice it is often convenient to modify this procedure. If the equation satisfies

the criterion of integrability, we can easily pick out terms which make M dx+N dy = 0,and get

M dx+ Y and N dy +X,where Y cannot contain x and X cannot contain y. Hence if we find M dy and N dx,the functions X and Y will be determined.

In the above equation, the only terms containing x and y are 2xy dx+ x2dy, whichobviously have been derived from x2y. Hence integration of these and the omitted termsgives the above result.

(2) Solve(

x2 − 4xy − y2)

dx+(

y2 − 4xy − 2x2)

dy = 0. Pick out terms in x and y,we get −

(

4xy + 2y2)

dx −(

4xy + 2x2)

dy = 0.Integrate. ∴ −2x2y − 2xy2 = constant.Pick out the omitted terms and integrate for the complete solution. We get,

∫

x2dx+

∫

y2dy − 2x2y − 2xy2 = x3 − 6x2y − 6xy2 + y3 = constant.

(3) Show that the solution of(

a2x+ x2)

dx+ b3 + z2x)dy = 0 is

a2xy + b3y + 13x

3 = C. Use (4).

(4) Solve(

x2 − y2)

dx− 2xy dy = 0. Ansr. 13x

2 − y2 = C/x. Use (4).

Equations made exact by means of integrating factors. As just pointedout, the reason any differential equation does not satisfy the criterion ofexactness, is because the “integrating factor” has been cancelled out duringthe genesis of the equation from its primitive.

If, therefore, the equation

M dx+N dy = 0,

does not satisfy the criterion of integrability, it will do so when the factor,previously divided out, is restored. Thus, the preceding equation is madeexact by multiplying through with the integrating factor µ. Hence,

µ (M dx+N dy) = 0,

satisfies the criterion of exactness, and the solution can be obtained asdescribed above.

291

§ 120. How to find Integrating Factors. Sometimes integrating fac-tors are so simple that they can be detected by simple inspection.

EXAMPLES.–(1) y dx− x dy = 0 is inexact. It becomes exact by multiplication witheither x−2, x−1y−1, or y−2.

(2) In (y − x) dy+y dx = 0, the term containing y dx−x dy is not exact, but becomesso when multiplied as in the preceding example.

∴dy

y− x dy − y dx

y2; or log y − y

x= C.

For the general theorems concerning the properties of integrating factors, the readermust consult some special treatise, say Boole’s A Treatise on Differential Equations,pages 55 et seq., 1865.

§ 25 isincludedin thisdocu-mentfollowingthe sup-plement.

We have already established, in § 25, that an integrating factor alwaysexists which will make the equation

M dx+N dy = 0,

an exact differential.Moreover, there is also an infinite number of such factors, for if the

equation is made exact when multiplied by µ, it will amain exact whenmultiplied by any function of µ.

The different integrating factors correspond to the various forms inwhich the solution of the equation may present itself. For instance, theintegrating factor x−1y−1, of y dx+x dy = 0, corresponds with the solutionlog x+ log y = C. The factor y−2 corresponds with the solution xy = C ′.

Unfortunately, it is of no assistance to know that every differential equa-tion has an infinite number of integrating factors. No general practicalmethod is known for finding them. Here are a few elementary rules appli-cable to special cases.

Rule I. Since

d (xmyn) = xm−1yn−1 (my dx+ nx dy) ,

an expression of the typemy dx+nx dy, has an integrating factor xm−1yn−1;or, the expression

xαyβ (my dx+ nx dy) = 0, (1)

has an integrating factorxm−1−αyn−1−β,

or more generally still,xkm−1−αykn−1−β, (2)

where k may have any value whatever.

EXAMPLE. –Find an integrating factor of y dx + x dy = 0. Here, α = 0, β = 0,m = 1, n = −1 ∴ y−2 is an integrating factor of the given equation.

292

If the expression can be written

xαyβ (my dx+ nx dy) + xα′

yβ′

(m′y dx+ n′x dy) , (3)

the integrating factor can be readily obtained, for

xkm−1−αykn−1−β; and xk′m′−1−α′

yk′n′−1−β′

,

are integrating factors of the first and second members respectively. Inorder that these factors may he identical,

km− 1− α = k′m′ − 1− α′; kn− 1− β = k′n′ − 1− β ′.

Values of k and k′ can be obtained to satisfy these two conditions by solvingthese two equations. Thus,

k =n′ (α− α′)−m′ (β − β ′)

mn′ −m′n; k′ =

n (α− α′)− n (β − β ′)

mn′ −m′n. (4)

EXAMPLES.–(1) Solve y3 (y dx− 2x dy) + x4 (2y dx + x dy) = 0. Hints. Show thatα = 0, β = 3, m = 1, n = −2, α′ = 4, β′ = 0, m′ = 2, n′ = 1; ∴ xk−1y−2k−4 isan integrating factor of the first, x2k

′−5yk′−1 of the second member. Hence, from (4),

k = −2, k′ = 1, ∴ x−3 is an integrating factor of the whole expression. Multiplythrough and integrate for 2x4y − y4 = Cx2.

(2) Solve(

y3 − 2yx2)

dx +(

2xy2 − x3)

dy. Ansr. x2y2(

y2 − x2)

= C. Integratingfactor deduced after rearranging the equation is xy.

Rule II. If the equation is homogeneous and of the form: M dx+N dy =0, then (Mx +Ny)−1 an integrating factor.

Let the expressionM dx+N dy = 0,

be of the mth degree and µ. an integrating factor of the nth degree,

∴ µM dx+ µN dy = du, (5)

is of the (m+ n)th degree, and the integral u is of the (m+ n + 1)th degree. § 22 isincludedin thisdocu-mentfollowingthe sup-plement.

By Euler’s theorem, § 22,∴ µMx+ µNy = (m+ n+ 1) u. (6)

Divide (5) by (6),

M dx+N dy

Mx+Ny=

1

m+ n+ 1

du

u

The right side of this equation is a complete differential, consequently, theleft side is also a complete differential. Therefore, (Mx+Ny)−1 has madeM dx+N dy = 0 an exact differential equation.

EXAMPLES.–(1) Show that(

x3y − xy3)−1

is an integrating factor of(

x2y + y3)

dx−2xy2dy = 0.

(2) Show that 1/(

x2 − nyx+ y2)

is an integrating factor of y dy+ (x− ny) dx = 0.The method, of course, cannot be used if Mx + Ny is equal to zero. In this case,

we may write y = Cx, a solution.

293

Rule III. If the equation is of the form,

f1 (x, y) y dx+ f2 (x, y)x dy = 0,

then (Mx−Ny)−1 is an integrating factor.

EXAMPLE. –Solve (1 + xy) y dx+(1− xy)x dy = 0. Hint. Show that the integratingfactor is 1/2xy. Divide out 1

2 . ∴∫

M dx = 1/xy + log x. Ansr. x = Ce−1/xy ←−If Mx − Ny = 0, the method fails and xy = C is then a solution of the equation.

E.g., (1 + xy) y dx+ (1 + xy) x dy = 0.

Rule IV. If 1N

(

∂M∂y− ∂N

∂x

)

is a function of x only, e∫f(x)dx is an inte-

grating factor. Or, if 1M

(

∂N∂x− ∂M

∂y

)

= f (y), then e∫f(y)dy is an integrating

factor. These are important results.

EXAMPLES.–(1) Solve(

x2 + y2)

dx− 2xy dy = 0. Ansr. x2 − y2 = Cx. Hint. Showf (x) = −2/x. The integrating factor is, therefore,

e−∫2dx/x = elog 1/x2

= 1/x2.

(Why?) Prove that this is an integrating factor, and solve as in the preceding section.(2) Solve

(

y4 + 2y)

dx +(

xy3 + 2y4 − 4x)

dy = 0. Ansr. xy3y4 + 2x = Cy2.(3) We may prove the rule for a special case in the following manner. The steps will

serve to recall some of the principles established in some earlier chapters.

Letdy

dx+ Py = Q, (7)

where P and Q are either constant or functions of x. Let µ be an integrating factorwhich makes

dy + (Py −Q) dx = 0, (8)

an exact differential.

∴ µ dy + µ (Py −Q)dx ≡ N dy +M dx.

∴∂N

∂x=∂µ

∂x;

∂M

∂y= (Py − q) ∂µ

∂y+ Pµ.

∴∂µ

∂x= (Py −Q)

∂µ

∂dx+ Pµ.

∴∂µ

∂xdx = (Py −Q)

∂µ

∂ydx+ Pµdx;

= −∂µ∂ydy + Pµdx.

∴∂µ

∂xdx+

∂µ

∂ydy = dµ = Pµdx

∴ P =1

µ

dµ

dx;

∫

P dx = logµ.

and since loge e = 1.

(∫

P dx

)

log e = logµ; µ = e∫P dx. (9)

This result will be employed in dealing with linear equations, § 122.

294

§ 121. The First Law of Thermodynamics. According to the dis-cussion at the end of the first chapter, one way of stating the first law ofthermodynamics is as follows:

dQ = dU + dW,

which means that when a quantity of heat, dQ, is added to a substance,one part of the heat is spent in changing the internal energy, dU , of thesubstance and another part, dW , is spent in doing work against externalforces. In the special case, when that work is expansion against atmosphericpressure, dW = p dv, as shown in § 91. See (11), page 524.

We know that the condition of a substance is completely defined byany two of the three variables p, v, θ, because when any two of these threevariables is known, the third can be deduced from the relation

pv = Rθ.

Hence it is assumed that the internal energy of the substance is completelydefined when any two of these variables are known.

Now let the substance pass from any state A to another state S (Fig.111). The internal energy of the substance in the state B

Figure 111:

completely determined by the coordinatesof that point, because U is quite inde-pendent of the nature of the transforma-tion from the state A to the state B. Itmakes no difference to the magnitude of Uwhether that path has been via APE orAQB. In this case U is said to be a single-valued function by the coordinates of thepoint corresponding to any given state. Inother words, dU is a complete differential.Hence

dU =∂U

∂xdx+

∂U

∂ydy,

is an exact differential equation, where x and y represent any pair of thevariables p, v, θ.

On the other hand, the external work done durmg the transformationfrom the one state to another, depends not only on the initial and fi-nal states of the substance, but also on the nature of the path describedin passing from the state A to the state B. For example, the substancemay perform the work represented by the area AQBB′A′ or by the areaAPBB′A′, in its passage from the

295

state A to the state B. In fact the total work done in the passage fromA to B and back again, is represented by the area APBQ (page 183). Inorder to know the work done during the passage from the state A. to thestate B, it is not only necessary to know the initial and final states of thesubstance as defined by the coordinates of the points A and B, but wemust know the nature of the path from the one state to the other.

Similarly, the quantity of heat supplied to the body in passing from onestate to the other, not only depends on the initial and final states of thesubstance but also on the nature of the transformation.

All this is implied when it is said that “dW and dQ are not perfectdifferentials.” Although we can write

∂2U

∂x∂y=

∂2U

∂y∂x,

we must put, in the case of W or Q,

∂2Q

∂x∂y≶

∂2Q

∂y∂x; or

∂2Q

∂x∂y=

∂2Q

∂y∂x.

Therefore the partial differentiation of x with respect to y, furnishes acomplete differential equation only when we multiply through with theintegrating factor µ, so that

µ dQ = µ∂Q

∂xdx+ µ

∂Q

∂ydy,

where x and y may represent any pair of the variables p, v, θ.The integrating factor is proved in thermodynamics to be equivalent to

the so-called Carnot’s function (see Preston’s Theory of Heat).To indicate that dW and dQ are not perfect differentials, some writers

superscribe a comma to the top right-hand corner of the differential sign.The above equation would then be written,

d ′Q = dU + d ′W.

§ 122. Linear Differential Equations of the First Order. A lineardifferential equation of the first order involves only the first power of thedependent variable y and of its first differential coefficients. The generaltype is,

dy

dx+ Py = Q, (1)

where P and Q may be functions of x, or constants.

296

We have just proved that e∫P dx is an integrating factor of (1), therefore

e∫P dx (dy + Py dy) = e

∫P dxQdx,

is an exact differential equation. The general solution is,

ye∫P dx =

∫

e∫P dxQdx+ C. (2)

The linear equation is one of the most important in applied mathematics.In particular cases the integrating factor may assume a very simple form.

In the following examples, remember that elog x = x, ∴ if∫

P dx = log x,

e∫P dx = x.

EXAMPLES.–(1) Solve(

1 + x2)

dy = (m+ xy) dx. Reduce to the form (1) and weobtain

dy

dx− x

1 + x2y =

m

1 + x2.

∴

∫

P dx = −∫

x dx

1 + x2= −1

2log(

1 + x2)

= − log√

(1 + x2).

Remembering log 1 = 0, log e = 1, the integrating factor is evidently,

log e∫P dx = log 1− log

√

1 + x2, or e∫P dx =

1√1 + x2

.

Multiply the original equation with this integrating factor, and solve the resulting exactequation as § 119, (4), or, better still, by (2) above. The solution: y = mx+C

√

(1 + x2),follows at once.

(2) Ohm’s law for a variable current flowing in a circuit with a coefficient of self-induction L (henries), a resistance R (ohms), and a current of C (amperes) and anelectromotive force E (volts), is given by the equation,

E = RC + LdC

dt.

This equation has the standard linear form (1). If E is constant, show that the solutionis,

C = E/R+Be−Rt/L,

where B is the arbitrary constant of integration (page 159). Show that C approximatesto E/B after the current has been flowing some time (t). Hint for solution. Integratingfactor is eRt/L.

(3) The equation of motion of a particle subject to a resistance varying directly asthe velocity and as some force which is a given function of the time, is

dv/dt+ kv = f (t) .

Show that v = Ce−kt + e−kt∫

ektf (t) dt.If the force is gravitational, say g,

v = Ce−kt + g/k

(4) Solve x dy + y dx = x3dx. Integrating factor= x. Ansr. y = 14x

3 + C/x.

Many equations may be transformed into the linear type of equation,by a change in the variable. Thus, in the so-called Bernoulli’s equation,

dy/dx+ Py = Qyn. (3)

297

Divide by yn, multiply by (1− n) and substitute yn−1 = v, in the result.Thus,

1− n

yndy

dx+ (1− n)Py1−n = (1− n)Q,

and dy/dx+ (1− n)Pv = Q (1− n),which is linear in v. Hence, the solution is

ve(1−n)∫P dx = (1− n)

∫

Qe(1−n)∫P dxdx+ C.

∴ y1−ne(1−n)∫P dx = (1− n)

∫

Qe(1−n)∫P dxdx+ C.

EXAMPLES.–(1) Solve dy/dx+y/x = y2. Treat as above, substituting v = 1/y. The

integration factor is e∫dx/x = elog x = 1/x.

Ansr. Cxy − xy log x = 1.

(2) Solve dy/dx + x sin 2y = x3 cos2 y. Divide by cos2 y. Put tan y = v. The

integration factor is e∫2x dx, i.e., ex

2

. Ansr. ex2

tan y − 12 e

x2 (

x2 − 1)

= C. Hint to

solve vex2

=∫

x3ex2

dx + C. Put x2 = z, ∴ 2x dx = dz, and this integral becomes12

∫

zezdz, or 12e

z (z − 1), etc.(3) Here is an instructive differential equation, which Harcourt and Esson encoun-

tered during their work on chemical dynamics in 1866.

1

y2dy

dx+K

y− K

x= 0.

I shall give a method of solution in full, so as to revise some preceding work. Theequation has the same form as Bernoulli’s. Therefore, substitute

v =1

y; i.e.,

dv

dx= − 1

y2dy

dx.

∴dv

dx−Kv + K

x= 0,

an equation linear in v. The integrating factor is

e∫P dx, or, e−Kx; Q, in (2), = −K/x,

therefore, from (2) ve−Kx = −∫

Kx e

−Kxdx + C.From § 108,

ve−Kx = −K∫

1

x

{

1− (Kx) +(Kx)2

1 · 2 −(Kx)3

1 · 2 · 3 + . . .

}

dx+ C.

∴ ve−Kx = −K∫ {

dx

x−K dx+

K2x dx

1 · 2 − K3x2dx

1 · 2 · 3 + . . .

}

+ C.

But v = 1/y. Multiply through with yeKx, and integrate.

1 = KeKx

{

C1 − log x+Kx− (Kx)2

1 · 2 +(Kx)

3

1 · 2 · 3 − . . .}

y.

We shall require this result on page 333.

§ 123. Differential Equations of the First Order and of the Firstor Higher Degree.– Solution by Differentiation.

Case i. The equation can be split up into factors. If the differential equation canbe resolved into n factors of the first degree, equate each factor to zero and solve eachof the n equations separately. The n solutions may be left either distinct, or combinedinto one.

298

EXAMPLES.–(1) Solve x(

dydx

)2

= y. Resolve into factors of the first degree,

dy

dx= ±

√

y

x.

Separate the variables and integrate,

√x±√y = ±

√C,

which, on rationalisation, becomes

(x− y)2 − 2C (x+ y) + C2 = 0

Geometrically this equation represents a system of parabolic curves each of whichtouches the axis at a distance C from the origin. The separate equations of the abovesolution merely represent different branches of the same parabola.

(2) Solve(

dydx

)2

−(

x2 − y2)

dydx − xy = 0. Ansr. xy = C, or x2 − y2 = C. Hint.

Factors (xp+ y) (yp− x), where p ≡ dydx .

(3) Solve(

dydx

)2

− 7 dydx + 12 = 0. Ansr. y = 4x+ C, or 3x+ C.

See pageS-13 ofthe sup-plementfor moredetailson thecase iimethod.KH

Case ii. The equation cannot be resolved into factors, but it can be solved for x,y, dy/dx, or y/x. An equation which cannot be resolved into factors, can often beexpressed in terms of x, y, dy/dx, or y/x, according to circumstances. The differentialcoeffcient of the one variable with respect to the other may be then obtained by solvingfor dy/dx and using the result to eliminate dy/dx from the given equation.

EXAMPLES. Solve dy/dx+ 2xy = x2 + y2. Since (x− y)2 = x2 − 2xy + y2,

y = x+√

dy/dx

Differentiatedy

dx= 1 +

1

2

(

d2y

dx2

)

/

(

dy

dx

)12

.

Separate the variables x and p, where p ≡ dy/dx, and solve for dy/dx,

x =1

2log

√p− 1√p+ 1

+ logC;

√

dy

dx=C + e2x

C − e2x .

∴ Ansr. y = x+(

C + e2x)

/(

C − e2x)

.

(2) Solve x (dy/dx)2 − 2y (dy/dx) + ax = 0. Ansr. y = 12C(

x2 + a/C)

. Hint. ←−Substitute for p. Solve for y and differentiate. Substitute p dx for dy, and clear offractions. The variables p and x can be separated. Integrate. p = xC. Substitute in thegiven equation for the answer.

(3) Solve x (dy/dx)2+ 2y (dy/dx) − y = 0. Ansr. (2x+ C). Hint. Solve for x.

Differentiate and substitute dy/p for dx, and proceed as in example (2). yp = C, etc.

Case iii. The equation cannot be resolved into factors, x or y is absent.If x is absent solve for dy/dx or y according to convenience; if y is absent,solve for dx/dy or x. Differentiate the result with respect to the absentletter if necessary and solve in the regular way.

299

EXAMPLES.–(1) Solve (dy/dx)2+ x (dy/dx) + 1 = 0. For the sake of greater ease,

substitute p for dx/dy. The given equation thus reduces to

x = p+ 1/p. (1)

Differentiate with regard to the absent letter y, thus,

p =(

1− 1/p2)

dp/dy; or dy/dp = 1/p+ 1/p2. (2)

Combining (1) and (2), we get the required solution.(2) Solve dy/dx = y + 1/y. Ansr. y2 = Ce2x − 1.(3) Solve dy/dx = x+ 1/x. Ansr. y = 1

2x2 + log x+ C.

§ 124. Clairaut’s Equation. The general type of this equation is,

y = xdy

dx+ f

(

dy

dx

)

; (1)

or, writing p = dy/dx, for the sake of convenience,

y = px+ f (p) . (2)

Many equations of the first degree in x and y can be reduced to this formby a more or less obvious transformation of the variables, and solved in thefollowing way:–

Differentiate (2) with respect to x, and equate the result to zero.

p = p+ xdp

dx+ f ′ (p)

dp

dx; or, {x+ f ′ (p)} dp

dx= 0.

Hence eitherdp

dx= 0; or, x + f (p) = 0.

If in the former,dp/dx = 0; ∴ p = C,

where C is an arbitrary constant. Hence,

dy = C dx; or, y = Cx+ f (C) ,

is a solution of the given equation.Again, p in x+ f (p) may be a solution of the given equation. To find

p, eliminate p between

y = px+ f (p) , or, x+ f ′ (p) = 0.

The resulting equation between x and y also satisfies the given equation. § 67 isincludedin thisdocu-mentfollowingthe sup-plement.

There are thus two classes of solutions to Clairaut’s equation.

EXAMPLES. –Find both solutions. in the following equations:–(1) y = px+ p2. Ansr. Cx+ C2 = y and x2 + 4y = 0.(2) (y − px) (p− 1). Ansr. (y = Cx) (C − 1) = C;

√y +√x = 1. Read over § 67. ←−

300

§ 125. Singular Solutions. Clairaut’s equation introduces us to a newidea. Hitherto we have assumed that whenever a function of x and ysatisfies an equation, that function plus an arbitrary constant, representsthe complete or general solution. We now find that functions of x and ycan sometimes be found to satisfy the given equation, which, unlike theparticular solution, are not included in the general solution.

This function must be considered a solution, because it satisfies thegiven equation. But the existence of such a solution is quite an accidentalproperty confined to special equations, hence their cognomen, singularsolutions.

To take the simple illustration of page 149,

y = px+a

p. (1)

Remembering that y has been written for dy/dx, differentiate with respectto x, we get, on rearranging terms,

(

x− a

p2

)

dp

dx= 0.

where either x− a

p2= 0; or,

dp

dx= 0.

If the latter,

p = C; or, y = Cx+a

C. (2)

If the former, p =√

a/x, which gives,.when substituted in (1), thesolution,

y2 = 4ax. (3)

This solution is not included in the general solution, but yet it satisfies thegiven equation. (3) is the singular solution of (1).

Equation (2), the complete solution of (1), has been shown to representa system of straight lines which differ only in the value of the arbitraryconstant C; equation (3), containing no arbitrary constant, is an equationto the common parabola. A point moving on this parabola has, at anyinstant, the same value of dy/dx as if it were moving on the tangent ofthe parabola, or on one of the straight lines of equation (2). The singularsolution of a differential equation is geometrically equivalent to the envelopeof the family of curves represented by the general solution. The singularsolution is distinguished from the particular solution, in that the latter iscontained in the general solution, the former is not.

Again referring to Fig. 78, it will be noticed that for any point on theenvelope, there are two equal values of p or dy/dx, one for the parabola,one for the straight line.

301

For ex-pandeddiscus-sion ofdiscrim-inants,see sup-plementpageS-17.KH

In order that the quadratic

ax2 + bx+ c = 0,

may have equal roots, it is necessary (page 388) that

b2 = 4ac; or, b2 − 4ac = 0. (4)

This relation is called the discriminant. From (1), since

y = px+ a/p; ∴ xp2 − yp+ a = 0. (5)

In order that equation (5) may have equal roots,

y2 = 4ax,

as in (4). This relation is the locus of all points for which two values of Cbecome equal, hence it is called the p-discriminant of (1).

In the same way if C be regarded as variable in the general solution (2),

y = Cx+ a/C; or, xC2 − yC + a = 0.

The condition for equal roots, is that

y2 = 4ax.

which is the locus of all points for which the value of C is the same. It iscalled the C -discriminant.

Before applying these ideas to special cases, we may note that the en-velope locus may be a single curve (Fig. 78) or several (Fig. 79). Foran exhaustive discussion of the properties of these discriminant relationsl must refer the reader to the numerous textbooks on the subject, or toCayley, Messenger of Mathematics, 2, 6, 1879. To summarise:

1. The envelope locus satisfies the original equation but is not in-cluded in the general solution (see xx′, Fig. 112).

Figure 112: –Nodal and Tac Loci

2. The tac locus is the locus passing through the several points wheretwo non-consecutive members of a family of curves touch. Such a locus isrepresented by the line AB (Fig. 79), PQ (Fig. 112). The tac locus doesnot satisfy the original equation, it appears in the p-discriminant, but notin the C-discriminant.

302

3. The node locus is the locus passing through the different pointswhere each curve of a given family crosses itself (the point of intersection– node – may be double, triple, etc.). The node locus does not satisfythe original equation, it appears in the C-discriminant but not in the p-discriminant. RS (Fig. 112) is a nodal locus passing through the nodesA,. . . , B,. . . , C,. . . , M .

Figure 113: –Cusp Locus

4. The cusp locuspasses through all the cusps(page 186) formed by themembers of a family ofcurves. The cusp locus doesnot satisfy the original equa-tion, it appears in the p- andin the C-discriminants. Itis the line Ox in Fig. 118.Sometimes the nodal or cusp loci coincides with the envelope locus.*

EXAMPLES.–Find the singular solutions and the nature of the other loci in thefollowing equations:

(1) xp2 − 2yp + ax = 0.For equal roots y2 = ax2. This satisfies the original equation and is not included inthe general solution: x2 − 2Cy + aC2 = 0. y2 = ax2 is thus the singular solution.

(2) 4xp2 = (3x− a)2. General solution: (y + C)2 = x (x− a)2.For equal roots in p, 4x (8x− a)2 = 0, or x(3x−a)2 = 0 (p-discriminant). For equal

roots in C, differentiate the general solution with respect to C. Therefore(x+ C) dx/dC = 0, or C = −x. ∴ x (x− a)2 = 0 (C-discriminant) is the condi-tion to be fulfilled when the C-discriminant has equal roots. x = 0 is common to thetwo discriminants and satisfies the original equation (singular solution); x = a satis-fies the C-discriminant but not the p-discriminant and, since it is not a solution of theoriginal equation, x = a represents the node locus; x = 1

3a satisfies the p- but not theC-discriminant nor the original equation (tac locus).

(3) p2 + 2xp − y = 0.

General solution:(

2x3 + 3xy + C)2

= 4(

x2 + y)3; p-discriminant: x2 + y = 0; C-

discriminant:(

x2 + y)3

= 0. The original equation is not satisfied by either of these

equations and, therefore, there is no singular solution. Since(

x2 + y)

appears in bothdiscriminants, it represents a cusp locus.

(4) Show that the complete solution of the equation, y2(

p2 + 1)

= a2, is

y2 + (x− C)2 = a2; that there are two singular solutions, y = ±a; that there is atac locus on the x-axis for y = 0 (Fig. 79, see also § 138).

* The second part of van der Waals’ The Continuity of the Gaseous and Liquid Statesof Aggregation – Binary Mixtures (1900) has some examples of the preceding “mathe-matics.”

303

§ 126. Trajectories. This section will serve as an exercise on somepreceding work. A: trajectory is a curve which cuts another system ofcurves at a constant angle. If this angle is 90◦ the curve is an orthogonaltrajectory.

EXAMPLES. (1) Let xy = C be a system of rectangular hyperbolas, to find theorthogonal trajectory, first eliminate C by differentiation with respect to x, thus weobtain,

x dy/dx+ y = 0

If two curves are at right angles(

12π = 90◦

)

, then from (17), § 32, 12π = (α′ − α), where (17) of

§ 32 es-tablishesthatslopes ofm and−1/marenormalto eachother.KH

α, α′ are the angles made by tangents to the curves at the point of intersection with thex-axis. But by the same formula,

tan

(

±1

2π

)

= (tanα′ − tanα) / (1 + tanα tanα′) .

Now tan± 12π =∞ and 1/∞ = 0,

∴ tanα = − cotα′; or, dy/dx = −dx/dy.*

The differential equation of the one family is obtained from that of the other bysubstituting dy/dx for −dx/dy. Hence the equation to the orthogonal trajectory ofthe system of rectangular hyperbolas is, x dx + y dy = 0, or x2 − y2 = C, a system ofrectangular hyperbolas whose axes coincide with the asymptotes of the given system.

(2) For polar coordinates show that we must substitute −dr/ (r · dθ) for r · dθ/dr.(3) Find the orthogonal trajectories of the system of parabolas y2 = 4ax. Ansr.

Ellipses, 2x2 + y2 = C2.(4) Show that the orthogonal trajectories of the equipotential curves, 1/r−1/r′ = C,

are the magnetic curves cos θ + cos θ′ = C.

§ 127. Symbols of Operation. It will be found convenient to denotethe symbol of the operation “dy/dx” by the letter “D”. If we assumethat the infinitesimal increments of the independent variable dx have thesame magnitude, whatever be the value of x, we can suppose D to have aconstant value. Thus

D,D2, D3, . . . , stand ford

dx,d2

dx2,d3

dx3, . . . ;

Dy,D2y,D3y, . . . , stand fordy

dx,d2y

dx2,d3y

dx3, . . .

The operations denoted by the symbols D, D2,. . . , satisfy the elementaryrules of algebra except that they are not commutative†with regard to thevariables. For example,

* No doubt the reader sees that in (18), § 12, dx/dy is the cotangent of the anglewhose tangent is dy/dx.† The so-called fundamental laws of algebra are: I. The law of association: The

number of things in any group is independent of the order. II. The commutative law: (a)Addition. The number of things in any number of groups is independent of the order.(b) Multiplication. The product of two numbers is independent of the

304

D (u+ v + . . . ) = Du+ dv + . . . , (distributive law).

D (Cu) = CD (u) , (commutative law),

where C is a constant. We cannot write D (xy) = D (yx). But,

DmDnu = Dm+nu (index law),

is true when m and n are positive integers. If

Du = v; u = D−1v; or, u =1

Dv;

∴ v = D ·D−1v; or, D ·D = 1;

that is to say, by operating with D upon D−1v, we annul the effect of the D−1 operator.It is necessary to remember later on, that if Dx = 1, 1

D2 = 12x

2; 1D3 = 1

2·3x3, . . .

In this notation, the equation

d2y

dx2− (α+ β)

dy

dx+ αβy = 0,

is written,

{

D2 − (α+ β)D + αβ}

y = 0; or, (D − α) (D − β) y = 0.

Now replace D with the original symbol, and operate on one factor with y, Thus,

(

d

dx− α

)(

d

dx− β

)

y = 0;

(

d

dx− αy

)(

dy

dx− βy

)

= 0.

By operating on the second factor with the first, we get the original equation back again.

§ 128. The Linear Equation of the nth Order.

(General Remarks.)

As a general rule the higher orders of differential equations are more difficult ofsolution than equations of the first order. As with the latter, the more expeditiousmode of treatment will be to refer the given equation to a set of standard cases havingcertain distinguishing characters. By far the most important class is the linear equation.

A linear equation of the nth order is one in which the dependent variable andits n derivatives are all of the first degree and are not multiplied together. The typicalform in which it appears is

dny

dxn+X1

dn−1y

dxn−1+ · · ·+Xny = X. (1)

order. III. The distributive law: (a) Multiplication. The multiplier may be distributedover each term of the multiplicand, e.g., m(a+b) = ma+mb. (b) Division. (a+b)/m =a/m + b/m. IV. The index law: (a) Multiplication. aman = am+n. (5) Division.am/an = am−n.

305

Or, in symbolic notation,

Dny +X1Dn−1y + · · ·+Xny = X,

where X,X1, . . . , Xn, are either constant magnitudes, or functions of theindependent variable x. If the coefficient of the highest derivative be otherthan unity, the other terms of the equation can be divided by this coeffi-cient. The equation will thus assume the typical form (1). We have studiedthe linear equation of the first order in § 123. For the sake of fixing ourideas, the equation

d2y

dx2+ P

dy

dx+Qy = r, (2)

of the second order, will be taken as typical of the class. P , Q, R have themeaning above attached to X1, X2, X .

The general solution of the linear equation is made my of two parts.1. The complementary function which is the most general solution

of the left-hand side of equation (2) equated to zero, or,

d2y

dx2+ P

dy

dx+Qy = 0. (3)

The complementary function involves two arbitrary constants.2. The particular integral which is any solution of the original equa-

tion (2), the simpler the better. In particular cases when the right-handside is zero, the particular integral does not occur.

To show that the general solution of (2) contains a general solution of(3). Assume that the complete solution of (2) may be written,

y = u+ v, (4)

where v is any function of x which satisfies (2), that is to say, v is theparticular integral* of (2), u is the general solution of (3), to be determined.Substitute (4) in (2).

d2u

dx2+ P

du

dx+Qu+

d2v

dx2+ P

dv

dx+Qv = R.

Butd2v

dx2+ P

dv

dx+Qv = R;

therefore,d2u

dx2+ P

du

dx+Qu = 0.

Therefore, u must satisfy (3).Given a particular solution* of the linear equation, to find the

* Not to be confused with the particular solution of page 289.

306

complete solution. Let y = v be a particular solution of the followingequation,

d2y

dx2+ P

dy

dx+Qy = 0,

where P and Q are functions of x. Substitute y = uv, See pageS-21 ofthe sup-plementfor thederiva-tionof thisequation.KH

vd2u

dx2+

(

2dv

dx+ Pv

)

du

dx= 0.

This equation is of the first order and linear with du/dx as the dependentvariable. Put du/dx = z and

vdz

dx+

(

2dv

dx+ Pv

)

z = 0;dz

z+ 2

dv

v+ P dx = 0;

logdu

dx+ 2 log v +

∫

P dx = 0; or,du

dx= C1

1

v2e−

∫P dx.

∴ u = C1

∫

1

v2e−

∫P dx + C2; or, C1v

∫

1

v2e−

∫P dx + C2v,

where C1 and C2 are arbitrary constants.

EXAMPLES.–(1) If y = eax is a particular solution of d2y/dx2 = a2y, show that thecomplete solution is y = C1e

ax + C2e−ax.

(2) If y = x is a particular solution of(

1− x2)

d2y/dx2 − x dy/dx + y = 0, the

complete solution is y = C1

√1− x2 + C2x.

If a particular solution of the linear equation is known, the order of theequation can be lowered by unity. This follows directly from the precedingresult. If y = v is a known solution, then, if y = tv be substituted in thefirst member of the equation, the coefficient of t in the result, will be thesame as if t were constant and therefore zero. t being absent, the resultwill be a linear equation in t but of an order less by unity than that ofthe given equation. It follows directly, that if n particular solutions of theequation are known, the order of the equation can be reduced n times.

For the description of a machine designed for solving (3), see Proceedingsof the Royal Society, 24, 269, 1876 (Lord Kelvin).

§ 129. The Linear Equation with Constant Coefficients. The in-tegration of these equations obviously resolves itself into finding the com-plementary function and the particular integral.

First, when the second member is zero, in other words, to find the com-plementary function of any linear equation with constant coefficients. Thetypical equation is,

d2y

dx2+ P

dy

dx+Qy = 0, (1)

307

where P and Q are constants. The particular integral does not appear inthe solution.

If the equation were of the first order, its solution would be, y = Ce∫mdx.

On substituting emx for y in (1), we obtain(

m2 + Pm+Q)

emx = 0,

provided m2 + Pm+Q = 0. (2)

This equation is called the auxillary equation. If m1 be one value ofm which satisfies (1), then y = em1x, is an integral of (1). But we must gofurther.

Case 1. When the auxillary equation has two unequal roots, say m1,andm2 the general solution of (1) may be written down without any furthertrouble. u = C1e

m1x + C2em2x. (3)

EXAMPLES.–(1) Solve(

D2 + 14D − 32)

y = 0. Assume y = Cemx is a solution. Theauxillary becomes, m2 + 14m− 32 = 0. The roots are m = 2, or m = 16. The requiredsolution is, therefore, y = C1e

2x + C2e−16x.

(2) Solve d2y/dx2 −m2y = 0. Ansr. y = C1emx + C2e

−mx (see page 319).(3) Show that y = C1e

3x + C2ex is a complete solution of ←−

d2y/dx2 + 4 dy/dx+ 3y = 0.

Case 2. When the two roots of the auxillary are equal. if m1 = m2 in(3), it is no good putting (C1 + C2) e

m1x as the solution, because C1+C2, isreally one constant. The solution would then contain one arbitrary constantless than is required for the general solution. To find the other particularintegral, it is usual to put

m1 = m2 + h,

where h is some finite quantity which will ultimately be made zero. Withthis proviso, we write the solution,

y = limh→0

C1em1x + C2e

(m2+h)x.

Hence, y = limh→0

em1x(

C1 + C2ehx)

.

Now expand ehx by Maclaurin’s theorem (page 230). See pageS-22 ofthe sup-plementfor an al-ternativedevel-opmentof thisgeneralsolution.KH

∴ y = limh→0

em1x{

C1 + C2

(

1 + hx+ 12h2x2 + . . .

)}

;

= limh→0

em1x{

C1 + C2 + C2

(

hx+ 12!h2x2 + . . .

)}

;

= limh→0

em1x(

A+Bx+ 12!C2h

2x2 + C2R)

,

where R denotes the remaining terms of the expansion of ehx, A = C1+C2,B = C2h. Therefore, at the limit,

y = em1x (A+Bx) . (4)

For the sake of uniformity, we shall still write the arbitrary integrationconstants C1, C2, C3, . . .

308

For an equation of a still higher degree, the preceding result may bewritten,

y = em1x(

C1 + C2x+ C3x2 + · · ·+ Cr−2x

r−1)

. (5)where r denotes the number of equal roots.

EXAMPLES.–(1) Solve d3y/dx3 − d2y/dx2 − dy/dx+ u = 0. Assume y = emx. Theauxillary equation is m3 −m2 −m+ 1 = 0. The roots are 1, 1, −1. Hence the generalsolution can be written down at sight:

y = C1e−x + (C2 + C − 3x) ex.

(2) Solve(

D3 − 3D2 + 4)

y = 0. Ansr. e2x (C1 + C2x) + C3e−x.

See S-66of thesupple-ment forhow case3 arisesin theequa-tionsfor theorbits ofplanets.KH

Case 3. When the auxillary equation has imaginary roots, all unequal.Remembering that imaginary roots are always found in pairs in equationswith real coefficients (page 386), let the two imaginary roots be

m1 = α + ιβ; and m2 = α− ιβ.

Instead of substituting y = emx in (3), we substitute these values of m in(3) and get

y = C1e(α+ιβ)x + C2e

(α−ιβ)x;

= eαx(

C1eιβx + C2e

−ιβx)

;

= eαxC1 (cos βx+ ι sin βx) + C2 (cos βx− ι sin βx) . (6)

(See the chapter on “Hyperbolic Functions.”) Separate the real and imag-inary parts, as in Ex. 3, p. 280,

y = eαx (C1 + C2) cos βx+ ι (C1 − C2) sin βx;

if we put C1 + C2 = A, ι (C1 − C2) = B.

y = eαx (A cosβx+B sin βx) . (7)

In order that the constants A and B in (7) may be real, the constants C1

and C2 must include the imaginary parts.

EXAMPLES.–(1) Show from (6) that

y = (coshαx + sinhαx) (A1 cosβx+B2 sinβx) .

(Exercise on Chapter VI.)(2) Integrate d2y/dx2 + dy/dx + y = 0. The roots are α = − 1

2 and β = 12

√3;

∴ y = e−x/2(

A cos 12

√3x+B sin 1

2

√3x)

.(3) The equation of a point vibrating under the influence of a periodic force, is,

d2x

dt2+ α2x = a cos 2π

t

T.

Find the complementary function. The roots are ±ια. From (7)

x = A cosαt+B sinαt.

(4) If(

D3 −D2 +D − 1)

y = 0, y = C1 cosx+ C2 sinx+ Cex.

Case 4. When some of the imaginary roots of the auxillary equationare equal. If a pair of the imaginary roots are repeated,

309

we may proceed as in Case 2, since, when m1 = m2, C1em1x + C2ee

m2x isreplaced by (A +Bx) em1x; similarly, when m3 = m4, C3e

m3x+C3em4x may

be replace by (C +Dx) em3x. If, therefore,

m1 = m2 = α + ιβ; and m3 = m4 = α− ιβ,

the solution

y = (C1 + C2x) e(α+ιβ)x + (C3 + C4x) e

(α−ιβ)x,

becomes y = eαx (A+Bx) cos βx+ (C +Dx) sin βx. (8)

EXAMPLES.–(1) Solve(

D4 − 12D3 + 62D2 − 156D+ 169)

y = 0. Given the rootsof the auxillary: 3 + 2ι, 3 + 2ι, 3− 2ι, 3− 2ι. Hence

y = e3x {(C1 + C2x) sin 2x+ (C3 + C4x) cos 2x} .

(2) If(

D2 + 1)2

(D − 1)2 y = 0, (A+Bx) sinx+ (C +Dx) cosx+ (E + Fx) ex.

Second, when the second member is not zero, that is to say to findboth the complementary function and the particular integral. The generalequation is,

d2y

dx2+ P

dy

dx+Qy = R, (9)

where P and Q are constant, R is a function of x. We have just shownhow to find one part of the complete solution of the linear equation withconstant coefficients, namely, by putting R, in (9), equal to zero. Theremaining problem is to find a particular integral of this equation. Themore useful processes will be described in the next section.

In the symbolic notation, (9) may be written,

f (D) y = R (10)

The particular integral is, therefore,

y = f (D)−1R; or y =R

f (D). (11)

The right-hand side of either of equations (11), will be found to give asatisfactory value for the particular integral in question.

Since the complementary function contains all the constants necessaryfor the complete solution of the differential equation, it follows that nointegration constant must be appended to the particular integral.

§ 130. How to find Particular Integrals. It will be found quickest toproceed by rule:Case 1 (General). When the operator f (D)−1 can be resolved into factors.

310

Thereare other(some-timeseasier)methodsof findingpar-ticularintegrals.See thesupple-mentstartingat thebottomof pageS-23 fordetails.KH

We have seen that the linear differential equation of the first order,

dy/dx− ay = R; or y = R/ (D − a) , (1)

is solved byy = eax

∫

e−axRdx. (2)

The term Ceax in the solution of (1), belongs to the complementary func-tion.

Suppose that in a linear equation of a higher order, say,

d2y/dx2 − 5 dy/dx− 6y = R,

the operator f (D)−1 can be factorised. The complementary function iswritten down at sight from,

(

D2 − 5D + 6)

y = 0; or (D − 3) (D − 2) y = 0.

namely, y = C1e3x + C2e

2x. (3)

The particular integral is

y1 =1

(D − 3) (D − 2)R =

(

1

D − 3− 1

D − 2

)

R;

= e3x∫

e−3xRdx− e2x∫

e−2xRdx, (4)

from (2). The general solution is the sum of (3) and (4),

∴ y = C1e3x + C2e

2x + e3x∫

e−3xRdx− e2x∫

e−2xRdx.

EXAMPLES.–(1) In the preceding illustration, put R = e4x and show that the generalsolution is, C1e

3x + C2e2x + 1

2e4x.

(2) If(

D2 − 4D + 3)

y = 2e3x, y = C1ex + C2e

3x + xe3x.

Case 2 (General). When the operator f (D)−1 can be resolved intopartial fractions with constant numerators. The way to proceed in thiscase is illustrated in the first example below.

EXAMPLES.–(1) Solve d2y/dx2 − 3 dy/dx+ 2y = e3x. In symbolic notation this willappear in the form,

(D − 1) (D − 2) y = e3x.

The complementary function is y = C1ex + C2e

3x. The particular integral is obtainedby putting

y =1

(D − 2) (D − 1)e3x;

(

1

D − 2− 1

D − 1

)

e3x,

according to the method of resolution into partial fractions. Operate with the firstsymbolic factor, as above,

y1 = e2x∫

e−2xe3xdx− ex∫

e−xe3xdx =1

2e3x,

The complete solution is, therefore, y = C1ex + C2e

2x + 12e

3x.

(2) Solve (D − 2)2y = ex. Ansr. y = C1e

2x + C2xe2x + ex.

Case 3 (Special). When R is a rational function of x, say xn. This caseis comparatively rare. The procedure is to expand. f (D)−1 in ascendingpowers of D as far as the highest power of x in R.

311

EXAMPLES.–(1) Solve d2y/dx2 − 4 dy/dx+ 4y = x2. The complementary functionis y = e2x (A+Bx); the particular integral is:

(2−D)−2

x2 =1

4

(

1 + 2D

2+ 3

D

22

)

x2 =1

8

(

2x2 + 4x+ 3)

.

(2) If d2y/dx2 − y = 2+ 5x, y = C1ex + C2e

−x − 5x− 2.

Case 4 (Special). When R contains an exponential factor, so that

R = eaxX,

where X may or may not be a function of x and a, has some constant value.i. When X is a, function of x. Since Dneax = aneax, where n is any positive integer

(page 38), we have (page 95)

D (eaxX) = eaxDX + aeaxX = eax (D + a)X,

and generally, as in Leibnitz’ theorem (page 49), See pageS-25 ofthe sup-plementfor betternotationof (5)KH

DneaxX = eax (D + a)nX ;

∴Dneax

(D + a)n = eaxX ; and

eax

(D + a)nX =

eax

DnX.

(5)

The operation∫

D−1eaxX is performed (when X is any function of x) by transplanting

eax from the right- to the left-hand side of the operator f (D)−1

and replacing D by(D + a). This will, perhaps, be better understood from the following examples:

EXAMPLES.–(1) Solve d2y/dx2 − 2 dy/dx + y = x2e3x. The complete solution by

page 308, is (C1 + xC2) ex + (D − 2D + 1)

−1x2e3x. From (5),

1

D2 − 2D + 1x2e3x =

1

(D − 1) (D − 1)x2e3x.

By rule: e3x may be transferred from the right to the left side of the operator providedwe replace D by D + 3.

e3x1

(D + 2)2 x

2.

We get e3x(

1

4x2 − 1

2x+

3

8

)

,

as the value of the particular integral.

(2) Evaluate (D − 1)−1ex log x. Ansr. xex log x/e.

ii. When X a constant. If X is constant, the operation (5) reduces to

1

f (D)eax =

1

f (a)eax. (6)

The operation f (D) eaxX is performed by replacing D + a by a.

EXAMPLES.–(1) Find the particular integral in(

D2 − 3D + 2)

y = e3x. Obviously,

1

D2 − 3D + 2e3x =

1

32 − 3 · 3 + 2e3x =

1

2e3x.

(2) Show that 14e

x, is a particular integral in

d2y/dx2 + 2 dy/dx+ 1 = ex.

312

An anomalous case arises when a, is a root of f (D) = 0. By thismethod, we should get for the particular integral of dy/dx− y = ex.

1

D − 1ex =

ex

1− 1=∞ ex.

The difficulty is evaded by using the method (5) instead of (6). Thus,

1

D − 1ex = ex

1

D· 1 = x ex

The complete solution is, therefore, y = Cex + x ex.Another mode of treatment is the following: Since a is a root of f (D) = 0,

by hypothesis, D − a is a factor of f (D) (page 386). Hence,

f (D) = (D − a) f ′ (D) ;

∴1

f (D)eax =

1

(D − a)· 1

f ′ (D)eax =

1

(D − a)· 1

f ′ (a)eax =

x eax

f ′ (a).

(7)

If the root a occurs r times in f (D) = 0, then D−a enters r times intof (D). Therefore,

1

f (D)eax =

1

(D − a)r· 1

f ′ (D)eax =

1

(D − a)r· 1

f ′ (a)eax =

xreax

rf ′ (a). (8)

EXAMPLES. – Find the particular integrals in, (1) (D + 1)3y = e−x. Ansr. 1

6x3e−x.

Hint. Replace D by D − 1. e−xD−3, etc. See page 312.

(2)(

D3 − 1)

y = x ex. Ansr. ex(

16x

2 − 13x)

. Hint. First get ex (D − 1)−1x, then ←−

ex (1 +D + . . . )x, etc.

Case 5 (Special). When R contains sine or cosine factors. By thesuccessive differentiation of sin (nx+ a),

(

D2)n

sin (nx+ a) =(

−n2)n

sin (nx+ a) . ∗where n and a are constants.

∴ f(

D2)

sin (nx+ a) = f(

−n2)

sin(nx+ a).

1

f (D2)sin (nx+ a) =

1

f (−n2)sin (nx+ a) . (9)

It can be shown in the same way that,

1

f (D2)cos (nx+ a) =

1

f (−n2)cos (nx+ a) . (10)

EXAMPLES.–(1) Find the particular integral of

d3y/dx3 + d2y/dx2 + dy/dx+ y = sin 2x.

Here,R

f (D)=

1

D3 +D2 +D + 1sin 2x =

1

(D2 + 1) +D (D2 + 1)sin 2x.

* The proof resembles a well-known result in trigonometry, § 19:–D (sinnx) = d (sinnx) /dx = n cosnx;

D2 (sinnx) = d2 (sinnx) /dx2 = −n2 cosnx, etc.

313

Substitute for D2 =(

−22)

as in (9). We thus get − 13 (D + 1)

−1sin 2x. Multiply by

D − 1 and again substitute D2 =(

−22)

in the result. Thus115 (D − 1) sin 2x, or 1

15 (2 cos 2x− sin 2x)

is the desired result.(2) Solve d2y/dx2 − k2y = cosmx. Ansr. C1e

kx + C2e−kx − (cosmx) /

(

m2 + k2)

.(3) If α and β are the roots of the auxillary equation derived from

d2y/dt2 +mdy/dt+ n2y = a sinnt,

(Helmholtz’s equation for the vibrations of a tuning-fork) show that,

C1eαt + C2e

βt − (a cosnt) /mn.

is the complete solution.

An anomalous case arises when D2 in D2 + n2 is equal to −n2. For in-stance, the complementary function of d2y/dx2+n2y = cosnx, is C1 cosnx+C2 sinnx, the particular integral is (D

2 + n2) cosnx. If the attempt is madeto evaluate this, by substituting D2 = −n2, we get (cos nx) / (−n2 + n2) =∞ cosnx. We were confronted with a similar difficulty on page 243. Thetreatment is practically the same. We take the limit of (D2 + n2) cosnx,when n becomes n + h and h converges towards zero.

1

−n2 + n2cosnx; or

1

− (n+ h)2+ n2

cos (n+ h)x.

∴ limh→0

1

− (n+ h)2 + n2cos (n+ h)x = lim

h→0

1

−2nh− h2 cos (nx+ hx) ;

= limh→0− 1

2nh+ h2(cosnx · coshx− sinnx · sinhx) ;

= limh→0− 1

2nh+ h2

{

cosnx

(

1− h2x2

2!+ . . .

)

− sinnx (hx− . . . )}

;

= limh→0− 1

2n+ h

(cosnx

h− x sinnx+ powers of h

)

.

But cosnx is contained in the complementary function and hence, when h = 0, weobtain, x sinnx

2n+ (a term in the complementary function) .

This latter may be disregarded when the particular integral alone is under consideration.The complete solution is, therefore,

y = C1 cosnx+ C2 sinnx+ (x sinnx) /2n.

EXAMPLES.–(1) Show that − 12x cos x, is the particular integral of(

D2 + 1)

y = sinx

(2) Evaluate(

D2 + 4)−1

cos 2x. Ansr. 14x sin 2x.

(3) Evaluate(

D2 + 4)−1

sin 2x. Ansr. 14x cos 2x.

(4) Solve d3y/dx3 − y = x sinx. The particular integral consists of two parts,12 {(x− 3) cosx− x sinx}. The complementary function is

C1ex + C2e

−x/2 sin(

12

√3x)

+ C3ex/2 cos

(

12

√3x)

.

But see next case.

Case 6. When R contains some power of x as a factor. Say,

R = nX,

314

whereX is any function of x. The successive differentiation of two products See pageS-31 ofthe sup-plementfor theLaplacemethod,which isa singleapproachto solv-ing all6 of thecasesin thissectionandmore.§ 20 isincludedfollowingthe sup-plement.KH

gives, § 20,DnxX = xDnX + nDn−1X.

∴ f (D) xX = xf (D)X + f ′ (D)X.

Substitute Y = f (D)X , where Y is any function of x. Operate withf (D)−1, we get

1

f (D)xY =

{

x− 1

f (D)· f ′ (D)

}

1

f (D)Y. (11)

EXAMPLES.–(1) Find the particular integral in d3y/dx3 − y = xe2x. From (11), theparticular integral is

=

{

x− 1

D3 − 1· 3D2

}

1

D3 − 1e2x =

{

x− 1

7· 3 · 4

}

1

7e2x, etc.

(2) Show in this way, that the particular integral of(

D4 − 1)

y = x sinx,

is 18

(

x2 cosx− 3x sinx)

.(3) Solve d2y/dx2 − y = x ex sinx.

Ansr. y = C1ex + C2e

−x − 125e

x {(10x+ 2) cosx+ (5x− 14) sinx}.(4) Integrate d2y/dx2−y = x2 cosx. Ansr. y = C1e

x+C2e−x+x sinx+ 1

2 cosx(

1− x2)

.

§ 131. The Linear Equation with Variable Coefficients.

Case 1. The homogeneous linear differential equation. The generaltype of this equation is:

xn dny

dxn+ a1x

n−1 dn−1y

dxn−1+ · · ·+ any = X, (1)

where X is a function of x; a1, a2, . . . , an are constants. This equation canbe transformed. into one with constant coefficients by the substitution of

x = ez; or z = log x.

we then have,

dy/dx = ez and therefore, x dy/dx = dy/dz. (2)

Just as we have found it very convenient to employ the symbol, “D” todenote the operation, “ d

dx,” so we shall find it even more convenient to

denote the operation, “x ddx,” by the symbol, “ϑ.” “ϑ” is treated in ex-

actly the same manner as we have treated “D”* in § 128 and subsequently.

* A little care is required in using this new notation. The operations of differentia-tion and multiplication by a variable are not commutative. The operation x2D2 is notthe same as ϑ2, or as xD · xD. But we must write,

xDy = ϑy;

x2D2y = ϑ (ϑ− 1) y;

x3D3y = ϑ (ϑ− 1) (ϑ− 2) y;

. . . . . .

xnDny = ϑ (ϑ− 1) (ϑ− 2) . . . (ϑ− n+ 1) y.

315

EXAMPLES.–(1) ϑ = xD = x ddx = d

dz .(2) Show that ϑxm = mxm

i. The complementary function. From the first of equations (2), wehave § 12, 9,

dy

dx=

dy

dz· dzdx

=1

x· dydx

;d2y

dx=

1

x

(

d2y

dx2− dy

dz

)

; etc.

Substitute these values in (1). The equation reduces to one with constantcoefficients which may be treated by the methods described in the precedingsections.

EXAMPLES.–(1) Solve x3 · d3y/dx3 + 3x · dy/dx− 3y = x2 + x.

∴ ϑ (ϑ− 1) (ϑ− 2) y + 3ϑy − 3y = e2z + ez.

∴ (ϑ− 1)(

ϑ2 + 3)

y = e2z + ez.

∴ y = C1ez + C2 cos

(

z√3)

+ C3 sin(

z√3)

+1

7e2z +

1

4zez.

∴ y = C1x+ C2 cos(√

3 log x)

+ C3 sin(√

3 log x)

+1

7x+

1

4x log x.

(2) Solve x2 · d2y/dx2 + x · dy/dx+ q2y = 0; i.e.,(

ϑ2 + q2)

y = 0.

Ansr. y = C1 sin (q log x) + C2 cos (q log x).

The linear equation with variable coefficients bears the same relationto x, that the equation with constant coefficients does to emx. Hence ifxm be substituted for y, the factor xm will divide out from the result andan equation in m will remain. The n roots of this latter equation willdetermine the complementary function.

EXAMPLES.–(1) Solve x · d2y/dx2 + 2x · dy/dx− 2y = 0. Put y = xm. We get

m (m− 1) + 2 (m− 1) = 0; or (m+ 2) (m− 1) = 0.

Hence from our preceding results, we can write down the complementary function atsight, y = C1x+ C2x

−2.(2) Solve x2 · d2y/dx2 + 4x · dy/dx+ 2y = 0. Ansr. y = C1/x+ C2/x

2.(3) Find the complementary function in {ϑ (ϑ− 1)− 3ϑ+ 4} y = x3.

Ansr. y = (C1 + C2 log x)x2.

(4) Integrate {ϑ (ϑ− 1)− 2} y = 0. Ansr. y = C1x2 + C2/x.

ii. The particular integral. We may use the operator ϑ, to obtain theparticular internal of linear equations with variable coefficients in the sameway that D was used to determine the particular integral of equations withconstant coefficients.

The symbolic form of the particular integral is,

y =R

f (ϑ).

The operator f (ϑ)−1 may be resolved into partial fractions or into factorsas in the case of D.

316

EXAMPLES.–(1) Show that y = C1x4 + C2/x + 1

5x4 log x a complete solution of

x2 · d2y/dx2 − 2x · dy/dx− 4y = x4.(2) Find the value of 1

ϑ2(ϑ−3) x3. Using the ordinary method just described we get

the indeterminate form 19 · x3

ϑ−3 . In this case we must adopt the method of page 312 andwrite

1

9x3

1

ϑ− 1· 1 =

1

9x3∫

dx

x=

1

9x3 log x.

(3) Solve x3 · d2y/dx2 + 7x · dy/dx+ 5y = x5. Write this

{ϑ (ϑ− 1) + 7ϑ} y = x5.

The particular integral is(

ϑ2 + 6ϑ+ 5)−1

x5, or x5/60. complementary function isC1x

−1 + C2x−5.

(4) Solve x2 · d2y/dx2 + 4x · dy/dx+ 2y = ex.

Ansr. y = C1/x+ C2/x2 + ex/x2.

(5) Solve x3 · d3y/dx3 + 2x2d2y/dx2 − x · dy/dx+ y = x+ x3.

Ansr. y = C1/x+ C2x+ C3x log x+ 14x (log x)

2+ x3/16.

(6) Solve x3 · d3y/dx3 + 2x2 · d2y/dx2 + 2y = 10x+ 10/x.Ansr. y = C1x cos (log x) + C2x sin (log x) + 5x+ C3/x+ (2 logx) /x.(7) Find the particular integral of the third example in the last set. Ansr. x3.(8) Equate example (2), of the preceding set, to 1/x instead of to zero, and show

that the particular integral is then (log x) /x.

Here the“latter”appearsto be(3) andthe firstexamplebut the“former”not tohave anexampleshown.Otherthanchoice ofsymbols,the twosubstitu-tions areequiv-alent.KH

Case 2. Legendre’s Equation. Type:

(a+ bx)ndny

dxn+ A1 (a + bx)n−1 d

n−1y

dxn−1+ · · ·+ Any = R, (3)

where A1, A2, . . .An are constants, R is any function of x. This sort ofequation is easily transformed into the homogeneous equation and, there-fore, into the linear equation with constant coefficients. To make the formertransformation, substitute z = ax+ b, for the latter, e = a+ bx.

EXAMPLES.–(1) Solve

(a+ bx)2 · d2y/dx2 + b (a+ bx) · dy/dx+ c2y = 0.

Ansr. y = C1 sin {(c/b) log (a+ bx)}+ C2 cos {(c/b) log (a+ bx)}.(2) Solve (x+ a)

2 · d2y/dx2 − 4 (x+ a) · dy/dx+ 6y = x.

Ansr. y = C1 (x+ a)2+ C2 (x+ a)

2+ 1

6 (3x+ 2a).

§ 132. The Exact Linear Differential Equation. A very simplerelation exists between the coefficients of an exact differential equationwhich may be used to test whether the equation is exact or not. Take theequation,

X0d3y

dx3+X1

d2y

dx2+X2

dy

dx+X3y = R. (1)

317

where X1, X2, . . . , R are functions of x. Let their successive differentialcoefficients be indicated by dashes, thus X ′, X ′′, . . .

SinceX0·d3y/dx3 has been obtained by the differentiation ofX0·d2y/dx2,this latter is necessarily the first term of the integral of (1). But,

d

dx

(

X0d2y

dx2

)

= X0d3y

dx3+X ′

0

d2y

dx2.

Subtract the right-hand side of this equation from (1),

(X1 −X ′0)

d2y

dx2+X2

dy

dx+X3 y = R. (2)

Again, the first term of this expression is a derivative of (X1 −X ′0) dy/dx.

This, therefore, is the second term of the integral of (1). Hence, by differ-entiation and subtraction, as before,

(X2 −X ′1 +X ′′

0 )dy

dx+X3 y = R. (3)

This equation may be deduced by the differentiation of (X2 −X ′1 +X ′′

0 ) y,provided the first differential coefficient of (X2 −X ′

1 +X ′′0 ) with respect to

x, is equal to X3 that is to say,

X ′2 −X ′′

1 +X ′′′0 = X3; or X3 −X ′

2 +X ′′1 −X ′′′

0 = 0. (4)

But if this is really the origin of (3), the original equation (1) has beenreduced to a lower order, namely,

X0d2y

dx2+ (X1 −X ′

0)dy

dx+ (X2 −X ′

1 +X ′′0 ) y =

∫

Rdx+ C. (5)

This equation is called. the first integral of (1), because the order of the See sup-plementpageS-39 forclarifi-cationof howthe firstintegralarises.KH

original equation has been lowered unity, by a process of integration.Condition (4) is a test of the exactness of a differential equation.If the first integral is an exact equation, we can reduce it, in the same

way, to another first integral of (1). The process of reduction may berepeated until an inexact equation appears, or until y itself is obtained.Hence, an exact equation of the nth order has n independent first integrals.

EXAMPLES.–(1) Is the equation

x5 · d3y/dx3 + 15x4 · d2y/dx2 + 60x3 · dy/dx+ 60x2y = ex exact?

From (4), X3 = 60x2; X ′2 = 180x2; X ′′

1 = 180x2; X ′′′0 = 60x2. Therefore,

X3 −X ′2 +X ′′

1 −X ′′′0 = 0 and the equation is exact.

(2) Solve the equation, x · d3y/dx3 +(

x2 − 3)

d2y/dx2 + 4x · dy/dx+ 2y = 0, as faras possible, by successive reduction. The process can be employed twice, the residue isa linear equation of the first order, not exact.

(3) Solve the equation given in example (1).

Ansr. x5y = ex + C1x2 + C2x+ C3.

318

There is another quick practical test for exact differential equations(Forsyth) which is not so general as the preceding. When the terms inX are either in the form of axm, or of the sum of expressions of this type,xmdny/dxn is a perfect differential coefficient, if m < n. This coefficientcan then be integrated whatever be the value of y. If m = n or m > n,the integration cannot be performed by the method for exact equations.To apply the test, remove all the terms in which m is less than n, if theremainder is a perfect differential coefficient, the equation is exact and theintegration may be performed.

EXAMPLES.–(1) Apply the test to

x3 · d4y/dx4 + x2 · d3y/dx3 + x · dy/dx+ y = 0.

x · dy/dx + y remains. This has evidently been formed by the operation D (xy), hencethe equation is a perfect differential.

(2) Apply the test to(

x3D4 + x2D3 + x2D + 2x)

y = sinx.

x2 · dy/dx + 2xy remains. This is a perfect differential, formed from D(

x2y)

. Theequation is exact.

e.g., ifbetweenthe 1stintegralandthe 1stintegralof the 1stintegralof theequa-tion...KH

If two independent first integrals are known the equation is sometimeseasily solved. The elimination of dy/dx between two first integrals will givethe complete solution.

§ 133. The Integration of Equations with Missing Terms. Differ-ential equations with missing letters are common.

First, the independent variable is absent. Type:

d2y/dx2 = qy; or d2y/dx2 = q f (y) . (1)

This equation is, in general, neither linear nor exact.Case 1. When f (y), in (1), is negative, so that

d2y

dx2+ q2x = 0, (2)

where the academic x and y have given place to t and x respectively, inorder to give the equation the familiar form of the equation of the motionof a particle under the influence of a central attracting force.

Multiply both sides of the equation by 2dx/dt, and integrate with re-spect to x,

2dx

dt· d

2x

dt2= −2q2x dx

dt; or

(

dx

dt

)2

= −q2x2 + C.

Separate the variables and integrate again,

dx√α2 − x2

= ±q dt; cos−1 x

α= ± (qt+ ε) ,

319

where ε is the integration constant and C = q2α2. The solution involvestwo arbitrary constants α and ε, which respectively denote the amplitudeand epoch of a simple harmonic motion, whose period of oscillation is 2π/q.Put C1 = −α cos ε, and C2 = −α sin ε.

x = C1 cos qt + C2 sin qt.

[Case 2.] When f (y) in (1) is positive, the solution assumes the form,

x = C1eqt + C2e

−qt; or x = A cosh qt+B sinh qt,

as on page 309. All these results are important in connection with alter-nating currents and other forms of harmonic motion.

Another way of treating equations of type (9), occurs with an equationlike

yd2y

dx2+

(

dy

dx

)2

− 2y2 = 0, (3)

which has the form of the standard. equation for the small oscillationsof a pendulum in air. Under this condition, the resistance of the air isnegligible. Let

p = dy/dx, ∴ d2y/dx2 = dp/dx = p · dp/dy.

Substitute these results in the given equation, multiply through with 2/y.

pydp

dy+ p2 = 2y2; or 2p

dp

dy+

2

yp2 = 4y.

Multiply by y2 and

d (y2p2)

dy= 4y3; or p2y2 = y4 + C4,

where C4 is an arbitrary constant. The rest is obvious.

EXAMPLES.–(1) The solution of equation (3) is sometimes written in the form

y2 = C21 sinh (2x+ C2) .

Verify this.(2) Solve d2x/dt2 + µx + ν = 0. Put x = x1 + ν/µ and afterwards omit the suffix.

Ansr. x = ν/µ+ C1 cos t√µ+ C2 sin t

õ.

(3) If the term µx in the preceding example had been of opposite sign, show thatthe solution would have been, x = ν/µ+C1 cosh t

õ+C2 sinh t

√µ, where µ is negative.

(4) Solve d2y/dx2 − a (dy/dx)2 = 0. Ansr. C1x+ C2 = e−ay.

(5) Solve 1 + (dy/dx)2= y d2y/dx2. Ansr. y = a cosh (x/a+ b).

(6) Fourier’s equation for the propagation of heat in a cylindrical bar, isd2V/dx2 − β2V = 0. Hence show that V = C1e

βx + C2e−βx.

Second, the dependent variable is absent. Type:

d2y/dx2 = x; or d2y/dx2 = f (x) . (4)

320

If these equations are exact, they may be solved by successive integration.If the equation has the form

d2y/dx2 + dy/dx+ x = 0.

Say,d2v

dr2+

1

r· dvdr

= −P

lµ,

a familiar equation in hydrodynamics, it is usually solved by substitutingy = dp/dx, ∴ dp/dx = d2p/dx2. The resulting equation is of the firstorder, integrable in the usual way.

EXAMPLES.–(1) The above equation represents the motion of a fluid in a cylindricaltube of radius r and length l. The motion is supposed to be parallel to the axis of thetube and the length of the tube very great in comparison with its radius r. P denotesthe difference of the pressure at the two ends of the tube. If the liquid wets the walls ofthe tube, the velocity is a maximum at the axis of the tube and gradually diminishes tozero at the walls. This means that the velocity is a function of the distance (r1) of thefluid from the axis of the tube. Solve the equation, remembering that µ is a constantdepending on the nature of the fluid.

Substitute p = dv/dr,

∴ dp/dr + p/r = −P/lµ;

rdp

dr+ p = −P

lµr, is pr = − P

2lµr2 + C1; (5)

∴dv

dr= − P

2lµr +

C1

r; v = − P

4lµr2 + C1 log r + C2.

To evaluate C1 in (5), note that at the axis of the tube r = 0. This means that ifC1, is a finite or an infinite magnitude the velocity will be infinite. This is obviouslyimpossible, therefore, C1, must be zero. To evaluate C2, note that when r = r1, vvanishes and, therefore, we get the final solution of the given equation in the form,v = 1

4P(

r21 − r2)

/lµ, which represents the velocity of the fluid at a distance r1, fromthe axis.

(2) Solve ad2y/dx2 =

√

1 + (dy/dx)2. Make the necessary substitutions and inte-

grate.a · dp/

√

(1 + p2); becomes x/a log(

p+√

p2 + 1)

+ C;

or, in the exponential form,

C1ex/a − p =

√

(1 + p2); and dy/dx =1

2C1e

x/a − e−x/a/2C1;

by squaring. On integration

y =1

2aC1e

x/a +1

2ae−x/a/C1 + C2.

(3) Some expressions can be reduced to the standard form by an obvious transfor-mation. Thus,

d5y/dx5 − d3y/dx3 = x.

Substitute p for d3y/dx3 and differentiate p = d3y/dx3 twice. Thus,

d2p/dx2 − p = x,

whence y can be obtained by successive integration as indicated above.(4) Solve d2V/dr2 + 2dV/r · dr = 0. This equation occurs in the theory of

321

potential. Put dV/dr for the independent variable and divide through. On integration

log (dV/dr) + 2 log r = logC1.

where log C1 is an arbitrary constant. Integrate again

dV/dr = C1/r, becomes V = C + 2− C1/r.

(5) If x · d2y/dx2 = 1, show that y = x log x+ C1x+ C2.

§ 134. Equations of Motion, chiefly Oscillatory Motion. By New-tons second law, if a certain mass (m) of matter is subject to a constantforce (F0) for a certain time; we have, in rational units,

F0 = (Mass) × (Acceleration of the particle).

If the motion of the particle is subject to friction, we must regard thefriction as a force tending to oppose the motion generated by the impressedforce. But friction is proportional to the velocity (v) of the motion of theparticle, and equal to the product of the velocity and a constant called thecoefficient of friction, written, say, µ. Let F1 denote the total force actingon the particle in the direction of its motion,

F1 = F0 − µv = md2s/dt2. (1)

If there is no friction, we have, for unit mass,

F0 = d2s/dt2. (2)

The motion of a pendulum in a medium which offers no resistance toits motion, is that of a material particle under the influence of a centralforce (F ) attracting with an intensity which is proportional to the distanceof the particle away from the centre of attraction. That is (Fig. 7),

F = −q2s. (3)

where q2 is to be regarded as a positive constant which tends to restore theparticle to a position of equilibrium – the so-called coefficient of restitution.It is written in the form of a power to avoid a root sign later on. Thenegative sign shows that the attracting force (F ) tends to diminish thedistance (s) of the particle away from the centre of attraction. If s = 1, q2

represents the magnitude of the attracting force unit distance away. From(2),

d2s

dt2= −q2s (4)

This is a typical equation of harmonic motion, as will be shown directly.One solution of (4) is

s = C cos (at + ε) . (5)

322

This equation is the simple harmonic motion of § 50, C denotes the am- § 50reviewsthe prop-erties ofsinusoidsas theycan beused todescribeperiodicsinu-soidalmotion.

plitude of the vibration. If ε = 0, we have the simpler equation,

s = C cos qt. (6)

When the particle is at its greatest distance from the central attract-ing force, qt = π, § 50, page 112. For a complete to and fro motion,2t = T0 = period of oscillation, hence

T0 = 2π/q. (7)

Equation (4) represents the small oscillations of a pendulum; also theundamped* oscillations of the magnetic needle of a galvanometer.

In the sine galvanometer, the restitutional force tending to restore the needle toa position of equilibrium, is proportional to the sine of the angle of deflection of theneedle. If J denotes the moment of inertia of the magnetic needle and G the directiveforce exerted by the current on the magnet, the equation of motion of the magnet, whenthere is no retarding force, is

Jd2φ

dt2= −G sinφ. (8)

For small angles of displacement, φ and sin θ are approximately equal. Hence,

d2φ

dt2= −G

Jφ. (9)

From (4), q =√

J/G, and therefore, from (9),

T0 = 2π√

J/G, (10)

a well-known relation showing that the period of oscillation of a magnet in the magneticfield, when there is no damping action exerted on the magnet, is proportional to thesquare root of the moment of inertia of the magnetic needle, and inversely proportionalto the square root of the directive force exerted by the current on the magnet. See page524.

In a similar manner, it can be shown that the period of the small oscillations ofa pendulum suspended freely by a string of length l, is 2π

√

l/g, where g denotes theacceleration of gravity.

Equation (4) takes no account of the resistance to which a particle issubjected as it moves through such resisting media as

* When an electric current passes through a galvanometer, the needle is deflected andbegins to oscillate about a new position of equilibrium. In order to make the needlecome to rest quickly, so that the observations may be made quickly, some resistanceis opposed to the free oscillations of the needle either by attaching mica or aluminumvanes to the needle so as to increase the resistance of the air, or by bringing a mass ofcopper close to the oscillating needle. The currents induced in the copper by the motionof the magnetic needle, react on the moving needle, according to Lenz’s law, so as toretard its motion. Such a galvanometer is said to be damped. When the damping issufficiently great to prevent the needle oscillating at all, the galvanometer is said to be“dead beat” and the motion of the needle is aperiodic. In ballistic galvanometers, thereis very much damping.

323

air, water, etc. This resistance is proportional to the velocity, and has anegative value. To allow for this, equation (4) must have an additionalnegative term. We thus get

d2s

dt2= −µ ds

dt− q2s,

where µ, is the coefficient of friction. For greater convenience, we may writethis 2f ,

d2s

dt2+ 2f

ds

dt+ q2s = 0. (11)

Before proceeding further, it will perhaps make things plainer to putthe meaning of this differential equation into words. The manipulation ofthe equations so far introduced, involves little more than an application ofcommon algebraic principles. Dexterity in solving comes by practice. Ofeven greater importance than quick manipulation is the ability to form aclear concept of the physical process symbolised by the differential equation.Some of the most important laws of Nature appear in the guise of an“unassuming differential equation”. The reader should spare no pains toacquire familiarity with the art. The late Professor Tait has said that “amathematical formula, however brief and elegant, is merely a step towardsknowledge, and an all but useless one until we can thoroughly read itsmeaning.”

In equation (11), the term d2s/dt2 denotes the relative change of thevelocity of the motion of the particle in unit time, § 7; 2f · ds/dt showsthat this motion is opposed by a force which tends to restore the body toa position of rest, the greater the velocity of the motion, the greater theretardation; q2s represents another force tending to bring the moving bodyto rest, this force also increases directly as the distance of the body fromthe position of rest. To investigate this motion further, we cannot do betterthan follow Professor Perry’s graphic method.

The first thing is to solve (11) for s. This is done by the method of§ 130. Put s = emt and solve the auxillary quadratic equation. We thusobtain

m = f ±√

f 2 − q2. (12)

And finally,

s = e−(α+β)t; or rather s = C1e−αt + C2e

−βt,

where α = −f +√

f 2 + q2 and β = −f −√

f 2 + q2. The solution of (11)thus depends on the relative magnitudes of f and q.

Suppose that we know enough about the moving system to be able todetermine the integration constants. When t = 0, let v = v0 and s = 0.

324

Case i. The roots of the auxillary equation are real and unequal. The condition forreal roots −α and −β, in (12), is that f is greater than q (page 388). In this case,

s = C1e−αt + C2e

−βt, (13)

solves equation (11). To find what this means, let us suppose that f = 3, q = 2, t = 0,s = 0, v = v0. From (12), therefore,

m = −3±√9− 4 = −3± 2.24 = −0.76 and − 5.24.

Substitute these values in (18) and differentiate for the velocity v or ds/dt. Thus,

s = C1e−5.24t + C2e

−0.76t; ds/dt = −5.24C1e−5.24t − 0.76C2e

−0.76t.

∴ −5.24C1 − 0.76C2 = 1.

From (18), when t = 0, s = 0 and C1 + C2 = 0, or −C1 = +C2 = 16 .

∴ s = 16

(

e−0.76t − e−5.24t)

. (14)

Assign particular values to t, and plot the corresponding values of s by means of TablesXXI and XXII. Curve No. 1 (Fig. 114) was obtained by plotting corresponding valuesof s and t obtained in this way.

Figure 114: (after Perry).

Case ii. The roots of the auxillary equation are real and equal. The condition forreal and equal roots is that f = q.

∴ s = (C1 + C2t) e−ft. (15)

As before, let f = 2, q = 2, t = 0, s = 0, v0 = 1. The roots of the auxillary are −2 and−2. Hence

s = (C1 + C2t)e−2t; and ds/dt = C2e

−2t − 2 (C1 + C2t) e−2t.

∴ C2 − 2C1 = 1, C1 = 0 and C2 = 1; or s = te−2t. (16)

Plot (16) in the usual manner. Curve 2 (Fig. 114) was so obtained.It wouldappearthat thetext isrefer-ring towhen theroots arepurelyimag-inary(multi-ples ofι) andoppositesign. KH

Case iii. The roots of the auxillary equation are real, equal and of opposite sign.For equal roots of opposite sign, say ±q, we must have f = 0. Then

s = C1 sin qt+ C2 cos qt. (17)

Let t = 0, s = 0, v0 = 1, q = 2, f = 0. Differentiate (17),

ds/dt = qC1 cos qt− qC2 sin qt.

325

Hence 1 = 2C1 × 1− 2× C2 × 0, or C1 = 12 ; ∴ C2 = 0. Hence the equation,

s = 12 sin 2t. (18)

A graph from this equation is shown in curve 4 (Fig. 114).Case iv. The roots of the auxillary equation are imaginary. For imaginary roots,

f +√

f2 − q2, or, say a± bι, it is necessary that f < q (page 388). In this case,

s = e−at (C1 sin bt+ C2 cos bt) . (19)

Let the coefficient of friction, f = 1, q = 2, t = 0, s = 0, v0 = 1. The roots of theauxillary are m = −1±

√1− 4 = −1±

√−3 = −1+1.7ι, where ι =

√−1. Hence a = 1,

b = 1.7. Differentiate (19),

ds/dt = −ae−at (C1 sin bt+ C2 cos bt) + be−at (C1 sin bt− C2 cos bt) .

From (19), C2 = 0 and, therefore, C1 = 1/b = −0.57. Therefore,

s = 0.57e−t sin 1.7t. (20)

Curve 3 (Fig. 114) was plotted from equation (20) in the usual way.

There are several interesting feature about the motions represented bythese four solutions of (11), shown graphically in Fig. 114. Curves Nos. 3and. 4 (Cases iv. and iii.) show the conditions under which the equationof motion (11) is periodic or vibratory. The effects of increased frictiondue to the viscosity of the medium, is shown very markedly by the lessenedamplitude and increased period of curve 3. The net result is a dampedvibration, which dies away at a rate depending on the resistance of themedium (2f · v) and on the magnitude of the oscillations (q2s). Such is themotion of a magnetic or galvanometer needle affected by the viscosity ofthe air and the electromagnetic action of currents induced in neighbouringmasses of metal by virtue of its motion; it also represents the naturaloscillations of a pendulum swinging in a medium whose resistance variesas ihe velocity. Curve 4 represents an undamped oscillation, curve 3 adamped oscillation.

Curves 1 and 2 (Cases i. and ii.) represent the motion when the retard-ing forces are so great that the vibration cannot take place. The needle,when removed from its position of equilibrium, returns to its position ofrest after the elapse of an infinite time. (What does this statement mean?Compare with page 329.) Raymond calls this an aperiodic motion.

To show that the period of oscillation is augmented by damping. Fromequation (19) we can show that

s = e−atA sin bt. (21)

§ 50. The amplitude of this vibration corresponds to that value of t forwhich s has a maximum or a minimum value. These values are obtained inthe usual way, by equating the first differential coefficient to zero, hence

e−at (b cos bt− a sin bt) = 0. (22)

326

If we now define the angle φ such that bt = φ, or

tanφ = a/b, (23)

φ, lying between 0 and 12π (i.e., 90◦), becomes smaller as a increases in value. We have

just seen that the imaginary roots of −f ±√

f2 − q2 are −a + bι, for values of f lessthan q. Let

a2 + b2 = q2. (24)The period of oscillation of an undamped oscillation is, by (7), T0 = 2π/q, of a

damped oscillation T = 2π/b.

∴ T 2/T 20 = q2/b2 =

(

a2 + b2)

/b2 = 1 + a2/b2,

∴ T/T0 =√

a2 + b2/b. (25)

which expresses the relation between the periods of oscillation of a damped and of anundamped oscillation. The period of vibration is thus augmented on damping.

It is easy to show by plotting that tanφ, of (28), is a periodic function such that

tanφ = tan (φ+ π) = tan (φ+ 2π) = . . .

Hence φ, φ+ π, φ+ 2π, . . .

satisfy the above equation. It also follows that

bt1, bt2 + π, bt3 + 2π, . . .

also satisfy the equation, where t1, t2, t3 . . . are the successive values of the time. Hence

bt2 = bt1 + π, bt3 = bt1 + 2π, . . . ;

∴ t2 = t1 +1

2T, t3 = t1 + T, . . .

Substitute these values in (21) and put s1, s2, s3, . . . for the corresponding displacements,

∴ s1 = Ae−at1 sin bt1; − s2 = Ae−at2 sin bt2; . . .where the negative sign indicates that the displacement is on the negative side. Hence

s1/s2 = e−a(t1−t2) = eaT/2;

s2/s3 = s3/s4 = · · · = eaT/2. (26)

The amplitude thus diminishes in a constant ratio. Plotting these successive

Figure 115: Damped Oscillation

values of s and t, weget the curve shown inFig. 115. This ratiois called the damping ra-tio, by Kohlrausch (“Damp-fungsverhaltnis”). It is writ-ten k. The natural loga-rithm of the damping ratio,is Gauss’ logarithmic decre-ment, written L (the ordi-nary logarithm of k, is writ-ten L). Hence

λ = log k = aT log e = aT = aπ/b, (27)and from (25),

T 2

T 20

= 1 +λ2

π2; or T = T0

(

1 +1

2· λ

2

π2+ . . .

)

. (28)

327

Hence, if the damping is small, the period of oscillation is augmented by a small quantityof the second order.

The following table contains six observations of the amplitudes of a sequence ofdamped oscillations:

ObservedDeflection k. λ. L.

69 1.438 0.3633 0.157848 1.434 0.3604 0.156533.5 1.426 0.3548 0.154123.5 1.425 0.3542 0.153816.5 1.435 0.3612 0.156911.5 1.438 0.3633 0.15788

Meyer, Maxwell, etc., have calculated the viscosity of gases from the rate at whichthe small oscillations of a vibrating pendulum are damped.

When the motion represented by equation (11) is subject to some pe-riodic impressed force which prevents the oscillations dying away, the re-sulting motion is said to be a forced vibration. The equation representingsuch an oscillation is

d2s

dt2+

ds

dt+ q2s = f (x) . (29)

When f (x) = 0, the equation refers to the natural oscillations of a vi-brating electrical or mechanical system. The impressed force is, therefore,mathematically represented by the particular integral of equation (29) (seeexample (3) below).

The subjoined examples’principally refer to systems in harmonic mo-tion.

Moderntextsuse thesymbol,i, forcurrentinsteadof C.Thelatteris cus-tomarilyused forelectricalcapac-itanceratherthancurrent.KH

EXAMPLES.–(1) Ohms law for a constant current is E = RC; for a variable currentof C amperes flowing in a circuit with a coefficient of self-induction of L henries, witha resistance of R ohms and an electromotive force of E volts, Ohms law is representedby the equation,

E = RC = L · dC/dt, (30)

where dC/dt evidently denotes the rate of increase of current per second, L is theequivalent of an electromotive force tending to retard the current.

(i.) When E is constant, the solution of (30) has been obtained in a preceding setof examples,

C = E/R+Be−Rt/L,

where B is the constant of integration. To find B, note that when t = 0, C = 0. Hence,

C = E(

1− e−Rt/L)

R. (31)

The second term is the so-called “extra current at make,” an evanescent

328

Modernengineer-ing textsuse theword,“tran-sient”ratherthan“evanes-cent.”KH

factor due to the starting conditions. The current, therefore, tends to assume the steadycondition: C = E/R, when t is very great.

(ii.) When C is an harmonic function of the time, say,

C = C0 sin qt; ∴ dC/dt = C0 q cos qt,

or, compounding these harmonic motions (§ 50),

E = RC0 sin qt+ LC0 q cos qt,

where ε = tan−1 (Lq/R), the so-called lag* of the current behind the electromotive force,

the expression√

R2 + L2q2 is the so-called impedance.(iii.) When E is a function of the time, say f (t),

C = Be−Rt/L =1

Le−Rt/L

∫

e−Rt/Lf (t) · dt,

The evanescent term e−Rt/L may be omitted when the current has settled down intothe steady state. (Why?)

(v.) When E is zero,C = Be−Rt/L.

Evaluate the integration constant B by putting C = C0, when t = 0.(2) The relation between the charge (q) and the electromotive force (E) of two plates

of a condensor of capacity C connected by a wire of resistance R, is

E = R · dq/dt+ q/C,

provided the self-induction is zero. Solve for q. Show that when Theseequa-tionsfollowfrom (2)on page311 orfrom thealter-nativemethodsdetailedin § 130.Reviewthatsection ifyou havedifficultyarrivingat thesesolu-tions.KH

E = f (t) , q =1

Re−t/RC

∫

et/RCf (t) · dt+Be−t/RC ;

E = 0, q = Q0 e−t/RC ; (Q0 is the charge when t = 0).

E = constant, q = CE +Be−t/RC ;

E = E0 sinωt, q = Be−t/RC + CE0 (sinωt+RCω cosωt) /(

1 +R2C2ω2)

.

(3) The equation of motion of a pendulum subject to a resistance which varies withthe velocity and which is acted upon by a force which is a simple harmonic function ofthe time, is

d2x

dt2+ 2f

dx

dt+ q2x = cos (qt+ ε) .

Show that the complementary function is

x = A cos (qt+ ε) +B sin (qt+ ε) .

* An alternating (periodic) current is not always in phase (or, “in step”) with theimpressed (electromotive) force driving the current along the circuit. If there is self-induction in the circuit, the current lags behind the electromotive force; if there is acondensor in the circuit, the current in the condenser is greatest when the electromotiveforce is changing most rapidly from a positive to a negative value, that is to say, themaximum current is in advance of the electromotive force, there is then said to be alead in the phase of the current.

329

Thispara-graphdemon-stratesthemethodof unde-terminedcoeffi-cients.Furtherexamplesof thismethodcan befound inthe sup-plementpagesS-24throughS-29.

To solve the equation, assume that

x = A cos (qt+ ε) +B sin (qt+ ε) ,

is a solution. Substitute in the given equation,

∴ −Aq2 + 2fBq + n2A = 1; ∴ −Aq2 + 2fBq + n2B = 0.

A =n2 − q2

(n2 − q2) + 4f2q2; B =

2fq

(n2 − q2) + 4f2q2.

Put A = R cos ε. B = R sin ε.

The solution of the given equation is then

x = R cos (qt+ ε− ε1) , (32)

where R = 1/√

(n2 − q2) + 4f2q2; tan ε = 2fq/(

n2 − q2)

.The forced oscillations due to the impressed periodic force, are thus determined by

(32). The complementary function gives the natural vibrations superposed upon these.(4) If the friction in the preceding example, is zero,

d2x

dt2+ n2x = cos (qt+ ε) . (33)

A particular integral is x = {f · cos (qt+ ε)} /(

n2 − q2)

. This fails when n = q. In thiscase, assume that x = Ct sin (nt+ ε) is a particular integral. (33) is satisfied. providedC = f/2n. The physical meaning of this is that when the pendulum is acted on bya periodic force “in step” with the oscillations of the pendulum, the amplitude of theforced oscillations will increase proportionally with the time, until, when the amplitudeexceeds a certain limit, equation (33) no longer represents the motion of the pendulum.

(5) When an electric current, passing through an electrolytic cell, has assumedthe steady state, show that the ionic velocity is proportional to the impressed force(electromotive force). By Newtons law, for a moving body,

(Impressed force) = (Mass)× (Acceleration) .

Friction is to be regarded as a retarding force acting in an opposite direction to theimpressed force; this frictional force is proportional to the velocity of the body.

∴ (Impressed force less friction) = (Mass)× (Acceleration) .

Express these facts in symbolic language. See (1) above. Integrate the result andevaluate the constant for v = 0, when t = 0.

∴ µv = F(

1− e−µt/m)

. (34)

For ionic motion, m is very small, µ, is very great. When t is great, show that theexponential term vanishes, and

F ∝ v.

§ 135. The Velocity of Simultaneous and Dependent ChemicalReactions. While investigating the rate of decomposition of phosphine(§ 88), we had occasion to point out that the action really takes place intwo stages:—

STAGE I. PH3 = P + 3H.

STAGE II. 4P = P4; 2H = H2.

330

The former change alone determines the velocity of the whole reaction.The physical meaning of this is that the speed of the reaction which occursduring stage II., is immeasurably faster than the speed of the first. Exper-iment quite fails to reveal the complex nature of the complete reaction.*

Suppose, for example, a substance A forms an intermediate compoundB, and this, in turn, forms a final product C. If the speed of the reaction

A = B, is one gram per1

100000second,

when the speed of the reaction

B = C, is one gram per hour,

the observed “order” of the complete reaction

A = C,

will be fixed by that of the slower reaction, B = C, because the meth-ods used for measuring the rates of chemical reactions are not sensitiveto changes so rapid as the assumed rate of transformation of A into B.Whenever the “order” of this latter reaction, B = C is alone accessible tomeasurement. If, therefore, A = C is of the first, second, or nth order, wemust understand. that one of the subsidiary reactions (A = B, or B = C)is

(1) an immeasurably fast reaction, accompanied by(2) a slower measurable change of the first, second or nth order, accord-

ing to the particular system under investigation.If, however, the velocities of the two reactions are of the same order

of magnitude, the “order” of the complete reaction will not fall under anysimple type (§§ 88, 89), and, therefore, some changes will have to be madein the differential equations representing the course of the reaction. Let usstudy some of the simpler cases.

Case i. In a given system, a substance A forms an intermediate sub-stance B, which finally forms a third substance C.

Let one gram molecule of the substance A be taken. At the end of a certain time t,the system contains x of A, y of B, z of C. The rate of diminution of x is evidently

−dxdt

= k1x, (1)

* Professor Walker illustrates this by the following analogy (“Velocity of GradedReactions,” Proc. Royal Soc. Edin., Dec., 1897): “The time occupied in the transmissionof a telegraphic message depends both on the rate of transmission along the conductingwire and on the rate of the messenger who delivers the telegram; but it is obviously thislast, slower rate that is of really practical importance in determining the total time oftransmission”. . . .

331

where k1 denotes the velocity constant of the transformation of A to B. The rate offormation of C is

dz

dt= k2y, (2)

where k2 is the velocity constant of the transformation of B to C. Again, the rate atwhich B accumulates in the system is evidently the difference in the rate of diminutionof x and the rate of increase of z, or

dy

dt= k1x− k2y. (3)

The speed of the chemical reactions,

A = B = C,

is fully determined by this set of differential equations. When the relations between aset of variables involves a set of equations of this nature, the result is said to be a systemof simultaneous differential equations.

In a great number of physical problems, the interrelations of the variables are rep-resented in the form of a system of such equations. The simplest class occurs when eachof the dependent variables is a function of the independent variable.

The simultaneous equations are said to be solved when each variable is expressedin terms of the independent variable, or else when a number of equations between thedifferent variables can be obtained free from differential coefficients.

To solve the present set of differential equations, first differentiate (2),

d2z

dt2− k2

dy

dt= 0;

Add and subtract k1k2y, substitute for dy/dt from (3) and for k2y from (2), we thusobtain

d2z

dt2+ (k1 + k2)

dz

dt− k1k2 (x+ y) = 0.

But from the conditions of the experiment,

x+ y + z = 1, ∴ z − 1 = − (x+ y) .

Hence, the last equation may be written,

d2 (z − 1)

dt2+ (k1 + k2)

d (z − 1)

dt+ k1k2 (z − 1) = 0. (4)

This linear equation of the second order with constant coefficients, is to be solved forz − 1 in the usual manner (§ 130). At sight, therefore,

z − 1 = C1e−k1t + C2e

−k2t. (5)

But z = 0, when t = 0,∴ C1 + C2 = 1. (6)

Differentiate (5). From (2), dz/dt = 0, when t = 0. Therefore, making the necessarysubstitutions,

−C1k1 − C2k2 = 0. (7)

From (6) and (7),C1 = k1/ (k2 − k1) ; C2 = k2/ (k2 − k1) .

The final result may therefore be written,

z − 1 =k2

k1 − k2e−k2t +

k1k2 − k1

e−k1t. (8)

332

Harcourt and Esson have studied the rate of reduction of potassium permanganateby oxalic acid.

2KMnO4 + 3MnSO4 + 2H2O = K2SO4 + 2H2SO4 + 5MnO2;

MnO2 +H2SO4 +H2C2O4 =MnSO4 + 2H2O + 2CO2.

By a suitable arrangement of the experimental conditions this reaction may be usedto test equations (5) or (8).

Let x, y, z, respectively denote the amounts of Mn2O7, MnO2, and MnO (incombination) in the system. The above workers found that C1 = 28.5; C2 = 2.7;e−k1 = 0.82; e−k2 = 0.98. The following table places the above suppositions beyonddoubt.

t min-utes.

z − 1. t min-utes.

z − 1.

Found. Calculated. Found. Calculated.

0.5 25.85 25.9 3.0 10.45 10.41.0 21.55 21.4 8.5 8.95 9.01.5 17.9 17.8 4.0 7.7 7.82.0 14.9 14.9 4.5 6.65 6.62.5 12.55 12.5 5.0 5.7 5.8

Case ii. A solution contains a gram molecules of each of A and C, the substance Agradually changes to B, which, in turn, reacts with C to form another compound D.

Let x denote the amount of A which remains untransformed after the elapse of aninterval of time t, y the amount of B, and z the amount of C present in the system afterthe elapse of the same interval of time t. Hence show that

−dxdt

= k1x; − dz

dt= k2yz. (9)

The rate of diminution of B is proportional to the product of its active mass y into theamount of C present in the solution at the time t, but the velocity of increase of y isequal to the velocity of diminution of x,

∴dy

dt= k1x− k2yz. (10)

If x, y, z, could be measured independently, it would be sufficient to solve these equationsas in case i., but if x and y are determined together, we must proceed a little differently.Note z = x+y. From the first of equations (9), and (10) by addition and the substitutionof dt = −dx/k1x from (9), and of z − x = y, we get

1

z2· dzdt

+K

z− K

x= 0. (11)

where K has been written in place of k2/k1. The solution of this equation has beenpreviously determined (page 298) in the form

Ke−Kx

{

C1 − log x+Kx− 1

1.22(Kx)

2+ . . .

}

z = 1. (12)

333

In some of Harcourt and Esson’s experiments, C1 = 4.68; k1 = 0.69; k2 = 0.006864.From the first of equations (9), it is easy to show that x = ae−k1t. Where does a comefrom? What does it mean? Hence verify the third column in the following table:

t min-utes.

z.

Found. Calculated.

2 51.9 51.68 42.4 42.94 85.4 85.45 29.8 29.7

After the lapse of six minutes, the value of x was found to be negligibly very small.The terms succeeding log x in (12) may, therefore, be omitted without committing anysensible error. Substitute x = ae−k1t in the remainder,

k2k1

(C1 log a+ k1t) t; or (C′1 + t) z =

1

k′2

where C′1 = C1/k1 − (log a) /k1. Harcourt and Esson found that C′

1 = 0.1, and1/k2 = 157. Hence, in continuation of the preceding table, these investigators ob-tained the results shown in the following table. The agreement between the theoreticaland experimental numbers is remarkable.

t min-utes.

z − 1. t min-utes.

z − 1.

Found. Calculated. Found. Calculated.

6 25.7 25.7 10 15.5 15.57 22.2 22.1 15 10.4 10.48 19.4 19.4 20 7.8 7.89 17.8 17.8 30 5.5 5.2

The theoretical numbers are based on the assumption that the chemical changeconsists in the gradual formation of a substance which at the same time slowly disappearsby reason of its reaction with a proportional quantity of another substance.

This really means that the so-called “initial disturbances” in chemical reactions, aredue to the fact that the speed during one stage of the reaction, is faster than during theother. The magnitude of the initial disturbances depends on the relative magnitudes ofk1 and k2. The observed velocity in the steady state depends on the difference betweenthe steady diminution −dx/dt and the steady rise dz/dt. If k2 is infinitely great incomparison with k1, (8) reduces to

z = a(

1− e−k1t)

.

334

which will be immediately recognised as another way of writing the familiar equation

k1 =1

tlog

a

a− z .

So far as practical work is concerned, it is necessary that the solutions of the differ-ential equations shall not be so complex as to preclude the possibility of experimentalverification.

Case iii. In a given system A combines with R to form B, B combines with R toform C, and C combines with R to form D.

In the hydrolysis of triaeetin,

C3H5 ·A3 +H ·OH = 3A ·H + C3H5 (OH) 3,

(Triacetin) (Glycerol)

where A has been written for CH3 · COO·, there is every reason to believe that thereaction takes place in three stages:

C3H5 · A3 +H ·OH = A ·H + C3H5 · A2 ·OH (diacetin);

C3H5 · A2 ·OH +H · OH = A ·H + C3H5 ·A · (OH)2 (monacetin);

C3H5 · A · (OH)2 +H ·OH = A ·H + C3H5 · (OH)3 (glycerol).

These reactions are interdependent. The rate of formation of glycerol is conditioned bythe rate of formation of monacetin; the rate of monacetin depends, in turn, upon therate of formation of diacetin. There are, therefore, three simultaneous reactions of thesecond order taking place in the system.

Let a denote the initial concentration (gram molecules per unit volume) of triacetin,b the concentration of the water; let x, y, z, denote the number of molecules of mono-,di- and triacetin hydrolysed at the end of t minutes. The system then contains a − zmolecules of triaoetin, z−y, of diacetin, y−x, of monacetin, and b−(x+ y + z) moleculesof water. The rate of hydrolysis is therefore completely determined by the equations:

dx/dt = k1 (y − x) (b− x− y − z) ; (13)

dy/dt = k2 (z − y) (b− x− y − z) ; (14)

dz/dt = k3 (a− z) (b− x− y − z) ; (15)

where k1, k2, k3, represent the velocity coefficients (§ 88) of the respective reactions.Geitel tested the assumption: k1 = k2 = k3. Hence dividing (15) by (13) and by

(14), he obtained Seeworkedsolutionof thisequationon pageS-42. KH

dz/dy = (a− z) / (z − y) ; dz/dx = (a− z) / (y − x) . (16)

From the first of these equations,

dy + ydz

a− z =z · dza− z .

which can be integrated as a linear equation of the first degree. The constant is equatedby noting that if a = 1, z = 0, y = 0. The reader might do this as an exercise on § 123.The answer is

y = z + (a− z) log (a− z) . (17)

Now substitute (17) in the second of equations (16), rearrange terms and integrate as afurther exercise on linear equations of the first order. The final result is,

x = z + (a− z) log (a− z)− a− z2{log (a− z)}2 . (18)

335

Geitel then assigned arbitrary numerical values to z (say from 0.1 to1.0), calculated the corresponding amounts of z and y from (17) and (18)and compared the results with his experimental numbers. For experimentaland other details the original memoir must be consulted (vide infra).

EXAMPLE. –Calculate equations analogous to (17) and (18) on the supposition thatk1 6= k2 6= k3.

A study of the differential equations representing the mutual conversionof red into yellow, and yellow into red phosphorus, will be found in a paperby Lemoine in the Annales de Chimie et de Physique [4], 27, 289, 1872.

There is also a series of interesting papers by Rud. Wegscheider bearingon this subject in Zeit. f. phys. Chem., 30, 593, 1899; ib., 34, 290, 1900;ib., 35, 513, 1900; Monatshefte fur Chemie, 22, 749, 1901.

The preceding discussion is based upon papers by Harcourt and Esson,Phil. Trans., 156, 198, 1866; Geitel, Journ. fur prakt. Chem. [2], 55,429, 1897; J. Walker, Proc. Roy. Soc. Adman., 22, 1897. It is somewhatsurprising that Harcourt and Esson’s investigation has not received moreattention from the point of view of simultaneous and dependent reactions.The indispensable differential equations, simple as they are, might perhapsaccount for this. But chemists, in reality, have more to do with this typeof reaction than any other. The day is surely past when the study of aparticular reaction is abandoned simply because it “won’t go” according tothe stereotyped velocity equations of § 88.

§ 136. Simultaneous Differential Equations. By way of practice itwill be convenient to study a few more examples of simultaneous equations.

For a complete determination of each variable there must be the samenumber of equations as there are independent variables. Quite an analogousthing occurs with the simultaneous equations of ordinary algebra.

I. Simultaneous equations with constant coefficients. The methods usedfor the solution of these equations are analogous to those employed forsimilar equations in algebra. The operations here involved are chiefly pro-cesses of elimination and substitution, supplemented by differentiation orintegration at various stages of the computation. The use of the symbol Doften shortens the work. Most of the following examples are from resultsproved in the regular textbooks on physics.

EXAMPLES.–(1) Solve dx/dt + ay = 0, dy/dt + bx = 0. Differentiate the first,multiply the second by a. Subtract and y disappears. Hence writing

x = C1emt + C2e

−mt; or, y = C2

√

b/a · e−mt − C1

√

b/a · emt.

336

We might have obtained an equation in y, and substituted it in the second. Thus fourconstants appear in the result. But one pair of these constants can be expressed in termsof the other two. Two of the constants, therefore, are not arbitrary and independent,while the integration constant is arbitrary and independent. It is always best to avoidan unnecessary multiplication of constants by deducing the other variables from the firstwithout integration. The number of arbitrary constants is always equal to the sum ofthe highest orders of the set of differential equations under consideration.

(2) Solve dx/dt + y = 3x; dy/dt − y = x. Differentiate the first. Subtract eachof the given equations from the result.

(

D2 − 4D + 4)

x remains. Solve as usual.x = (C1 + C2x) e

2t. Substitute this value of x in the second of the given equationsand y = (C1 − C2 + C3t) e

2t.(3) The equations of rotation of a particle in a rigid plane, are

dx/dt = µy; dy/dt = −µx.

To solve these, differentiate the first, multiply the second by µ, etc. Finallyx = C1 cosµt + C2 sinµt; y = C′

1 cosµt + C′2 sinµt. To find the relation between these

constants, substitute these values in the first equation and

= −µC1 sinµt+ µC2 cosµt = µC′1 cosµt+ µC′

2 sinµt,

or C′1 = C2 and C′

2 = −C1.(4) Solve d2x/dt2 = −n2x; d2y/dt2 = −n2y.

x = C1 cosnt+ C2 sinnt; y = C′1 cosnt+ C′

2 sinnt.

Eliminate t so that

(C′1x− C1y)

2+ (C′

2x− C2y)2= (C1C

′2 − C2C

′1)

2, etc.

The result represents the motion of a particle in an elliptic path, subject to a centralgravitational force.

(5) Solve dx/dt + by + cz = 0; dy/dt + a1x + c1z = 0; dz/dt + a2x + b2z = 0.Operate on the first with D2 − b2c1 on the second with b2c − bD, on the third withbc1 − cD. Add. The terms in y and z disappear. The remaining equation has theintegral,

x = C1eαt + C2e

βt + C3eγt,

where α, β, γ, are the roots of

←−z3 + (a1b + a2c+ b2c1) z + a1b2c+ a2bc1 = 0.

The values of y and z are easily obtained from that of x by proper substitutions in theother equations.

(6) If two adjacent circuits have currents i1 and i2 then, according to the theory ofelectromagnetic induction,

Mdi1dt

+ L2di2dt

+R2i2 = E2; Mdi2dt

+ L1di1dt

+R1i1 = E1,

(see J. J. Thomsons Elements of Electricity and Magnetism, p. 882), where R1, R2

denote the resistances of the two circuits, L1, L2 the coefficients of self-induction, E1,E2 the electromotive forces of the respective circuits and M the coefficient of mutualinduction. All the coefficients are supposed constant.

First, solve these equations on the assumption that E1 = E2 = 0. Assume that

i1 = aemt and i2 = bemt,

337

satisfy the given equations. Differentiate each of these variables with respect to t andsubstitute in the original equation

aMm+ b (L2m+R2) = 0; bMm+ a (L1m+R1) = 0.

multiply these equations so that

(L1L2 −M)m2 + (L1R2 +R1L2)m+R1R2 = 0.

For physical reasons, the induction L1L2 must always be greater than M ; The roots ofthis quadratic must, therefore, be negative and real (page 388), and

i1 = a1e−m1t, or a2e

−m2t; i2 = b1e−m1t, or b2e

−m2t.

Hence, from the preceding equation,

a1Mm1 + b1L2m1 +B1R2 = 0; or a1/b1 = (L2m1 +R2) /Mm1;

similarly a2/b2 =Mm2/ (L1M2 +R1).Combining the particular solutions for i1 and i2, we get

i1 = a1e−m1t + a2e

−m2t; i2 = b1e−m1t + b2e

−m2t,

the required solutions.Second, if E1 and E2, have some constant value,

i1 = E1/R1 + a1e−m1t + a2e

−m2t; i2 = E2/R2 + b1e−m1t + b2e

−m2t,

are the required solutions.

II. Simultaneous equations with variable coefficients. The general typeof simultaneous equations of the first order, is

P1dx+Q1dy +R1dz;P2dx+Q2dy +R2dz, . . .

}

(1)

where the coefficients are functions of x, y, z. These equations can oftenbe expressed in the form dx

P=

dy

Q=

dz

R, (2)

which is to be looked upon as a typical set of simultaneous equations ofthe first order. If one of these equations involves only two differentials, theequation is to be solved in the usual way, and the result used to deducevalues for the other variables, as in the first of the subjoined examples.

When the members of a set of equations are symmetrical, the solutioncan often be simplified by taking advantage of a well-known theorem* inalgebra (ratio). According to this,

* Perhaps it is best to state the proof. Let

dx/P = dy/Q = dz/R = k, say; then,

dx = Pk; dy = Qk; dz = Rk;

or, l dx− lPk; mdy −mQk; n dz = nRk.

Add these results,

l dx+mdy + n dz = k (lP +mQ+ nR) .

∴l dx+mdy + n dz

lP +mQ+ nR= k =

dx

P=dy

Q=dz

R.

338

dx

P=

dy

Q=

dz

R=

l dx+mdy + n dz

lP +mQ + nR=

l′ dx+m′ dy + n′ dz

l′P +m′Q+ n′R= (3)

where l, m, n, l′, m′, n′,. . . are sets of multipliers such that

lP +mQ + nR = 0; l′P +m′Q+ n′R = 0, etc. (4)

hence, l dx+mdy + n dz = 0, etc. (5)

The same relations between x, y, z, that satisfy (5), satisfy (2).If (4) be an exact differential equation, equal to say du, direct integra-

tion gives the integral of the given system, viz.,

u = a, (6)

where a, denotes the constant of integration.In the same way, if

l dx+mdy + n dz = 0,

is an exact differential equation, equal to say dv, then, since dv is also equalto zero,

v = b, (7)

is a second solution. These two solutions must be independent.

EXAMPLES.–(1) Solve dx/y = dy/x = dz/z. The relation between dx and dycontains x and y only, the integral, y2 − x2 = C1, follows at once. Use this result toeliminate x from the relation between dy and dz. The result is

dz/z = dy/√

y2 − C1; or, y +√

y2 + C1 = C2z.

These two equations, involving two constants of integration, constitute a complete so-lution.

(2) Solve dx/ (mz − ny) = dy/ (nz − lz) = dz/ (ly −mx). l, m, n and x, y, z forma set of multipliers satisfying the above condition. Hence,

l dx+mdy + n dz = 0; x dx+ y dy + z dz = 0.

The integrals of these equations are

u = lx+my + nz = C1; v = x2 + y2 + z2 = C2,

which constitute a complete solution.(3) Solve dx/

(

x2 − y2 − z2)

= dy/2xy = dz/2xz. From the two last equationsy = C1z. Substituting x, y, z for l, m, n, each of the given ratios is equal to

(x dx + y dy + z dz) /(

x2 + y2 + z2)

. ∴ x2 + y2 + z2 = c2z.

is another solution.

§ 137. Partial Differential Equations. Equations obtained by thedifferentiation of functions of three or more variables are of two kinds:

1. Those in which there is only one independent variable, such as

P dx+Qdy +Rdz = S dt

which involves four variables – three dependent and one independent. Theseare called total differential equations.

339

2. Those in which there is only one dependent and two or more inde-pendent variables, such as,

P∂z

∂x+Q

∂z

∂y+R

∂z

∂t= 0,

where z is the dependent variable, x, y, t the independent variables. Theseequations are classed under the name partial differential equations.The former class of equations are rare, the latter very common. We shallconfine our attention to partial differential equations.

In the study of ordinary differential equations, we have always assumedthat the given equation has been obtained by the elimination of constantsfrom the original equation. In solving, we have sought to find this primitiveequation.* Partial differential equations, however, may be obtained by theelimination of arbitrary functions of the variables as well as of constants. page

56 isincludedin thisdocu-mentfollowingthe sup-plement.

It can be shown from Euler’s theorem (page 56) that if

u = xnf(y

x,z

x, . . .

)

,

be a homogeneous function,

x∂u

∂x+ y

∂u

∂y+ z

∂u

∂z+ · · · = nxmf

(y

x,z

x, . . .

)

= nu,

where the arbitrary function has disappeared.† Again, if

u = f(

ax3 + by3)

,

is an arbitrary function of x and y.

←−∂u

∂x= af ′

(

ax3 + by3)

;∂u

∂y= bf ′

(

ax3 + by3)

; ∴ b∂u

∂x− a

∂u

∂y= 0.

* Physically, the differential equation represents the relation between the dependentand the independent variables corresponding to an infinitely small change in each of theindependent variables.

The reader will, perhaps, have noticed that the term “independent variable” isan equivocal phrase. (1) If u = f (z), u is a quantity whose magnitude changes whenthe value of a changes. The two magnitudes u and z are mutually dependent. Forconvenience, we fix our attention on the effect which a variation in the value of z hasupon the magnitude of u. If need be we can reverse this and write z = f (u), so that unow becomes the “independent variable”. (2) If v = f (x, y), x and y are “independentvariables” in that x and y are mutually independent of each other. Any variation inthe magnitude of the one has no effect on the magnitude of the other. x and y arealso “independent variables” with respect to v in the same sense that a has just beensupposed the “independent variable” with respect to u.† This is usually proved in the textbooks in the following manner:

Let u = xnf (y/x, z/x, . . . ). Put y/x = Y , z/x = Z,. . .

∴ ∂Y/∂x = −y/x2, ∂Z/∂x = −z/x2 . . . ; ∂Y/∂y = 1/x, ∂Z/∂y = 0, . . .

Let v = f (Y, Z, . . . ), for the sake of brevity, therefore, since u = xnv,

340

EXAMPLES.–(1) If y − bu = f (bx− ay), b ∂u∂x + b ∂u

∂y = 1.

(2) If 1/z − 1/x = f (1/y − 1/x), x2∂z/∂x+ y2∂z/∂y = z2.(3) If z = a (x+ y), ∂z/∂x− ∂z/∂y = 0.

For this reason an arbitrary function of the variables is added to theresult of the integration of a partial differential equation instead of theconstant hitherto employed for ordinary differential equations.

If the number of arbitrary constants to be eliminated is equal to thenumber of independent variables, the resulting differential equation is ofthe first order. The higher orders occur when the number of constants tobe eliminated, exceeds that of the independent variables.

If u = f (x, y), there will be two differential coefficients of the first order,namely, ∂u/∂x and ∂u/∂y; three of the second order, namely, ∂2u/∂x2,∂2u/∂x∂y, ∂2u/∂y2. . .

§ 138. What is the Solution of a Partial Differential Equation?Ordinary differential equations have two classes of solutions – the completeintegral and the singular solution. Particular solutions are only varietiesof complete integral. Three classes of solutions can be obtained from somepartial differential equations, still regarding the particular solution as aspecial case of the complete integral. These are indicated in the followingexample.

The equation of a sphere in three dimensions is,

x2 + y2 + z2 = r2, (1)

when the centre of the sphere coincides with the origin of the coordinateplanes and r denotes the radius of the sphere. If the centre of the sphere liessomewhere on the xy-plane at a point (a, b), the above equation becomes

∂u

∂x= nxn−1f (Y, Z, . . . ) + xn

(

∂v

∂Y· ∂Y∂x

+∂v

∂Z· ∂Z∂x

+ . . .

)

;

by the method for the differentiation of a function of a function, 6 and 9, § 12. Therefore,

= nxn−1f (Y, Z, . . . )− xn−2

(

y∂v

∂Y+ z

∂v

∂Z+ . . .

)

.

∂u

∂y= xn

∂v

∂dY· ∂Y∂y

= xn−1 ∂v

∂Y;

∂u

∂z= xn

∂v

∂Z· ∂Z∂z

= xn−1 ∂v

∂Z; . . .

Now multiply by x, y, z,. . . respectively, and add,

x∂u

∂x+ y

∂u

∂y+ z

∂u

∂z+ · · · = nxnf (Y, Z, . . . ) = nu.

341

(x− a)2 + (y − b)2 + z2 = r2. (2)

When a and b are arbitrary constants, each or both of which may haveany assigned magnitude, equation (2) may represent two infinite systemsof spheres of radius r. The centre of any member of either of these twoinfinite systems (called a double infinite system) must lie somewhere on thexy-plane.

Differentiate (2) with respect to x and y.

x− a+ z∂z

∂x= 0; y − b+ z

∂z

∂y= 0. (3)

Substitute for x− a and y − b in (2). We obtain

z2

{

(

∂z

∂x

)2

+

(

∂z

∂y

)2

+ 1

}

= r2. (4)

Equation (2), therefore, is the complete integral of (4). By assigning anyparticular numerical value to a or b, a particular solution of (4) will beobtained, such is

(x− 1)2 + (y − 79)2 + z2 = r2. (5)

If (2) be differentiated with respect to a and b,

∂

∂a

{

(x− a)2 + (y − b)2 + z2 = r2}

;∂

∂b

{

(x− a)2 + (y − b)2 + z2 = r2}

.

or, x− a = 0, and y − b = 0.Eliminate a and b from (2),

z = ±r. (6)

This result satisfies equation (4), but, unlike the particular solution, is notincluded in the complete internal (2). Such a solution of the differentialequation is said to be a singular solution.

Geometrically, the singular solution represents two plane surfaces touchedby all the spheres represented by equation (2). The singular solution is thusthe envelope of all the spheres represented by the complete integral. If AB(Fig. 79) represents a cross section of the xy-plane containing spheres ofradius a, CD and EF are cross sections of the plane surfaces representedby the singular solution.

If the one constant is some function of the other, say,

a = b,

(2) may be written

(x− a)2 + (y − a)2 + z2 = r2. (7)

342

Differentiate with respect to a. We find

a =1

2(x+ y) .

Eliminate a from (7). The resulting equation

x2 + y2 + 2z2 − 2xy = r2.

is called a general integral of the equation.Geometrically, the general integral is the equation to the tubular en-

velope of a family of spheres of radius r and whose centres are along theline x = y. This line corresponds with the axis of the tube envelope. Thegeneral integral satisfies (4) and is also contained in the complete integral.

Instead of taking a = b as the particular form of the function connectinga and b, we could have taken any other relation, say a = 1

2. The enve-

lope of the general integral would then be like a tube surrounding all thespheres of radius r whose centres were along the line x = 1

2y. Had we put

a2 − b2 = 1, the envelope would have been a tube whose axis was an hy-perbola x2 − y2 = 1.

A particular solution is one particular surface selected from the doubleinfinite series represented by the complete solution. A general integral is theenvelope of one particular family of surfaces selected from those comprisedin the complete integral A singular solution is the complete envelope ofevery surface included in the complete integral.*

Theoretically an equation is not supposed to be solved completely untilthe complete integral, the general integral and the singular solution havebeen indicated. In the ideal case, the complete integral is first determined;the singular solution obtained by the elimination of arbitrary constants asindicated above; the general integral then determined by eliminating a andf (a).

Practically, the complete integral is not always the direct object ofattack. It is usually sufficient to deduce a number of particular solutionsto satisfy the conditions of the problem and afterwards to so combine thesesolutions that the result will not only satisfy the given conditions but alsothe differential equation.

* The study of Gibbs’ “Surfaces of Dissipated Energy,” “Surfaces of Dissociation,”“Surfaces of Chemical Equilibrium,” as well as “van der Waals’ Surfaces,” is the naturalsequence of § 68, 126 and the present section. But to enlarge upon this subject wouldnow cause a greater digression than is here convenient. Airy’s little book, in ElementaryTreatise on Partial Differential Equations, will repay careful study in connection withthe geometrical interpretation of the solutions of partial differential equations.

343

Of course, the complete integral of a differential equation applies to anyphysical process represented by the differential equation. This solution, Most

moderntextsuse thephrase,“bound-arycondi-tions”ratherthan“limitingcondi-tions”.KH

however, may be so general as to be of little practical use. To representany particular process, certain limitations called limiting conditions haveto be introduced. These exclude certain forms of the general solution asimpossible. See examples at the end of Chapter VIII.; also example (1)last set § 138, and elsewhere.

The more important varieties of partial differential equations from thepoint of view of this work are the linear equations of the second and higherorders.

§ 139. The Solution of Partial Differential Equations of the FirstOrder. For the ingenious general methods of Lagrange, Charpit, etc., thereader will have to consult the special textbooks, say, Forsyth’s A Treatiseon Differential Equations (Macmillan & Co., 1888).

Type I. The variables do not appear directly. The general form is,

f (∂z/∂x, ∂z/∂y) = 0. (I.)

The solution isz = ax+ by + C,

provided a and b satisfy the relation

f (a, b) = 0, or b = f (a) .

The complete integral is, therefore,

z = ax+ yf (a) + C. (1)

EXAMPLES.–(1) Solve (∂z/∂x)2 + (∂z/∂y)2 = m2. The solution is

z = ax+ by + C′,

provided a2+ b2 = m2. The solution is, therefore, ax+ y√m2 − a2+C. For the general

integral, put C = f (a). Eliminate a between the two equations,

z = ax+ y√

m2 − a2 + f (a) ; and x− ay√m2 − a2

+ f ′ (a) = 0.

in the usual way.(2) Solve pq = 1. Ansr. z = ax+ y/a+ f (a).NOTE.–We shall sometimes write, for the sake of brevity,

∂z/∂x = p; ∂z/∂y = q.

(3) Solve a (p+ q) = z. Sometimes, as here, when the variables do appear in theequation, the function of x, which occurs in the equation, may be associated with ∂z/∂x,or a function of y with ∂z/∂y, by a change in the variables. We may write the givenequation ap/z + aq/z = 1. Put dz/z = dZ; dy/a = dY , dx/a = dX ; hence, ∂Z/∂Y +∂Z/∂X = 1, the form required.

344

(4) Solve x2p2+ y2q2 = z2. Put X = log x, Y = log y, Z = log z. Proceed as before.

Ansr. z = Cxay√1−a2

.

If it is not possib1e to remove the dependent variable z in this way, theequation will possibly belong to the following class:

Type II. The independent variables x and y are absent. The generalform is,

f (z, ∂z/∂x, ∂z/∂y) = 0. (II.)Assume as a trial solution, that

∂z/∂y = a · ∂z/∂x.Let ∂z/∂x be some function of z obtained from II., say y = φ (z). Substitutethese values in

dz = p · dx+ q · dy.We thus get an ordinary differential equation which can be readily inte-grated.

dz = φ (z) · dz + aφ (z) · dy.

∴ x+ ay =

∫

dz/φ (z) + C. (2)

EXAMPLES.–(1) Solve p2z + q2 = 4. Here,

←−

(

a2 + z)

(dz/dx)2= 4.

√

a2 + z · dz/dx = 2,

∴ x+ C =

∫

√

a2 + z · dz = 12

(

a2 + z)3/2

. Ansr. 2(

a2 + z)3

= 3 (x+ ay + C)2.

(2) Solve p (1 + q)2= q (z − a). Ansr. 4 (bz − ab− 1) = (x+ by + C)

2.

If z does not appear directly in the equation, we may be able to referthe equation to the next type.

Type III. z does not appear directly in the equation, but x and ∂z/∂xcan be separated from y and ∂z/∂y. The leading type is

f1 (x, ∂z/∂x) = f2 (y, ∂z/∂y) . (III.)

Assume as a trial solution, that each member is equal to an arbitraryconstant a, so that ∂z/∂x and ∂z/∂y can be obtained in the form,

∂z/∂x = φ1 (x, a) ; ∂z/∂y = φ2 (y, a) .

dz = p · dx+ q · dy.

then assumes the form

dz = φ1 (x, a) dx+ φ2 (y, a) dy. (3)

EXAMPLES. –Solve the following equations:(1) q − p = x− y. Put ∂z/∂x+ x = ∂z/∂y + y = a. Write

∂z/∂x = a− x, etc.; ∂z/∂y = a− y.Hence, z = − 1

2 (a− x)2 − 1

2 (a− y)2 + C.

(2) q2 + p2 = x+ y. Ansr. z = 23 (x+ a)

3/2+ 2

3 (y − a)3/2

+ C.

(3) q = 2yp2. Ansr. z = ax+ a2y2 + C.

345

Type IV. Analogous to Clairault’s equation. The general type is

z = p · x+ q · y + f (p, q) . (IV.)

The complete integral is See sup-plementpageS-49for themethodof findingsingularsolu-tions toClairaultanalogequa-tions.KH

z = ax+ by + f (a, b) (5)

EXAMPLES.–Solve the following equations:(1) z = px+ qy + pq. Ansr. z = ax+ by + ab. Singular solution z = −xy.(2) z = px+ qy + r

√

1 + p2 + q2. Ansr. z = ax+ by + r√1 + a2 + b2.

Singular solution, x2 + y2 + z2 = r2. The singular solution is, therefore, a sphere; r, ofcourse, is a constant.

(3) z = px+ qy − n n√pq. Ansr. z = ax+ by − n n

√ab. Singular solution,

z = (2− n) (xy)1/(2−n).

§ 140. Partial Differential Equations of the nth Order. These arethe most important equations that occur in physical mathematics. Thereare no general methods for their solution, and it is only possible to performthe integration in special cases. The greatest advances in this direction havebeen made with the linear equation. Before proceeding to this importantequation, it appears convenient to solve some simpler types.

EXAMPLES.–Integrate the following equations.

(1)∂2z

∂x∂y= a. If

∂z

∂x= p and

∂z

∂y= q. Integrate with regard to y and we get

p = ay + f ′ (x). It is very possible that f ′ (x) is a function of y. Integrate with respectto x and z =

∫

{ay + f ′ (x)} dx = axy + f1 (x) + f2 (y).

(2)∂2z

∂x∂y− x

y= a. Ansr. z = 1

2x2 log y + f1 (x) + f2 (y).

(3)∂2z

∂x∂y+∂z

∂xf (x) = ψ (y). Ansr. z=

∫[

e−yf(x){∫

eyf(x)ψ (y)dy + f1 (x)}]

dx+f2 (y).

There are many points of analogy between the partial and the ordinarylinear differential equations. Indeed, it may almost be said that every or-dinary differential equation between two variables is analogous to a partialdifferential in the same class. The solution is in each case similar, but withthese differences:

First, the arbitrary constant of integration in the solution of an ordinarydifferential equation is replaced by a function of a variable or variables.

Second, the exponential form, Cemx, of the solution of the ordinary

linear differential equation assumes the form emx ∂∂y

φ(y).

346

The expression, emx ∂∂y

φ(y), is known as the symbolic form of Taylor’s theorem.

Having had considerable practice in the use of the symbol of operation D for ∂∂x , we

may now use D′ to represent the operation ∂∂y .

By Taylor’s theorem,

φ (y +mx) = φ (y) +mx∂φ (y)

∂y+m2x2

2!· ∂

2φ (y)

∂y2+ . . . ,

where x is regarded as constant.

∴ φ (y +mx) =

(

1 +mx∂

∂y+m2x2

2!

∂2

∂y2+ . . .

)

φ (y) .

The term in brackets is clearly an exponential series (page 280) equivalent to emx ∂∂y

φ(y),

or, writing D′ for ∂∂y ,

φ (y +mx) = emxD′φ(y) (1)

The general form of the linear equation is,

A0∂2z

∂x2+ A1

∂2z

∂x∂y+ A2

∂2z

∂y2+ A3

∂z

∂x+ A4

∂z

∂x+ A5z = A, (2)

where A1, A2,. . . , A, may be constants, or functions of x and y.As with ordinary linear equations,

Complete Solution = Particular Integral + Complementary function.

The complementary function is obtained by solving the left-hand sideof equation (2), equated to zero. We may write (2) in symbolic form,

(

A0D2 + A1DD′ + A2D

′2 + A3D + A4D

′ + A5

)

z = 0. (3)

where D is written for ∂∂x; D′ for ∂

∂y; DD′ for ∂2

∂x∂y. Sometimes we under-

standF (D,D′) z = 0. (4)

in place of (2).

§ 141. Linear Partial Equations with Constant Coefficients.

A. Homogeneous equations. Type:

A0∂2z

∂x2+ A1

∂2z

∂x∂y+ A2

∂2z

∂y2= R, (5)

where R is a function of x.To find the complementary function, put R = 0, and. instead of as-

suming, as a trial solution, that y = emx, as was the case with the ordinaryequation, suppose that

z = φ (y +mx) , (6)

347

is a trial solution. Differentiate (6), with respect to x and p, we thus obtain,

∂z

∂x= mf ′ (y +mx) ;

∂z

∂y= f ′ (y +mx) ;

∂2z

∂x∂y= mf ′′ (y +mx) ;

∂2z

∂x2= m2f ′′ (y +mx) ;

∂2z

∂y2f ′′ (y +mx) ;

Substitute these values in equation (5) equated to zero, and divide out thefactor f ′ (y +mx). The auxillary equation,

A0m2 + A1m+ A2 = 0. (7)

remains. If m is a root of this equation, f ′ (y +mx) = 0, is a part of thecomplementary function. If α and β are the roots of (7), then

z = eαxD′φ1(y) + eβxD

′φ2(y), (8)

as in (3), § 130. From (6), therefore,

z = f1 (y + αx) + f2 (y + βx) (9)

since α and β are the roots of the auxillary equation (7), we can write (5)in the form,

(D + αD′) (D + βD′) = 0. (10)

EXAMPLES.–Solve the following equations:

(1)∂2z

∂x2− ∂2z

∂y2= 0. Ansr. z = f1 (x+ y) + f2 (y − x).

(2)∂2z

∂x2− 4

∂2z

∂x∂y+ 4

∂2z

∂y2= 0. Ansr. z = f1 (y − 2x) + f2 (y − 2x). ←−

(3) 2∂2z

∂x2− 3

∂2z

∂x∂y− 2

∂2z

∂y2= 0. Ansr. z = f1 (2y − x) + f2 (y + 2x).

(4)∂2u

∂t2= a2

∂2u

∂x2. Ansr. u = f1 (at+ x) + f2 (at− x). This most important equa-

tion, sometimes called d’Alembert’s equation, represents the motion of vibrating strings,the law for small oscillations of air in narrow tubes (organ pipes), etc.

We cannot say much about the undetermined functions f1 (at+ x) and f2 (at− x)in the absence of data pertaining to some specific problem. Consider a vibrating harpstring, where no force is applied after the string has once been put in motion. Let x = ldenote the length of the string under a tension equal to the weight of a length L ofthe same kind of string. In order to avoid a root sign later on, a2 has been written inplace of gL, where g represents the constant of gravitation. Further, let u represent thedisplacement of any part of the string we please, and let the ordinate of one end of thestring be zero. Then, whatever value we assign to the time t, the limiting conditions areu = 0, when x = 0; and u = 0, when x = l.

∴ f1 (at) + f2 (at) = 0; f1 (at+ l) + f2 (at− l) = 0,

are solutions of dAlemberts equation. From the former, it follows that

f1 (at) must always be equal to − f2 (at) ;∴ f1 (at+ l)− f1 (at− l) = 0.

348

But at may have any value we please. In order to fix our ideas, suppose that at −l = q,∴ at+ l = q + 2l, where q has any value whatever.

∴ f1 (q + 2l) = f1 (q) .

The physical meaning of this solution is that when q is increased or diminished by 2l, thevalue of the function remains unaltered. Hence, when at is increased by 2l, or, what isthe same thing, when t is increased by 2l/a, the corresponding portions of the string willhave the same displacement. In Other words, the string performs at least one completevibration in the time 2l/a. Hence, we conclude that d’Alemberts equation represents afinite periodic motion, with a period of oscillation 2l/c.

EXAMPLE.–Show that12 {f1 (at+ l) + f1 (at− l)} = 0.

is a solution of d’Alembert’s equation, and interpret the result.A further study of d’Alembert’s equation would require the introduction of Fourier’s

series, Chapter VIII.

When two of the roots are equal, say α = β. We know that the solutionof

(D − a)2 z = 0, z = emx (C1x+ C2) , § 130;by analogy, the solution of

(D − aD′)2z = 0; z = emx {xf1 (y) + f2 (y)} ,

or, z =xf1 (y + ax) + f2 (y + ax) . (11)

EXAMPLES.–Solve:(1) ∂2z

∂x2 + 2 ∂2z∂x∂y + ∂2z

∂y2 = 0. Ansr. z = x f1 (y + x) + f2 (y + x). ←−(2)

(

D3 − 3D2D′ +DD′2 +D′3)

z = 0. ←−Ansr. z = xf1 (y − x) + f2 (y − x) + f3 (y + x) .

The particular integral will be discussed after.B. Non-homogeneous equations. Type:

A0∂2z

∂x2+ A1

∂2z

∂x∂y+ A2

∂2z

∂y2+ A3

∂z

∂x+ A4

∂z

∂y+ A5z = 0. (12)

If the non-homogeneous equation can be separated into factors, the integralis the sum of the integrals corresponding to each symbolic factor, so thateach factor of the form D −mD′, appears in the solution as a function ofy+mx, and every factor of the form D−mD′− a, appears in the solutionin the form z = eaxf (y +mx).

EXAMPLES.– (1) Solve∂2z

∂x2− ∂2z

∂y2+∂z

∂x+∂z

∂y= 0.

Factors, (D +D′) (D −D′ + 1) z = 0. Ansr. z = f1 (y − x) + e−xf2 (y + x).

(2) Solve∂2z

∂x2− ∂2z

∂x∂y− ∂z

∂x− ∂z

∂y= 0.

Factors, (D + 1) (D −D′) z = 0. Ansr. z = e−xf1 (y) + f2 (x− y).

349

It is, however, not often possible to represent the solutions of theseequations in this manner. When this is so, it is customary to take the trialsolution,

x = eαx+βy, (13)

and substitute for z in the given equation (12). Then,

∂z

∂x= αz;

∂z

∂y= βz;

∂2z

∂x∂y= αβz;

∂2z

∂x2= α2z;

∂2z

∂y2= β2z.

Equate the resulting auxillary equation to zero. We thus obtain

(

A0α2 + A1αβ + A2β

2 + A3α + A4β + A5

)

z = 0. (14)

This may be looked upon as containing a bracketed quadratic in α andβ. For any value of β, we can find the corresponding value of α, or thevalue of α, for any assigned value of β. There is thus an infinite number ofparticular solutions of this differential equation.

If u1, u2, u3,. . . , use particular solutions of any partial differential equa-tion, each solution can be multiplied by an arbitrary constant and the re-sulting products are also solution of the equation.

Similarly, it is not difficult to see that the sum of any member of par-ticular solutions will also be a solution of the given equation.

It is usually not very difficult to find particular solutions, even whenthe general solution cannot be obtained. The chief difficulty lies in thecombining of the particular solutions in such a way, that the conditions ofthe problem under investigation are satisfied. Plenty of illustrations willbe found at the end of the next chapter.

If the above quadratic is solved for α in terms of β, and if the resultingf (α, β), is homogeneous, we shall have the roots in the form,

α = m1β; α = m2β; , . . . , α = mnβ.

The equation will, therefore, be satisfied by any expression of the form, Where mand Care notsets ofdiscretenumbers,this sumbecomesanintegraltaken dmwhere Cis any in-tegrablefunctionof m.KH

z = ΣCeβ(y+mx), (15)

where m has any value m1, m2,. . . and C may have any value C1,C2,. . . The symbol “Σ” indicates the sum of the infinite series, obtainedby giving m and C all possible values.

The above solution (15), may be put in a simpler form when β is alinear function of α, say, β = aα+ b. This applies to equation (12). Again,we can sometimes solve the equation z = eαx+βy = 0, for α, in terms of β.In order to fix these ideas, let us proceed to the following examples.

350

EXAMPLES.–(1) Solve(

D2 −D′) z = 0. Hence α2 − β = 0. Put α = 12 , α = 1,

α = 2,. . . and we get the particular solutions

e14(2x+y), ex+y, e2x+4y, . . .

Now the difference between any two terms of the form eαx+βy, is included in the abovesolution, it follows, therefore, that the first differential coefficient of eαx+βy, is also anintegral, and, in the same way, the second, third and higher derivatives must be integrals.Thus we have the following particular solutions:–

Deαx+α2y = (x+ 2αy) eαx+α2y.

D2eαx+α2y = {(x+ 2αy) + 2y} eαx+α2y.

D3eαx+α2y = {(x+ 2αy) + 6y (x+ 2αy)} , etc.

If α = 0, we get the special case,

z = C1x+ C2

(

x2 + 2y)

+ C3

(

x3 + 6xy)

+ . . .

(2) Solve ∂2z∂x2 − ∂2z

∂y2 − 3 ∂z∂x + 3 ∂z

∂y = 0. Put z = Ceαx+βy. Note

(α− β) (α+ β + 3) = 0. ∴ β = α and β = 3− α.

Hence z = ΣC1eα(x+y) + e3yΣC1e

α(x−y); z = f1 (y + x) + e3yf2 (y − x).(3) Solve ∂2z

∂x∂y + a ∂z∂x + b ∂z∂y + abz = 0. Ansr. z = e−ayf1 (x) + e−bxf2 (y).

(4) Solve(

D2 −D′2 +D + 3D′ − 2)

z = 0. Ansr. z = exf1 (y − x)+e−2xf2 (y + x).

§ 142. The Particular Integral of Linear Partial Equations. Thefollowing methods for finding the particular integral of homogeneous ornon-homogeneous equations, are deduced by processes analogous to thoseemployed for the particular integrals of the ordinary equations.

The complementary function of the ordinary linear equation

(D −m) z = 0, is, Cemx;

so, for the partial equation(D −mD′) z = 0, we have emxD′φ(y), or the equivalent φ (y +mx).

This analogy extends to the particular integrals. In the former case,the particular integral of

(D −m) z = R, is, z = (D −m)−1R;

while for F (D,D′) z = R, we have z = F (D,D′)−1R.Case 1 (General). When F (D,D) can be resolved into factors, so that,

z = (D −mD′) f (x, y) . (16)

It is now necessary to find a value for this symbol (16). First show that

DeαxR = (D + α)R,

351

by putting mD in place of α, and f (x, y), in place of R (Case 4, § 131).

∴1

D −mD′f (x, y) =

1

D −mD′emxD′

e−mxD′

f (x, y) ;

= emxD′ 1

Df (x, y −mx) . (17)

The value sought. The particular integral may, therefore, be found by thefollowing series of operations:

(1) Subtract mr from p in the f (x, y), to be operated upon. (2) Inte-grate the result with respect to dx. (3) Add mx to y, after the integration.

If there is a succession of factors, the rule is to be applied to each oneseriatim, beginning on the right.

EXAMPLES.–Find particular integrals in the following examples. It is well to becareful about the signs of the different terms added and subtracted. It is particularlyeasy to err by want of attention to this.

(1) ∂2z∂x2 −a2 ∂2z

∂y2 = xy. The particular integral is 1D−aD′

· 1D+aD′

xy. Now xy becomes

x (y − ax). This, on integration with respect to dx, becomes 12x

2y − 13ax

3, and finally12x

2y − 12ax

3 − 13ax3 . Hence,

1

D − aD′ ·1

D + aD′ xy =1

D − aD′ ·x2y

2+ax3

6.

Subtract −ax from y, for 12x

2 (y + ax) + 16ax

3. Integrate and add −ax to the result.16x

3y remains. This is the required result.

(2)(

D2 + 3DD′ + 2D′2)−1x+ y. Ansr. 1

2x2y − 1

3x3.

Case 2 (Special). When R has the form f (ax+ by). Multiply F (D,D′) zby Dn and get F (D,D′)

must behomoge-neous inpowersin orderthatφ (D′/D)will be afunctionof asinglevariable.KH

Dmφ (D′/D) z = f (ax+ by) .

Operate on ax+ by with D′ and D respectively,

D′

Df (ax+ by) =

b

a.

As on page 313, the particular integral is

z =1

Dn· 1

φ (D′/D)f (ax+ by) .

=1

φ (b/a)

∫∫

. . .

∫

f (ax+ by) dxn.

How to use this formula will appear from the examples.

Numberof inte-grationsis equalto thehighestpowerof D inF (D,D′).KHEXAMPLES.–Find particular integrals in, (1)

(

D2 +DD′ − 2D′2) z = sin (x+ 2y).The particular integral is

1

D2 (1 +D′/D − 2D′2/D2)sin (x+ 2y) =

1

1 + 2− 8

∫∫

sin (x+ 2y)dx2;

= 15 sin (x+ 2y) .

(2)(

D2 + 5DD′ + 6D′2) z = 1/ (y + 2x). Ansr. 120 {(y + 2x) log (y + 2x)− y − 2x}.

352

The above process cannot be employed when F (D,D′), or F (a, b) hasthe same form as R, because a vanishing factor then appears in the result.In such a case, use the above method for all factors which do not vanishwhen a is put for D, b for D. The solution is then completed by means ofthe formula: 1

D −mD′f (y +mx) = xf (y +mx) . (18)

For afullyworkedexampleof thistype ofequation,see thetop ofS-53 ofthe sup-plement.KH

EXAMPLE.–Evaluate the particular integral in

(D −D′) (D + 2D′) z = x+ y.

For the first factor, use the above method and then

=1

D −D′ ·1

3D(x+ y) =

1

3DexD

′

e−xD′

(x+ y) ;

=1

3DexD

′ 1

Dy =

1

3DexD

′

xy =1

9x3 +

1

6x2y.

Case 3 (Special). When R has the form of sin (ax+ by), or cos (ax+ by).Proceed. as on page 313, when

1

F (D2, DD′, D′2)sin (ax+ by) =

1

(−a2,−ab,−b2) sin (ax+ by) , (19)

and in the same way for the cosine.

EXAMPLES.–Find the particular integrals(1)

(

D2 +DD′ +D′ − 1)

z = sin (x+ 2y).

∴1

D2 +DD′ +D′ − 1sin (x+ 2y) =

1

−1− 2 +D′ − 1sin (x+ 2y)

=D′ + 4

D′2 − 16sin (x+ 2y) = − 1

20 (D′ + 4) sin (x+ 2y) = − 1

10 {cos (x+ 2y) + 2 sin (x+ 2y)} .

(2) (D +DD′ − 2D′) z = sin (x− y) + sin (x+ y). Find the particular integral for ←−Thisexampleis moreeasilysolvedusing themethodof unde-terminedcoeffi-cients.See sup-plementS-53. KH

sin (x− y), then for sin (x+ y). Add the two results together.Ansr. 1

2 sin (x− y) + 13 cos (x+ y).

For the anomalous case proceed as in § 131.

Case 4 (Special). When R has the form eax+by, proceed as directed onpage 312,

1

F (D,D′)eax+by =

1

F (a, b)eax+by, (20)

that is to say, put a for D and b for D′.

EXAMPLES.–Find particular integrals in the following:(1)

(

D2 −DD′ − 2D′2 + 2D + 2D′) z = e2x+3y.

Ansr.(

D2 −DD′ − 2D′2 + 2D + 2D′)−1e2x+3y = − 1

10e2x+3y.

(2) (DD′ + aD + bD′ + ab) z = emx+ny. Ansr. emx+ny/ (m+ a) (n+ b).

If F (a, b) = 0, proceed as on page 313,

z =1

F ′a (a, b)

eax+by; or, z =1

F ′b (a, b)

eax+by, (21)

353

where F ′a or F ′

b, denotes the first differential coefficient with respect to thesubscript. The two results agree with each other.

EXAMPLE.–Solve(

D2 −D′2 − 3D + 3D′) z = ex+2y.

Ansr. f1 (x+ y) + e3yf2 (y − x) − yex+2y.

Case 5 (Special). When R has the form xrys, where r and s are positiveintegers. Operate with F (D,D′)−1 on xrys expanded in ascending powersof r and s.

EXAMPLES.–(1) Find the particular integral in:(

D2 +DD′ +D − 1)

z = x2y.

= −{

1−(

D2+DD′+D′)}−1=−

{

1+(

D2+DD′+D′)+(

D2+DD′+D′)2 + . . .}

x2y.

= −(

1 +D2 +DD′ +D′ + 2D2D′)x2y = x2y − 2y − 2x− x2 − 4.

The expansion is not usually carried higher than the highest power of the highest powerin f (x, y). See S-54

of thesupple-ment forwhy theabovestate-mentfails toapply toexample2. KH

(2) Evaluate(

D2 −D′2 − 3D + 3D′)−1xy. Ansr.

− 1

18x3 − 1

9x2 − 2

27x− 1

9xy − 1

6x2y.

(3)(

D2 − a2D′2) z = x.

=1

D2

(

1 + a2D′2

D2+ . . .

)

x =1

D2x =

∫∫

x dx2 =1

6x3.

Case 6 (Special). When R has the form eax+byX , where X is a functionof x or y. Use

F (D,D′)−1

eax+byX = eax+byF (D + a,D + b)−1X, (22)

derived as on page 315.Use themethodin case 5on thispage toapply theoperatorto thepoly-nomialafteryou haveappliedcase 6to thisexample.KH

EXAMPLE.–Find the particular integrals in

∂2z/∂x2 − ∂z/∂y = xeax+a2y.

Ansr. eax+a2y 1

D2 + 2aD −D′x = eax+a2y 1

2a· 1D

(

1 +D

2a

)−1

x

= eax+a2y 14

(

x2/a− x/a2)

.

§ 143. The Linear Partial Equation with Variable Coefficients.These may sometimes be solved by transforming them into a form withconstants. E.g.,

(i.) Any term xrys ∂r+sz∂xr∂ys

, may be reduced to the form with constantcoefficients, by substituting u = log x, v = log y.

EXAMPLES.–Solve the equations:

(1) x2∂2z

∂x2− y2 ∂

2z

∂y2− y ∂z

∂y+ x

∂z

∂x= 0. This reduces to ∂2z

∂u2 − ∂2z∂v2 = 0.

354

Hence the solution of this equation, z = φ1 (u+ v) + φ2 (u− v), must be re-convertedinto the form in x and y, thus, z = f1 (xy) + f2 (x/y).

(2) x2∂2z

∂x2+ 2xy

∂2z

∂x∂y+∂2z

∂y2= 0. Ansr. z = f1 (y/x) + xf2 (y/x).

(3) (x+ y)∂2z

∂x∂y−a∂z

∂x= 0. Put ∂z = v·∂x. Ansr. z = f1 (y)+

∫

(x+ y)a f ′2 (x)·dx.

(ii.) The transformation may be effected by substituting ϑ = x ∂∂x

andϑ′ = y ∂

∂y, and treating the result as for constant coefficients.

EXAMPLES.–(1) Solve the first two examples of the preceding set in this way.

(2) x2∂2z

∂x2+ 2xy

∂2z

∂x∂y+ y2

∂2z

∂y2− nx∂z

∂x− ny ∂z

∂y+ nz = 0.

∴ {ϑ (ϑ− 1) + 2ϑϑ′ + ϑ′ (ϑ′ − 1)− nϑ− nϑ′ + n} z = 0.

Ansr. z = xnf1 (y/x) + xf2 (y/x).

§ 144. The Integration of Differential Equations in Series. Whena function can be developed in a series of converging terms, arranged inpowers of the independent variable, an approximate value for the dependentvariable can easily be obtained. The degree of approximation attainedobviously depends on the number of terms of the series included in thecalculation. The older mathematicians considered this an underhand way ofgetting at the solution but, for practical work, it is invaluable. As a matterof fact, solutions of the more advanced problems in physical mathematicsare nearly always represented in the form of an abbreviated infinite series.Finite solutions are the exception rather than the rule.

EXAMPLES.–(1) Evaluate the integral in f (x) = 0. Assume that f (x) can bedeveloped in a converging series of ascending powers of x, that is to say,

f (x) = a0 + a1x+ a2x2 + a3x

3 + . . . (1)

By integration

∫

f (x) dx =

∫

(

a0 + a1x+ a2x2 + . . .

)

dx;

=

∫

a0dx+

∫

a1x dx +

∫

a2x2dx+ . . . ;

= a0x+ 12a1x

2 + 13a2x

3 + . . . ;

= x(

a0 +12a1x+ 1

3a2x2 + . . .

)

+ C. (2)

(2) It is required to find the solution of dy/dx = y, in series. Assume that y = f (x),has the form (1) above, and substitute in the given equation.

(a1 − a0) + (2a2 − a1) x+ (3a3 − a2)x2 + · · · = 0. (3)

355

This equation would be satisfied, if

a1 = a0; a2 =12a1 =

12a0; a3 =

13a2 =

13!a0;. . .

Hence,y = a0φ (x) ,

where φ (x) = 1 + x+ 12!x2 + 1

3!x3 + · · · = ex.

Put a for the arbitrary function,∴ y = aex.

That this is a complete solution, is proved by substitution in the original equation.Write the original equation in the form

y = vφ (x) ,

where v is to be determined. Hence,

←−dy

dxφ (x) + v′ {φ′ (x)− φ (x)} = 0,

since φ (x) satisfies the original equation,

dy

dx= 0, or v is constant.

For equations of higher degree, we must proceed a little differently. For example:

(3) Solved2y

dx2− xdy

dx− cy = x2. (4)

(i.) The complementary function. As a trial solution, put y = axm. The auxillaryequation is

m (m− 1) a0xm−2 − (m+ c)xm = 0. (5)

This shows that the difference between the successive exponents of x in the assumedseries, is −2. The required series is, therefore,

y = a0xm + a1x

2m−2 + · · ·+ an−1xm+2n−2 + anx

m+2n,

which is more conveniently written

y =

∞∑

0

anxm+2n. (6)

In order to completely determine this series, we must know three things about it.Namely, the first term, the coefficients of x and the different powers of x that make upthe series.

Substitute (6) in (4),∞∑

0

(m+ 2n) (m+ 2n− 1)anxm+2n−2 − (m+ 2n+ c) anx

m+2n = 0, (7)

where n, has all values from zero to infinity. If x is a solution of (4), the coefficient ofxm+2n−2 must vanish with respect to m. Hence by equating the coefficient of xm+2n−2

to zero,*(m+ 2n) (m+ 2n− 1) an − (m+ 2n− 2 + c) an−1 = 0. (8)

if n = 0, m = 0, or m = 1.When n is greater than zero,

an =m+ 2n− 2 + c

(m+ 2n) (m+ 2n− 1). (9)

This formula allows us to calculate the relation between the successive coefficients of xby giving n all integral values 1, 2, 3,. . .

* If we take the other part of the auxillary a diverging series is obtained, useless for ourpurpose.

356

First, suppose m = 0, then we can easily calculate from (9),

a1 =c

1 · 2 a0; a2 =c

3 · 4a1 =c (c+ 2)

4!a0; . . .

∴ y′ = a0

{

1 + cx2

2!+ c (c+ 2)

c4

4!+ . . . (10)

Next put m = 1, and, to prevent confusion, write b, in (9), in place of a,

bn =c+ 2n− 1

2n (2n+ 1)bn−1;

proceed exactly as before to find successively b1, b2, b3. . .

∴ y′′ = b0

{

x+ (c+ 1)x3

3!+ (c+ 1) (c+ 3)

x5

5!+ . . . (11)

The complete solution of the equation, is the sum of series (10) and (11), or if ay1 = y′,by2 = y′′,

y = ay1 + by2,

which contains the two arbitrary constants a and b.(ii.) The particular integral. By the above procedure we obtain the complementary

function. For the particular integral, we must follow a somewhat similar method. E.g.,equate (8) to x2 instead of to zero. The coefficient of m− 2, in (5), becomes The

m−2 = 2equationcomesfromequatingthe ex-ponents.KH

m (m− 1)a0xm−2 = x2;

∴ m− 2 = 2 and m (m− 1) a0 = 1;

∴ m = 4; a0 = 12.

From (9) an =c+ 2n+ 2

2 (n+ 2) (2n+ 3)an−1.

Substitute successive values of n = 1, 2, 3, . . . in the assumed expansion, and we obtain

(particular integral) = a0xm + a1x

m+2 + a2xm+4 + . . . ,

where a0, a1, a2, . . . and m have been determined.(4) Solve d2y/dx2 + xy = 0.

Ansr. y = a

(

1− 1

3!x3 +

1 · 46!

x6 − . . .)

+

(

x− 2

4!x4 +

2 · 57!

x7 − . . .)

.

The so-called Riccati’s equation,

dy

dx+ by2 = cxn

has attracted a lot of attention in the past. Otherwise it is of no particular interesthere. It is easily reduced to a linear form of the second order. Its solution appears as aconverging series, finite under certain conditions.

Forsyth (l.c.) or Johnson (l.c.) must be consulted for fuller details. A detailedstudy of the more important series employed in physical mathematics follows naturallyfrom this point. These are mentioned in the next section along with the titles of specialtextbooks devoted to their use.

§ 145. Harmonic Analysis. One of the most important equations in physicalmathematics, is

∂2V

∂x2+∂2V

∂y2+∂2V

∂z2=

1

κ

∂V

∂t. (1)

It has practically the same form for problems on the conduction of heat, the motion offluids, the diffusion of salts, the vibrations of elastic solids and

357

flexible strings, the theory of potential, electric currents and numberless other phenom-ena. x, y, z are the coordinates of a point in space, t denotes the time and V may denotetemperature, concentration of a solution, electric and magnetic potential, the Newto-nian potential due to an attracting mass, etc., κ is a constant. If the second member iszero, we have Laplace’s equation, if the second number is equated to 4πρ; where ρ is afunction of x, y, z, the result is known as Poissons equation.

∂2V

∂x2+∂2V

∂y2+∂2V

∂z2= 0, is Laplace’s equation.

∂2V

∂x2+∂2V

∂y2+∂2V

∂z2= 4πρ, is Poisson’s equation.

The first member is written ∇2V by some writers, ∆2V by others. The equation is Moderntextsalmostuniver-sally use∇2V ,which isspokenas “del-squaredvee.” KH

often more convenient to use in polar coordinates, viz.,

∇2V =∂2V

∂r2+

1

r2· ∂

2V

∂θ2+

2

r· ∂V∂r

+cot θ

r2· ∂V∂θ

+1

r sin2 θ, (2)

where the substitutions are indicated in (11), § 48.*Any homogeneous algebraic function of x, y, z, which satisfies equation (1), is said

to be a solid spherical harmonic. These functions are chief used for finding the potentialon the surface of a sphere, due to forces which are not circularly symmetrical.†

Particular solutions of (1) give rise, under. special conditions, to the so-called surfacespherical harmonics, tesseral harmomics and toroidal harmonics. The series See S-58

for devel-opmentof theseries ex-pansionof J0.KH

xn

2nΓ (n+ 1)

{

1− x2

22 (n+ 1)+

x4

24 · 2! (n+ 1) (n+ 2)− . . .

}

,

is called a Cylindrical Harmonic or a Bessel’s function of the nth order. The symbolJn (x) is used for it. The series is a particular solution of Bessel’s equation.

d2y

dx2+

1

x· dydx

+

(

1− n2

x2

)

y = 0.

If n = 0, the series is symbolised by J0 (x) and called a Bessel’s function, of the zerothorder. These functions are employed in physical mathematics when dealing with certainproblems connected with equation (1). Another particular solution is Most

moderntexts useY ratherthanK forBesselfunctionsof thesecondkind,with Kreservedfor mod-ifiedBesselfunc-tions.KH

J0 (x) log x+x2

22− x4

22 · 42(

1

1+

1

2

)

+ . . . ,

called a Bessel’s function of the second kind (of the zeroth order), symbolised by K0 (x).Similarly, the solution of Legendre’s equation

(

1− x2) d2y

dx2− 2x

dy

dx+m (m+ 1) y = 0,

is the series

1− m (m+ 1)

2!x2 +

m (m− 2) (m+ 1) (m+ 3)

4!x4 − . . .

* This transformation is described in the regular textbooks. But possibly the readercan do it for himself.† A point is said to be circularly symmetrical, when its value is not affected by

rotating it through an angle about the axis.

358

written, for brevity, Pm (x). This furnishes the so-called Surface Zonal Harmonics,Legendre’s coefficients, or Legendrians. Another particular solution,

x− (m− 1) (m+ 2)

3!x3 +

(m− 1) (m− 3) (m+ 2)

5!x5 − . . .

written Qm (x), gives rise to Surface Zonal Harmonics of the second kind. Both seriesare extensively employed in physical problems connected with equation (1).

The equation

(

x2 − b2) (

x2 − c2) d2y

dx2+x

(

x2 − b2 + x2 − c2) dy

dx−{

m (m+ 1)(

x2 − b2 + c2)

p}

y = 0,

called Lame’s equation, has “series” solution which furnishes Lames functions orEllipsoidal Harmonics, users in special problems connected with the ubiquitous equation

∇2V +1

κ

dV

dt

The so-called hypergeometric or Gauss’ series,

1 +ab

1!cx+

a (a+ 1) b (b + 1)

2!c (c+ 1)x2 + . . . ,

appears as a solution of certain differential equations of the second order, say,

x (1− x) d2y

dx2+ {c− (a+ b+ 1)x} dy

dx− aby = 0,

(Gauss’ equation), where a, b, c, are constants.The application of these series to particular problems constitutes that branch of

mathematics known as Harmonic Analysis.But we are getting beyond the scope of this work; for more practical details, the

reader will have to take up some special work such as Byerly’s Fourier’s Series andSpherical Harmonics. Weber and Riemann’s Die Partiellen Differential-Gleichungen derMathematischen Physik is the textbook for more advanced work. Gray and Mathewshave A Treatise on Bessel’s Functions and their Application, to Physics (Macmillan &Co., 1895).

359

Supplemental Notes, including

Worked Solutions to Selected Examples

These notes by Karl Hahn are furnished under Creative Commons 3.0 license. Seehttp://creativecommons.org/licenses/by/3.0/ for details. You may redistributethese notes freely or incorporate them into other works provided Karl Hahn is attributedas their author and the distributed copies reference Creative Commons 3.0 license.

Page 283 (§ 117. Separable Equations), example 2: Problem: solve

vdv

dx+

µ

x2= 0.

Separating variables: v dv+µ dxx2 = 0. Integrating: 1

2v2− µ

x = C, from which the book’sanswer follows immediately.

example 3: Solve(

1 + x2)

dy =√y·dx. Separating variables: dy/√y = dx/

(

1 + x2)

or

equivalently dy/√y − dx/

(

1 + x2)

= 0. Integrating both sides yields the book answer.

example 4: Solve y − x · dy/dx = a (y + dy/dx). Gathering terms:(1− a) y = (x+ a) dy/dx. Multiply by dx and divide by y (x+ 1):(1− a) dx/ (x+ 1) = dy/y. Integrate: (1− a) log (x+ 1) + C′ = log y, which is the logof the book answer if you allow that C = eC

′

.

example 5: Word problem. Let a be the constant of proportionality. Since the chargeis being dissipated, it time-derivative is negative. So we say dE

dt = −aE. Separatingvariables: dE/E = −a dt. Integrating: logE = −at+ C, or equivalently, E = E0e

−at,where E0 = eC .

Page 284 (§ 117. Separable Equations continued), example 7: The book appearsto be mistaken on this one. Solving the problem stated, it sets up as:

dy

dx= −x

y

Multiplying by y dx to separate variables yields y dy = −x dx or equivalentlyy dy+ x dx = 0. Integrating gives the equation of a circle centered at the origin with anarbitrary radius. To get the book’s answer, it would have to say, “What curves have slope−y/x to the x-axis?” That sets up to y/dy+x/dx = 0 or equivalently log y+log x = C′,which is the log of the book’s answer if you allow C = eC

′

.

Supplementary problem – the cooling of an object: This is a common problemassigned in classes on differential equations. An object cools at a rate that is proportionalto the difference between its temperature and the ambient temperature. Suppose suchan object starts at 100◦C. After 5 minutes its temperature is 60◦C and after 10 minutesits temperature is 50◦C. What is its temperature after 20 minutes? The differentialequation is dT

dt= (Ta − T )k,

where Ta is the constant ambient temperature, and k is the rate of proportionality.Separating variables: dT

T − Ta= −k dt,

whose solution, when integrated, is log (T − Ta) = −kt+ C. Taking the antilog:

T = Ta +∆T0 e−kt,

where ∆T0 = eC is the initial difference between temperature of the object and ambienttemperature. Observe that there are three unknowns in the above, Ta, ∆T0, and k. Thethree conditions given in the problem correspond to

S-1

100◦C = Ta +∆T0, (a)

60◦C = Ta +∆T0u, (b)

50◦C = Ta +∆T0u2, (c)

where u = e−k×5 minutes. Subtracting (a) from (b) and (a) from (c) eliminates Ta.

−40◦C = ∆T0 (u− 1) , (d)

−50◦C = ∆T0(

u2 − 1)

.

Taking the quotient of these two eliminates ∆T0.

4

5=

u− 1

u2 − 1,

which leads to the quadratic, 4u2 − 5u + 1 = 0, hence u = 1 or u = 14 . The former

is impossible because it leads to a zero denominator when you try to solve (d) for∆T0. The latter yields ∆T0 = 160

3◦C. Further back-substitution yields Ta = 140

3◦C. At

t = 20 minutes you have e−kt = u4 = 1256 . Hence

T (20 minutes) = 1403

◦C + 160768

◦C = 3758

◦C = 46.875◦C.

Page 284 (§ 117. Equations with homogeneous powers), example 1: The bookdoes steps up to integrating the substituted equation to get t

1−t +log (1− t)+logx = C.Back-substituting t = y

x ,

1

1− yx

+ log(

1− y

x

)

+ log x = C,

x

x− y + log (x− y) + log x− log x = C,

which, after you cancel the log x terms, becomes the log of the book’s answer.

example 2: The equation is equivalent to (y − x) dydx + y = 0. Substituting y = tx and

dydx = t+ x dt

dx

(tx− x)(

t+ xdt

dx

)

+ tx = 0,

(t− 1)

(

t+ xdt

dx

)

+ t = 0,

t2 − t+ (t− 1)xdt

dx+ t = 0,

∫

(t− 1) dt

t2+

∫

dx

x= C′,

log t+1

t+ log x = C′.

Back-substituting t = y/x:

log y− log x+x

y+ log x = C′,

log y = C′ − x

y,

which is the log of the book’s answer, if you let C = eC′

.

exmple 3: Convert the equation to x2 dydx − xy− y2 = 0. Making the same substitution

as in the last example:

S-2

x2(

t+ xdt

dx

)

− x2t− x2t2 = 0,

(

t+ xdt

dx

)

− t− t2 = 0,

xdt

dx− t2 = 0,

dt

t2− dx

x= 0,

−1

t− log x = C′.

Back-substituting and rearranging:

log x = C′ − x

y,

If you take the antilog of the above, you will see that it disagrees with the book’s answer.Yet if you take the differential of the above, you do indeed get back the original equation,so the error seem to be in the book. The final answer should be x = Ce−x/y, whereC = eC

′

.

example 4: Dividing by dx then making the substitution and integrating

x2 + t2x2 = 2x2t

(

t+ xdt

dx

)

,

1 + t2 = 2t

(

t+ xdt

dx

)

,

1 + t2 = 2t2 + 2txdt

dx,

1− t2 = 2txdt

dx,

dx

x=

2t dt

1− t2 ,

log x+ log(

1− t2)

= C′.

Back-substituting and simplifying

log x+ log

(

1− y2

x2

)

= C′,

log x+ log(

x2 − y2)

− 2 logx = C′,

log(

x2 − y2)

− log x = C′,

x2 − y2x

= C,

from which the book’s answer follows immediately.

Page 285 (§ 117. Equations that can be made homogeneous in powers),example 2: This equation is exact, so using the method outlined in this paragraph isvery much the long road to its solution. Using the method explained in § 119, you cansolve this one on inspection. Nevertheless we shall do it the hard way here. We assumewe have h and k such that x = v + h and y = w + k yields the equation, (2v − w) dv +(2w − v) dw = 0. The sequence to the solution to this (using the substitution, w = tv)is

S-3

(2v − w) + (2w − v) dwdv

= 0,

(2v − tv) + (2tv − v)(

t+ vdt

dv

)

= 0,

2− t+ (2t− 1)

(

t+ vdt

dv

)

= 0,

2− t+ 2t2 − t+ (2t− 1)

(

vdt

dv

)

= 0,

2− 2t+ 2t2 + (2t− 1)

(

vdt

dv

)

= 0,

dv

v+

1

2

(2t− 1) dt

t2 − t+ 1= 0,

log v + 12 log

(

t2 − t+ 1)

= C′,

Back-substituting and simplifying

log v + 12 log

(

w2/v2 − w/v + 1)

= C′,12 log

(

w2 − wv + v2)

= C′,

w2 − wv + v2 = C,

where C = e2C′

. Replacing v with x − h and w with y − k, this becomes our solutionequation,

(y − k)2 − (y − k) (x− h) + (x− h)2 = C,whose differential is

2 (y − k) dy − (y − k) dx− (x− h) dy + 2 (x− h) dx = 0.

For h and k to be such that this is equivalent to the original equation,(2y − x− 1) dy + (2x− y + 1) dx, you must have −2h − k = 1 from the dx terms and−2k− h = −1 from the dy terms. Solving yields h = −1 and k = 1. Replacing h and kwith these values in the solution equation produces the book’s answer.

Page 285 (§ 117. Equations where h and k cannot be established), secondexample 1: Substituting z = 2x+3y, you also have dz/dx = 2+3dy/dx or equivalentlydy/dx = 1

3 (dz/dx− 2).

(2x+ 3y − 5)dy

dx+ (2x+ 3y − 1) = 0,

1

3(z − 5)

(

dz

dx− 2

)

+ z − 1 = 0,

(z − 5)

(

dz

dx− 2

)

+ 3z − 3 = 0,

(z − 5)dz

dx− 2z + 10 + 3z − 3 = 0,

(z − 5)dz

dx+ z + 7 = 0,

(z − 5)dz

z + 7+ dx = 0.

Since z−5z+7 = 1− 12

z+7 , integrating followed by back-substitution yields

z − 12 log (z + 7) + x = C,

2x+ 3y − 12 log (2x+ 3y + 7) + x = C,

3x+ 3y − 12 log (2x+ 3y + 7) = C.

which, when divided by 3, is the book’s answer.

S-4

Page 285, second example 2: The substitution is again z = 2x + 3y, so again,dy/dx = 1

3 (dz/dx− 2).

(3y + 2x+ 4)− (4x+ 6y + 5)dy

dx= 0,

(z + 4)− 1

3(2z + 5)

(

dz

dx− 2

)

= 0,

3z + 12− (2z + 5)

(

dz

dx− 2

)

= 0,

3z + 12− (2z + 5)dz

dx+ 4z + 10 = 0,

7z + 22− (2z + 5)dz

dx= 0,

dx− (2z + 5)dz

7z + 22= 0.

Since 2z+57z+22 = 2

7 − 97

17z+22 , integration and back-substitution is

x− 27z +

949 log (7z + 22) = C,

7x− 2 (2x+ 3y) + 97 log (14x+ 21y + 22) = C,

3x− 6y + 97 log (14x+ 21y + 22) = C.

The above disagrees with the book’s answer by a scalar of the terms outside the log.Yet taking the differential of the above does produce the original equation, so it appearsthat the book is incorrect.

Page 288 (§ 118. Eliminating parameters), example 1: Divide by x, then differ-entiate twice

y

x= a+ bx,

1

x

dy

dx− y

x2= b,

1

x

d2y

dx2− 1

x2dy

dx− 1

x2dy

dx+ 2

y

x3= 0.

Multiply by x3, then gather like terms to arrive at the book’s answer.

example 2: Divide by x and differentiate

y2

x= 4m,

2y

x

dy

dx− y2

x2= 0.

Multiplying by x2 and dividing by y yields 2x · dy/dx− y = 0. This disagrees with thebooks answer. Yet if you substitute y =

√4mx and dy/dx =

√

m/x into this solution,it works. So the book appears to be in error.

example 3:y = α cosx +β sinx

+y′′ = −α cosx −β sinxy + y′′ = 0

S-5

example 4: First eliminate α:y′ = αaeax +βbebx

−ay = −αaeax −βaebxy′ − ay = β (b− a) ebx

Then use that result to eliminate β:y′′ − ay′ = βb (b− a) ebx−by′ + aby = −βb (b− a) ebx

y′′ − (a+ b) y′ + aby = 0

example 5: Divide by a− x and differentiate

dx

dt

1

a− x = k,

d2x

dt21

a− x +

(

dx

dt

)21

(a− x)2= 0,

(a− x) d2x

dt2+

(

dx

dt

)2

= 0. (a)

The general solution to the original equation is x = a+ Ce−kt. The implication is thatgiven arbitrary k and C, this function always satisfies (a). Observe that x (t) is, amongother things, the position of the end of a relaxing spring that pulls on a damper, wherethe fully relaxed position is at x = a. In this application, our result indicates that thesquare of the spring-end’s velocity is equal to the product of its acceleration times itsdistance from the fully relaxed position, regardless of the spring constant or dampingconstant. Similar relationships exist for analogous systems, such as the current in anelectric circuit having a resistor and inductor in series, or the temperature of an objectcooling to ambient temperature.

Page 290 (§ 119. Exact Equations) example 3: M = a2y+ x2 and N = b3 + a2x.Hence ∂M

∂y= a2 and

∂N

∂x= a2.

example 4: M = sin y + y cosx and N = sinx+ x cos y. Hence

∂M

∂y= cos y + cosx and

∂N

∂x= cosx+ cos y.

Page 291, example 3: M = a2y + x2 and N = b3 + a2x. Hence∫

M dx = a2xy + 13x

3 + g (y) and

∫

N dy = b3y + a2xy + h (x) .

Assuming the two integrals are equal, we infer g (y) = b3y and h (x) = 13x

3. Either wayyou end up with the book’s answer.

example 4: M =(

x2 − y2)

and N = −2xy. Hence∫

M dx = 13x

3 − xy2 + g (y) and

∫

N dy = −xy2 + h (x) .

Assuming the two integrals are equal, we infer g (y) = 0 and h (x) = 13x

3. This givesthe solution, 1

3x3 − xy2 = C, from which the book’s answer follows immediately.

S-6

Page 293 (§ 120. Finding integrating factors, Rule I) example 2: Multiplyingthis out and extracting factors of x dy and y dx,

y2 (y dx)− x2 (2y dx) + y2 (2x dy)− x2 (x dy) = 0.

Rearranging according to the factors outside the parentheses,

x2 (−2y dx− x dy) + y2 (y dx+ 2x dy) = 0.

In this arrangement, α = 2, β = 0, m = −2, n = −1, α′ = 0, β′ = 2, m′ = 1, andn′ = 2. Hence the equations for k and k′ are

−2k − 1− 2 = k′ − 1 and − k − 1 = 2k′ − 1− 2.

Multiply the second by 2 and subtract−2k − 3 = k′ − 1−2k − 2 = 4k′ − 6

−1 = −3k′ + 5

Hence k′ = 2 and k = −2. Using the formulae for finding the exponents of x and y inthe integrating factor: km− 1−α = 4− 1− 2 = 1 and kn− 1− β = 2− 1 = 1. A quickcheck finds that using the primed symbols you get the same exponents, as expected.Recasting the original equation using the integrating factor, xy

(

xy4 − 2y2x3)

dx+(

2x2y3 − x4y)

dy = 0,

which is indeed exact.∫

M dx = 12x

2y4 − 12y

2x4 + g (y) and

∫

N dy = 12x

2y4 − 12x

4y2 + h (x) .

Clearly g (y) = 0 and h (x) = 0. The solution, x2y4−x4y2 = C, that follows is equivalentto the book’s answer. Observe that in other such equations, it is not necessary for k ork′ (or m, m′, n, n′, α, etc.) to be integers.

The original equation in this example,(

y3 − 2yx2)

dx +(

2xy2 − x3)

dy = 0, is homo-geneous in powers, so it can also be solved using the method introduced in § 117.Substituting y = tx and dy/dx = t+ x dt/dx:

(

t3x3 − 2tx3)

+(

2t2x3 − x3)

(

t+ xdt

dx

)

= 0,

(

t3 − 2t)

+(

2t2 − 1)

(

t+ xdt

dx

)

= 0,

t3 − 2t

2t2 − 1+ t+ x

dt

dx= 0,

3t3 − 3t

2t2 − 1+ x

dt

dx= 0,

(

2t2 − 1)

dt

3t3 − 3t+dx

x= 0,

(

1

t+

t

t2 − 1

)

dt

3+dx

x= 0,

log t

3+

log(

t2 − 1)

6+ log x = C,

2 logy

x+ log

(

y2

x2− 1

)

+ 6 logx = C,

2 log y − 2 logx+ log(

y2 − x2)

− 2 logx+ 6 log x = C,

y2(

y2 − x2)

x2 = C,

which is also equivalent to the book’s answer.

S-7

Page 293 (§ 120. Finding integrating factors, Rule II) example 1: Note thatby supposition, all equations susceptible to rule II are homogeneous in powers andtherefore can also be solved by the method introduced in § 117. In this example (whichis equivalent to page 284, example 4), M = x2y + y3 and N = −2xy2. Hence

Mx+Ny = x3y + xy3 − 2xy3 = x3y − xy3.If the reciprocal of the above is an integrating factor, as the book suggests, then

x2y + y3

x3y − xy3 dx−2xy2

x3y − xy3 dy = 0

is exact. Cancelling common factors

x2 + y2

x3 − xy2 dx−2y

x2 − y2 dy = 0.

Letting M ′ be the left-hand function of the above and N ′ be the right, then

∂M ′

∂y=

(

x3 − xy2)

(2y)−(

x2 + y2)

(−2xy)(x3 − xy2)2

=2x3y −���2xy3 + 2x3y +���2xy3

x2 (x2 − y2)2=

4x�3y

��x2 (x2 − y2)2,

∂N ′

∂x= (−2y) −2x

(x2 − y2)2,

proving that the equation has indeed been made exact, hence µ =(

x3y − xy3)−1

is anintegrating factor of the original equation. Integrating

∫

M ′ dx = log∣

∣x2 − y2∣

∣− log x+ g (y) and

∫

N ′ dy = log∣

∣y2 − x2∣

∣+ h (x) .

So g (y) = 0, h (x) = − log x, and the solution is x2 − y2 = Cx.

Page 293, example 2: In this equation, M = x − ny and N = y. So the integratingfactor is

µ = (Mx+Ny)−1

=1

x2 − nxy + y2.

This leads to the modified equation:

µM dx+ µN dy =x− ny

x2 − nxy + y2dx+

y

x2 − nxy + y2dy = 0.

Letting M ′ = µM and N ′ = µN , we have

∂M ′

∂y=

(

x2 − nxy + y2)

(−n)− (x− ny) (−nx+ 2y)

(x2 − nxy + y2)2 ,

=���−nx2 +���n2xy − ny2 +��nx2 − 2xy −���n2xy + 2ny2

(x2 − nxy + y2)2.

∂N ′

∂x=

−2xy + ny2

(x2 − nxy + y2)2 ,

Hence the modified equation is indeed exact and µ =(

x2 − nxy + y2)−1

is an inte-grating factor for the original equation. Showing that

∫

M ′ dx = f (x, y) + g (y) andN ′ dy = f (x, y) + h (x), with the same f (x, y) for both, is messy and difficult for arbi-trary n, but the reader is encouraged to try it for the easy case when n = 2. Use thesubstitution method in § 117 to show that

(

y −(

n2 + α

)

x

y −(

n2 − α

)

x

)

12α

= Cx

is a solution for n 6= 2, where α2 = 14n

2 − 1.

S-8

Page 294 (§ 120. Rule III) example: From the problem we have M = (1 + xy) yand (1− xy)x. Then

Mx−Ny =��xy + x2y2 −��xy + x2y2 = 2x2y2, hence µ =1

2x2y2.

The modified equation is

(1 + xy) y

2x2y2dx+

(1− xy)x2x2y2

= 0.

Again letting M ′ = µM and N ′ = µN , we find that

∂M ′

∂y=

2x3y − (1 + xy)(

2x2)

4x4y2

= ���2x3y − 2x2 −���2x3y

4x4y2,

∂N ′

∂x=−2xy3 − (1− xy)

(

2y2)

4x2y4

=����−2xy3 − 2y2 +���2xy3

4x2y4,

confirming that the modified equation is exact and that µ = 1/2x2y2 is an integratingfactor.

∫

M ′ dx =

∫

dx

2x2y+

∫

xy

2x2ydx = − 1

2xy+ 1

2 log x+ g (y) ,

∫

N ′ dy =

∫

dy

2xy2−∫

xy

2xy2dy = − 1

2xy− 1

2 log y + h (x) ,

Clearly g (y) = − 12 log y and h (x) = 1

2 log x. So the solution is

− 1

2xy− 1

2 log y +12 log x = C. (a)

Multiplying by 2, then moving everything except the log x term to the right-hand sideand taking the exponential of both sides,

x = Cye1xy .

This solution differs from the book’s answer by the sign of the exponent. Yet if you takethe differential of (a), you can readily return to the original equation. So it appears thatthe book is in error.

Page 294 (§ 120. Rule IV) example 1: In this equation, M = x2 + y2 andN = −2xy. Hence

1

N

(

∂M

∂y− ∂N

∂x

)

= − 1

2xy(2y − (−2y)) = − 2

x.

So the integrating factor isµ = e−2 log x = x−2.

The modified equation is

µM dx+ µN dy =

(

1 +y2

x2

)

dx− 2y

xdy = 0,

which is easily verified as being exact. Integrating∫

µM dx = x− y2

x+ g (y) and

∫

µN dy = −y2

x+ g (x) .

After you infer g and h, the solution is x− y2

x = C, from which the book’s answer followsimmediately.

S-9

Page 294, example 3: In this equation, M = y4+2y and N = xy3 +2y4− 4x. Hence

1

M

(

∂N

∂x− ∂M

∂y

)

=y3 − 4− 4y3 − 2

y4 + 2y=−3y3 − 6

y4 + 2y= −3

y.

So the integrating factor isµ = e−3 log y = y−3.

The modified equation is

µM dx+ µN dy =

(

y +2

y2

)

dx+

(

x+ 2y − 4x

y3

)

dy = 0,

which is easily verified as being exact. Integrating∫

µM dx = xy +2x

y2+ g (y) and

∫

µN dy = xy + y2 +2x

y2+ h (x) .

After you infer g and h, the solution is xy+ 2xy2 + y2 = C, from which the book’s answer

follows when you multiply through by y2.

Page 297 (§ 122. 1st Order Linear Equations), example 1: The book alreadydemonstrates the origin of the integrating factor,

µ =1√

1 + x2

Applying equation (2) of § 122 yields

yµ =

∫

µmdx

1 + x2+ C =

y√1 + x2

=

∫

mdx

(1 + x2)32

+ C.

The remaining integral can be found by substituting tanu = x, which yieldsy√

1 + x2=

mx√1 + x2

+ C.

from which the book’s answer follows.

Page 297, example 2: The symbology used in this problem is likely to be veryconfusing to students of electrical engineering. Virtually all modern text books usethe symbol, i, to represent electrical current rather than C. So a modern text wouldrepresent the equation to be solved here as E = iR+L di

dt . I will use today’s symbologyin the solution of this example.

iR

L+di

dt=E

L.

Hence P = RL and Q = E

L . The integrating factor is

µ = e∫

RLdt = e

RtL

Applying equation (2) of § 122,

ieRtL =

∫

E

Le

RtL dt+B,

ieRtL =

E

Re

RtL +B,

i =E

R+Be−

RtL ,

which is equivalent to the book’s answer.

S-10

Page 297, example 4: The equation is equivalent to

dy

dx+y

x= x2

So P = 1x and Q = x2. The integrating factor is given by µ = e

∫dx/x = x. So according

to equation (2) of § 122

xy =

∫

x3dx+ C,

xy = 14x

4 + C,

y = 14x

3 +C

x,

which agrees with the book’s answer.

Page 298 (§ 122. Bernoulli’s and similar equations), example 1: Lettingv = 1/y, as the book suggests, we have

dv

dt= − 1

y2dy

dx.

Dividing the equation by y2 and multiplying by −1 yields the modified equation,

− 1

y2dy

dx− 1

xy= −1.

Substituting yields dv

dx− v

x= −1.

This equation is first-order linear, with P = −x−1 and Q = −1. Hence the integratingfactor is µ = e−

∫dx/x = x−1. Applying equation (2) of § 122,

v

x= −

∫

dx

x+ C,

v

x= − logx+ C,

v = − logx+ Cx,

1

y= −x log x+ Cx,

1 = −xy log x+ Cxy,

which is equivalent to the book’s answer.

Page 298, example 2: The original text had x sin2 x rather than x sin 2x as the secondterm. Although the book’s misprint is not obvious on first glance, it becomes obviouswhen you begin following the procedure the book gives for solving it. Using the originalversion of the equation in the book, you end up with a nonlinear equation when you makethe substitution – an equation that is not susceptible by any of the methods coveredso far. So my rendering of the text makes the correction I have already indicated. Bytrigonometric identity, the equation becomes

dy

dx+ 2x sin (y) cos (y) = x3 cos2 y.

Dividing by cos2 y, as the book suggests, and then applying more trig identities,

dy

dx

(

1 + tan2 y)

+ 2x tan y = x3.

The book also suggests the substitution, v = tan y, whose derivative is dvdx =

(

1 + tan2 y)

dydx .

The equation becomes dv

dx+ 2xv = x3.

S-11

Hence P = 2x and Q = x3. The integrating factor is µ = e∫2x dx = ex

2

. Applyingequation (2) of § 122 yields

ex2

v =

∫

x3ex2

dx+ C.

To integrate the remaining integral, let u = x2 and du = 2x dx, then integrate by parts.

ex2

v = 12

∫

ueudu+ C,

ex2

v = 12 (u− 1) eu + C,

ex2

v = 12

(

x2 − 1)

ex2

+ C,

ex2

tan y = 12

(

x2 − 1)

ex2

+ C,

which is equivalent to the book’s answer. If you really needed y as a function of x, you

could rearrange this into y = tan−1(

12

(

x2 − 1)

+ Ce−x2)

.

Comments on page 298, example 3: All the steps to solution are shown in the text,arriving at

e−Kx

y= −K

∫

e−Kx

xdx+ C.

Handbook of Mathematical Functions, edited by Milton Abramowitz and Irene Stegun(U.S. Department of Commerce, National Bureau of Standards, Applied MathematicsSeries · 55), defines the function

E1(x) =

∫ ∞

x

e−t

tdt

in chapter 5 on the exponential integral function (by Walter Gautschi and William F.Cahill). According to this definition, example 3 is solved by

e−Kx

y= K E1(Kx) + C.

This is the first example we have encountered so far in which the solution cannot beexpressed in closed form. In this example there is no function that can be expressed asany finite combination of powers, roots, logs, exponentials, trig functions and inverse trigfunctions whose derivative is e−Kx/x. The E1 (x) function can only be approximatedusing power series or other convergent algorithms. In general, we say we have solved adifferential equation if we can find a function solving the equation whose derivative isexpressible in elementary functions, even if the solution itself cannot be.

Page 299 (§ 123. Solution by Differentiation), case i, example 2: Substitutingthe symbol, p, for dy/dx, the equation becomes xyp2−

(

x2 − y2)

p−xy = 0. This factorsinto (xp+ y) (yp− x) = 0. This results in factored differential equation

(

dy

dx+y

x

)(

dy

dx− x

y

)

= 0.

Hence eitherdy

dx= − y

xor

dy

dx=x

y.

Both of these are separable resulting in

dy

y= −dx

xor y dy = x dx.

Integrating yields log y = log x+ C or 12y

2 = 12x

2 + C. The book’s answer follows fromthese.

Page 299, case i, example 3: Substituting p for dy/dx, the equation is p2−7p+12 = 0.This factors into (p− 4) (p− 3) = 0. Back-substitute and solve the two factors sepa-rately, and the book’s answer follows immediately.

S-12

Page 299 case ii, comments: The book is overly brief on explaining this method ofsolution. The steps are to isolate either y or x and then differentiate with respect to theother independent variable. This eliminates the isolated variable. Now replace dy/dxor dx/dy with p and the second derivatives of y or x with the first derivative of p. Ineach of the examples, the resulting first order differential equation for p turns out tobe separable. Solving that equation results in a solution containing an undeterminedconstant, C. Once you have that solution for p, you integrate that function to findthe y or x function that is the solution to the original equation. This integration picksup an additional constant of integration, K. But unlike C, this second constant is notundetermined. In the earlier step where you took the derivative of the equation toeliminate one of the variables, you also lost a part of the equation. The constant, K,is the recovery of that lost part. So the final step of this method is to solve for K. Todo this, substitute your solution function, including its added K, back into the originalequation. You will find that not every value of K will make it work. Solving for thevalue or values of K that does make it work completes your solution.

Page 299, case ii, example 1: The book shows the step of isolating y and taking thederivative of the resulting equation with respect to x. Making the substitution of p fordy/dx

p = 1 +1

2

dpdx√p,

(p− 1)dx =1

2

dp√p,

dx =1

2

dp

p32 − p 1

2

,

which can be integrated by substituting u2 = p and 2u du = dp. The integral is shownin the book as x being a function of p, which then must be manipulated into p as afunction of x.

x =1

2log

(√p− 1√p+ 1

)

+ logC,

e2x = C

(√p− 1√p+ 1

)

,

e2x (√p+ 1) = C (

√p− 1) ,

√p(

e2x − C)

= −C − e2x,√

dy

dx=C + e2x

C − e2x ,

dy

dx=

(

C + e2x)2

(C − e2x)2. (a)

This function can be integrated by substituting v = e2x and dv/(2v) = dx, then decom-posing the resulting integrand into partial fractions.

y =

∫

(

C + e2x)2

(C − e2x)2dx =

2C

C − e2x + x+K. (b)

We solve for K by substituting this expression for y into the original equation. To keepthings tidy, we start by letting y = y0 +K. Then the substituted original equation is

S-13

d

dx(y0 +K) + 2x (y0 +K) = x2 + (y0 +K)

2,

dy

dx+ 2xy0 + 2xK = x2+y20 + 2Ky0 +K2.

Now replace y0 with 2CC−e2x + x and dy/dx with the right-hand side of (a).

(

C + e2x)2

(C − e2x)2+ 2x

(

2C

C − e2x + x

)

+ 2Kx =

x2 +

(

2C

C − e2x + x

)2

+ 2K

(

2C

C − e2x + x

)

+K2.

Carefully multiplying all of this out and taking the cancellations

C2 + 2Ce2x + e4x

(C − e2x)2+����4Cx

C − e2x +��2x2 +���2Kx =

��x2 +

4C2

(C − e2x)2+����4Cx

C − e2x +��x2 +

4KC

C − e2x +���2Kx+K2.

Now multiply what remains by(

C − e2x)2

C2 + 2Ce2x + e4x = 4C2 + 4KC(

C − e2x)

+K2(

C − e2x)2,

C2 + 2Ce2x + e4x = 4C2 + 4KC2 − 4KCe2x +K2C2 − 2K2Ce2x +K2e4x. (c)

Clearly e4x = K2e4x, hence K = 1 or K = −1. Trying these values of K on theremaining terms of (c) quickly reveals that only K = −1 works for the entire equation.Now replace K with −1 in (b)

y =2C

C − e2x + x− 1.

y =2C

C − e2x + x− C − e2xC − e2x ,

y =C + e2x

C − e2x + x,

which is the book’s answer.

Page 299, case ii, example 2: First substitute p for dy/dx and solve for y

xp2 − 2yp+ ax = 0,

x

2

(

p+a

p

)

= y.

Taking the derivative using the product rule (remembering that p is the derivative of y)

1

2

(

p+a

p

)

+x

2

(

1− a

p2

)

dp

dx= p.

Before solving, subtract p from both sides and divide out the common factor

1

2

(

−p+ a

p

)

+x

2

(

1− a

p2

)

dp

dx= 0,

p

2

(

−1 + a

p2

)

+x

2

(

1− a

p2

)

dp

dx= 0,

−p+ xdp

dx= 0,

log p = log x+ logC.

S-14

Hence p = Cx andy =

∫

p dx = C

∫

x dx = 12Cx

2 +K.

We substitute this solution for y back into the original equation and solve for K.

x (Cx)2 − 2

(

12Cx

2 +K)

(Cx) + ax = 0,

C2x3 − C2x3 − 2CKx+ ax = 0,

K =a

2C.

The resulting solution, y = 12

(

Cx2 + a/C)

, does not exactly match the book’s answer,but this solution does work when substituted into the original equation, and the book’sdoes not. Furthermore, page 303 lists a solution for this same equation that is equivalentto the solution shown here (with the solution given on page 303, solve for y and replaceC with 1/C′).

Page 299, case ii, example 3: The book instructs to solve for x and take the derivativewith respect to y.

2xdy

dx= y

(

1−(

dy

dx

)2)

,

x =y

2

(

dx

dy− dy

dx

)

, (d)

x =y

2

(

p− 1

p

)

,

where p is the derivative of x with respect to y. Taking the derivative with respect to yof the above eliminates x

p =1

2

(

p− 1

p

)

+y

2

(

1 +1

p2

)

dp

dy,

1

2

(

p+1

p

)

=y

2

(

1 +1

p2

)

dp

dy,

p = ydp

dy,

log p = log y + C,

p =dx

dy= Cy.

Integrating the last line of the above to get x as a function of y

x =

∫

p dy = C

∫

y dy = 12Cy

2 +K.

Substituting this solution into (d)

12Cy

2 +K =y

2

(

Cy − 1

Cy

)

,

K = − 1

2C.

So the solution is x = 12Cy

2 − 12C . Letting C′ = 1/C

2x =y2

C′ − C′,

2x+ C′ =y2

C′ ,

C′ (2x+ C′) = y2.

which is equivalent to the book’s answer.

S-15

Page 300, case iii, example 2: Both this example and the next succumb readily tothe method of separation of variables (see § 117). Indeed using the method explainedin § 123, case iii to do these seems almost an exercise in masochism, as it involves anorder of magnitude more work. But to illustrate the method, here is example 2 doneboth ways. First by separation of variables.

dy

y + 1y

= dx,

y dy

y2 + 1= dx,

12 log

(

y2 + 1)

= x+ 12 logC,

y2 + 1 = Ce2x.

from which the book’s answer follows immediately. But case iii would have you solvefor y then differentiate. Letting p = dy/dx,

p =y +1

y, (a)

py =y2 + 1,

y2 − py + 1 = 0,

y = 12

(

p±√

p2 − 4)

,

dy

dx= p =

1

2

dp

dx± p dp

dx

2√

p2 − 4.

This last equation is separable. Divide both sides by p, multiply by 2dx to get

2

∫

dx =

∫

dp

p±∫

dp√

p2 − 1,

2x+ logC = log p± log(

p+√

p2 − 4)

.

I will use the difference-solution of the ± and leave it up to the reader to try the sum-solution to confirm that it also arrives at the same final solution.

Ce2x =p

p+√

p2 − 4,

(

p+√

p2 − 4)

Ce2x = p,

Ce2x√

p2 − 4 =(

1− Ce2x)

p,

C2e4x(

��p2 − 4

)

=(

1− 2Ce2x +���C2e4x

)

p2,

−4C2e4x =(

1− 2Ce2x)

p2, (b)

2Ce2x√2Ce2x − 1

= p =dy

dx. (c)

Integrate (c) by substituting u = 2Ce2x and du2u = dx to get

√2Ce2x − 1 + K = y.

Letting K = 0 (which you can confirm is correct by substituting y =√2Ce2x − 1 +K

into the original equation), squaring both sides, and replacing 2C with C′ yields ananswer equivalent to the book’s. You could also have substituted (a) into (b), solvedfor y, and obtained the same result, but doing that algebra is more difficult than takingintegral of (c).

S-16

Page 300 (§ Clairaut’s equation), example 1: The equation given is already inClairaut form. With p used to represent dy/dx, take the derivative of both sides:

�p = �p+ xdp

dx+ 2p

dp

dx,

which factors into(x+ 2p)

dx

dp= 0.

Either x = −2p or dp/dx = 0. If the former, then substitute p = −x/2 into the originalequation.

y = −x2

2+x2

4= −x

2

4,

from which one of the book’s answers follows. From dp/dx = 0 we have p = C andy = Cx + K. Substituting that into the original equation yields K = C2. Hence theother solution is y = Cx+ C2.

Page 300 (§ 124), example 2: To put this equation into Clairaut form, divide byp− 1 and then add px to both sides.

y = px+p

p− 1. (a)

Taking the derivative

�p = �p+ xdp

dx− 1

(p− 1)2

dp

dx,

0 =

(

x− 1

(p− 1)2

)

dp

dx.

If the first factor is zero, then 1√x+ 1 = p.

Substitute that into (a)

y =

(

1√x+ 1

)

x+

1√x+ 1

1√x

=√x+ x+ 1 +

√x = x+ 2

√x+ 1 =

(√x+ 1

)2.

Hence√y = ± (

√x+ 1). Regardless of which sign you choose from the ±, it still differs

from the book’s answer by a sign. The solution, y = x+2√x+1 works when substituted

into the original equation. The book’s answer does not. The solution that results fromthe factor, dy/dx = 0, is again y = Cx+K. Substituting that into (a) yields K = C

C−1 ,from which the book’s other solution follows.

Page 302 (§ 125. Singular solutions), Comments on p-discriminants and C-discriminants: My opinion is that Mellor does not expand enough on the relationshipbetween p-discriminants, C-discriminant, and singular solutions. For one thing, he sug-gests that these discriminants exist only when the differential equation is a quadratic inp or when the solution is a quadratic in C. Chapter A2, section 10 of Ordinary Differen-tial Equations and their Solutions by George M. Murphy (van Nostrand, 1960) providesa much broader method for finding discriminants. A p-discriminant occurs whenever xand y in the differential equation can conspire to cause p to have a double root. You findsuch double roots in the general case by taking the partial derivative of the equationwith respect to p, solving the result for p, and substituting that expression for p backinto the original equation. Mellor uses the example, xp2 − yp + a = 0, and, using thequadratic formula, arrives at the p-discriminant of y2 = 4ax.

S-17

Solving it Murphy’s way, we have

∂

∂p(xp2 − yp+ a) = 0,

2xp− y = 0,

p =y

2x.

Substituting p into the original equation,

x( y

2x

)2

− y2

2x+ a = 0,

− y2

4x+ a = 0,

y2 = 4ax.

Observe that Murphy’s method exploits the fact that a function has a double rootonly when both the function itself and its first derivative are simultaneously zero. Thesame method will also find the C-discriminant when applied to this equation’s generalsolution, xC2 − yC + a = 0.

The existence of a p-discriminant or a C-discriminant is a necessary but not suffi-cient condition for the existence of a singular solution. Both p-discriminants and C-discriminants of more complicated equations (an equation that is cubic in p for example)may be factorable, in which case there is a possibility for multiple singular solutions –that is each factor may be a singular solution. Murphy states without proof that if afactor of a p-discriminant is a singular solution, then there will be an identical factorof the C-discriminant, and vice versa. Mellor seems to suggest the same, but neverstates it outright. Both books are clear, however, that when a discriminant (or factorof a discriminant) is found, it must be tested by substituting it back into the originalequation to see whether or not it is indeed a singular solution.

Graph showing lines, y = Cx+C2,enveloping the parabola, y = − 1

4x2

Envelopment of singular solutions: As anexample, the equation, px − y + p2 = 0, has asits general solution, y = Cx + C2. This solutionhas a C-discriminant of y = − 1

4x2. The figure

shows how the family of curves specified by thegeneral solution forms an envelope for the par-ticular solution (general solution is shown withC = −0.2,−0.4, . . . − 2.0). To find an envelopepoint, we find the general solution for C and alsofor C + δC, then we find where those two linesintersect:

y = (C + δC)x + (C + δC)2

−y = −Cx − C2

0 = δCx + 2C δC + δC2

Dividing out δC from the difference gives0 = x + 2C + δC. Taking the limit as δC goes to zero gives x = −2C. Substitutingfor x into the general solution gives y = −C2. So any point of the form,

(

−2C,−C2)

,is a point on the envelope. Observe that the relationship between the expressions for xand y is precisely y = − 1

4x2, that is, the singular solution to the original equation. You

could also have solved the equation, x = −2C for C and substituted into the generalsolution for C to obtain the same result. Observe also that solving for the intersectionof neighboring lines in the family as we have done here is no different from Murphy’smethod for finding the C-discriminant. By subtracting the neighbors from each other,

S-18

dividing out δC, then taking the limit as δC goes to zero, we have effectively taken thepartial derivative of the general solution with respect to C. After that we substitutedthat into the general solution, which is the same as the last step of Murphy’s method.This shows, at least for C-discriminants, how finding the conditions for C’s double rootis identical to finding points on the curve that the general solution envelopes.

Page 303 (Singular solutions), example 2: You arrive at the general solution bysolving the differential equation for p, and integrating both sides. Use the substitution,u = x2 and du = 2x dx to integrate. Apply the constant of integration, C, to they-side of the result, then square both sides. The book provides details for finding thep-discriminant and C-discriminant using the equal quadratic roots method. ApplyingMurphy’s method to finding the p-discriminant, you have 4xp2− (3x− a)2 = 0. Takingthe partial with respect to p gives 8xp = 0. Hence either x = 0 or p = 0. Note thatthis p-discriminant is factorable. Substituting that into the original equation yields(3x− a)2 = 0. So p-discriminants are x = 0 and x = 1

3a. Using Murphy’s methodto find the C discriminant, we take the partial of the general solution with respect toC and get 2(x + C) = 0, hence C = −x. Substituting into the general solution gives

0 = x (x− a)2. Again the discriminant is factorable. x = 0 is the only common factorof both the p-discriminant and C-discriminant. It satisfies the original equation in thatif you divide the original equation by p2 (3x− a)2 and allow 1/p = dx/dy = 0 for theline defined by x = 0.

Page 303 (Singular solutions), example 4: The general solution emerges as follows:

y2(

p2 + 1)

= a2,

y =±a

√

p2 + 1.

Take the derivative of both sides to eliminate y:

p =±pa

(p2 + 1)32

dp

dx,

dx =±a

(p2 + 1)32

dp,

integrating yields x− C =±ap

√

p2 + 1,

a2p2 = (x− C)2(

p2 + 1)

,

0 =(

u2 − a2)

p2 + u2,

where u = x− C and du = dx. Hence

p2 = − u2

u2 − a2 ,

p = ± u√a2 − u2

.

Integrating to establish y,

y = ±√

a2 − u2 +K = ±√

a2 − (x− C)2 +K.

Substituting back into the original equation yields K = 0. Hence y2 = a2 − (x− C)2,from which the book’s general solution follows.

Rearrange the original equation into y2p2 + y2 − a2 = 0. Using the quadratic method,the p-discriminant is y2(y2 − a2) = 0. This discriminant factors into y = 0 or y = ±a.Rearrange the general solution into C2 − 2xC + x2 + y2 − a2 = 0. Using the quadratic

S-19

method, the C-discriminant is 4x2 = 4x2 + 4y2 − 4a2, or equivalently, y = ±a, which isin common with a p-discriminant factor. When y = ±a, then p = dy/dx = 0. So thisfactor works as a singular solution.

Page 304 (§ 126. Trajectories), example 2: The transformation between Cartesianand polar coordinates is x = r cos θ and y = r sin θ. Taking the derivatives of these withrespect to r,

dx

dr= cos θ − sin θ

(

rdθ

dr

)

anddy

dr= sin θ + cos θ

(

rdθ

dr

)

.

By the chain rule we have

dy

dx=

(

dydr

)

(

dxdr

) =sin θ + cos θ

(

r dθdr)

cos θ − sin θ(

r dθdr) . (a)

In Cartesian coordinates we find orthogonal trajectories by replacing dy/dx with−dx/dy.Hence

−dxdy

=− cos θ + sin θ

(

r dθdr)

sin θ + cos θ(

r dθdr) . (b)

Replacing r dθ/dr with −dr/ (r dθ), the right-hand side of (a) becomes

sin θ − cos θ(

1rdrdθ

)

cos θ + sin θ(

1rdrdθ

) .

Graph showing orthogonal trajec-tories to y2 = 4ax

Multiplying top and bottom by r dθ/dr yields thesame expression as we got in (b) for −dx/dy, whichcompletes the proof.

Page 304 (§ 126. Trajectories), example 3:The curve family is defined by y2 = 4ax. Solvingthis for a gives

y2

4x= a.

Taking the derivative using the quotient rule,

8xy dydx − 4y2

16x2= 0.

Replacing dy/dx with −dx/dy and multiplyingthrough by 4x2,

−2xy dxdy− y2 = 0.

An equivalent differential equation is2x dx + y dy = 0. This is immediately integrableto x2 + 1

2y2 = C2, which is equivalent the book’s

answer.

Page 304 (§ 126. Trajectories), example 4:Place two point charges of q and −q on the hor-izontal axis symmetrically about the origin at adistance of L apart, with the positive charge onthe right. Allowing Lq to remain constant, we let L go to zero and q go to infinity tocreate an idealized dipole. Then the potential function, V , in polar coordinates is givenby

V =kLq cos θ

r2,

where k is Coulomb’s constant. You could take the derivative with respect to r of thisfunction and apply the transformation indicated in example 3 above, and you wouldarrive at the correct trajectory. You are encouraged to try that. You would find that,in the end, you would have to take the exponential of a function expressed in logs. So

S-20

here I will pass into logs before taking the derivative and solve it that way. The aboveequation becomes

logV = log (kLq) + log (cos θ)− 2 log r.Taking the derivative with respect to r,

0 = − tan θdθ

dr− 2

r,

0 = − tan θ

(

rdθ

dr

)

− 2,

Equipotentials and trajectories ofan ideal dipole.

Now make the transformation from example 3:

0 = tan θ

(

1

r

dr

dθ

)

− 2,

0 =dr

r− 2 cot θ dθ,

0 = 2 cot θ dθ − dr

r.

Integrating the above,

logC = 2 log (sin θ)− log r,

C =sin2 θ

r.

Page 307 (§ 128. Linear equations, when a particular solution is known),derivation of order reduction: We begin with the assumption that v is a knownsolution to the equation,

d2y

dx2+ P

dy

dx+Qy = 0.

where P and Q are functions of x. We let y = uv. Then, by the product rule,

dy

dx= u

dv

dx+ v

du

dxand

d2y

dx2= u

d2v

dx2+ 2

du

dx

dv

dx+ v

d2u

dx2.

Substituting

ud2v

dx2+ 2

du

dx

dv

dx+ v

d2u

dx2+ P

(

udv

dx+ v

du

dx

)

+Quv = 0. (a)

But because v is assumed to solve the original equation,

d2v

dx2+ P

dv

dx+Qv = 0 = u

d2v

dx2+ Pu

dv

dx+Quv. (b)

Removing from (a) the terms that occur to the right of the second equal in (b), whatremains is

vd2u

dx2+

(

2dv

dx+ Pv

)

du

dx= 0.

Page 307 (§ 128. Linear equations), example 1: y = eax is assumed to be asolution to d2y

dx2= a2y, or equivalently,

d2y

dx2− a2y = 0.

Hence P = 0, Q = −a2 and v = eax. Substituting these into the formula on page 307,

logdu

dx+ 2 log (eax) = C,

du

dx= Ce−2ax,

S-21

u =C

−2ae−2ax + C2,

y = uv = − C2ae−ax + C2e

ax.

Letting C1 = − C2a yields the book’s answer.

Page 307 (§ 128. Linear equations), example 2: y = x is assumed to be a solutionto(

1− x2) d2y

dx2− x dy

dx+ y = 0, or equivalently,

d2y

dx2− x

1− x2dy

dx+

y

1− x2 = 0.

Hence P = −x1−x2 , Q = 1

1−x2 , and v = x. We also have∫

P dx = 12 log

(

1− x2)

. Usingthe formula on page 307,

logdu

dx+ 2 logx+ 1

2 log(

1− x2)

= logC,

du

dx=

C

x2√1− x2

,

u = −C√1− x2x

+ C2.

Because v = x, we have y = uv = ux = −C√1− x2 + C2x. Letting C1 = −C gives

book’s answer.

Page 308 (§ 129. Linear equations with constant coefficients), example 2:The auxillary equation is D2 − m2 = (D +m) (D −m) = 0, which has roots at ±m.The book’s answer follows.

Page 308 (§ 129. Linear equations with constant coefficients), example 3:The auxillary equation is D2 + 4D + 3 = (D + 3) (D + 1) = 0, which has roots at −3and −1. The book’s answer is missing minus signs in the exponents.

Page 308 (§ 129. Linear equations with constant coefficients), case 2 com-ments: Rather than using the Maclaurin series to establish the general solution forthe case of a double root, you can also use the formula on page 307 that establishes ageneral solution if a particular solution is known. The general case of a second orderlinear equation with constant coefficients that has a double root is

d2y

dx2− 2a

dy

dx+ a2y = 0,

whose auxillary equation has its double root at a. Because a is a root, we know thaty = eax is a particular solution. So P = −2a, Q = a2 and v = eax. We also have∫

P dx = −2ax. Applying the formula

logdu

dx+ 2 log (eax)− 2ax = logC1,

logdu

dx+��2ax−��2ax = logC1,

logdu

dx= logC1,

du

dx= C1,

u =C1x+ C2,

y = uv = (C1x+ C2) eax.

S-22

Page 309 (§ 129. linear equations with constant coefficients, case 3, nonrealroots) example 1: All that’s needed is to prove that eαx = coshαx + sinhαx. Bydefinition

coshαx =eαx + e−αx

2and sinhαx =

eαx − e−αx

2,

hencecoshαx+ sinhαx =

eαx +���e−αx + eαx −���e−αx

2= eαx.

Page 309 (§ 129. linear equations with constant coefficients, case 3, nonreal

roots) example 2: The auxillary equation is D2 +D+1 = 0. Its roots are − 12 ± ι

√32 .

This means that y = C1e(−1+ι

√3)x/2 + C2e

(−1−ι√3)x/2 is the general solution. To turn

this into a real-valued function, let C1 = 12 (A− ιB) and C2 = 1

2 (A+ ιB) be complexconjugates of each other. Then

y = 12 (A− ιB) e(−1+ι

√3)x/2 + 1

2 (A+ ιB) e(−1−ι√3)x/2,

y = 12 (A− ιB) e−x/2eι

√3x/2 + 1

2 (A+ ιB) e−x/2e−ι√3x/2, and by Euler’s formula,

y =

(

A

2− ιB

2

)

e−x/2

(

cos

√3x

2+ ι sin

√3x

2

)

+

(

A

2+ ι

B

2

)

e−x/2

(

cos

√3x

2−ι sin

√3x

2

)

,

y =A

2e−x/2 cos

√3x

2+��������ιA

2e−x/2 sin

√3x

2−��������ιB

2e−x/2 cos

√3x

2+B

2e−x/2 sin

√3x

2+

A

2e−x/2 cos

√3x

2−��������ιA

2e−x/2 sin

√3x

2+��������ιB

2e−x/2 cos

√3x

2+B

2e−x/2 sin

√3x

2,

y =Ae−x/2 cos

√3x

2+Be−x/2 sin

√3x

2.

Page 309 (§ 129. linear equations with constant coefficients, case 3, nonrealroots) example 4: The auxillary equation factors into (D − 1)

(

D2 + 1)

= 0. Thishas roots at 1 and ±ι. The book’s answer follow immediately from these roots.

Page 311 (§ 130. Finding particular solutions) case 1, example 1: The illus-tration has you find the particular integral, p (x), by integrating (see equation (4) onpage 311)

p (x) = e3x∫

e−3xe4x dx − e2x∫

e−2xe4x dx = e3x∫

exdx− e2x∫

e2x dx.

The result is p (x) = e4x− 12e

4x, which, when simplified and combined with the comple-mentary solution, yields the book’s answer.

Page 311 (§ 130. Finding particular solutions) case 1, example 2: Auxillaryequation factors into (D − 3) (D − 1) = 0. The decomposition into partial fractions is

1

(D − 3) (D − 1)=

12

D − 3−

12

D − 1.

Integrating using equation (4) on page 311,

p (x) = 12e

3x

∫

2e−3xe3xdx− 12e

x

∫

2e−xe3xdx = e3x∫

dx− ex∫

e2xdx.

This yields p (x) = xe3x − 12e

3x. The second summand is linearly dependent uponcomplementary solution, so we can eliminate it from the particular solution, leavingp (x) = xe3x. Combining that with the complementary solution, C1e

x + C2e3x, yields

the book’s answer.

Page 311 (§ 130. Finding particular solutions) case 2, example 2: In thiscase, R = ex. In problems of finding particular solutions, the expression in the R-position is commonly referred to as the forcing function. When the forcing function is

S-23

an exponential that is linearly independent of the complementary solution, there is aneasy way to find the particular solution. When these conditions exist, the particularsolution must be a constant multiple of the forcing function. So we let p (x) = λR. Inthis case that means p (x) = λex. We then substitute that into the original equationand solve for λ. In this case, where the original equation is

(

D2 − 4D + 4)

y = ex,

λex −���4λex +���4λex = ex.

Hence λ = 1 and p (x) = ex. So the general solution is y = C1e2x + C2xe

2x + ex.

This method also works when the forcing function is the sum of exponentials, providedthat none of the exponentials in that sum are linearly dependent upon the complemen-tary solution. So, for example, if we were to solve

(

D2 − 4D + 4)

y = e3x − 2e−3x, thenp(x) = λ1e

3x + λ2e−3x. Substituting into the original equation,

9λ1e3x + 9λ2e

−3x − 12λ1e3x + 12λ2e

−3x + 4λ1e3x + 4λ2e

−3x = e3x − 2e−3x.

By classifying terms, this separates into

9λ1 − 12λ1 + 4λ1 = λ1 = 1 and 9λ2 + 12λ2 + 4λ2 = 25λ2 = −2.Hence λ1 = 1 and λ2 = − 2

25 and p (x) = e3x − 225e

−3x.

The procedures shown above for finding particular solutions for both R = ex andR = e3x − 2e−3x are examples of the method of undetermined coefficients.

The principle of superposition: In the example above, we had a forcing functionof R = R1 + R2, where R1 = e3x and R2 = e−3x. If you solve the particular integralfor just R1 by itself to get p1 (x) and for just R2 by itself to get p2 (x), you will findthat p (x) = p1 (x) + p2 (x) is the particular integral we got for R = R1 + R2. Thisis the principle of superposition. It asserts that whenever the forcing function is asum of functions, you can find the particular integral for each summand separately,and the particular integral for the sum will be the sum of those particular integrals.In addition, for any constant scalar multiplier, α, the particular integral for a forcingfunction, αR, will be αp (x), where p (x) is the particular integral for R by itself. Theprinciple of superposition of forcing functions and the assertion on scalar multiples offorcing functions are both generally true for all linear differential equations.

Page 312 (§ 130. Finding particular solutions) case 3, example 1: This exampletoo can be done using the method of undetermined coefficients. We observe that if theforcing function, R, is a polynomial of degree n, then the particular integral must alsobe a polynomial of degree n. In the example, n = 2. So we let p (x) = λ2x

2 + λ1x+ λ0and substitute into into the original equation.(

D2 − 4D + 4) (

λ2x2 + λ1x+ λ0

)

= 2λ2 − 8λ2x+ 4λ2x2 − 4λ1 + 4λ1x+ 4λ0 = x2.

Separating terms according to powers of x,

4λ2x2 = x2,

− 8λ2x + 4λ1x = 0,2λ2 − 4λ1 + 4λ0 = 0.

Dividing out the powers of x from the first two lines and solving the linear system yieldsλ0 = 3

8 , λ1 = 12 , and λ2 = 1

4 , which is the same as the book’s answer.

Page 312 (§ 130. Finding particular solutions) case 3, example 2: This onecan be done by inspection using the method of undetermined coefficients. The forcingfunction, R, is the first degree polynomial, 2 + 5x. Since the second derivative of anyfirst degree polynomial is zero, then if p (x) is the particular integral, substituting it intothe original equation gives −p (x) = 2 + 5x, or equivalently, p (x) = −2− 5x. Summingthat particular solution with the complementary solution yields the book’s answer.

S-24

Were you to do it by the method in the book, you would first resolve the reciprocaloperator into partial fractions.

1

D2 − 1=

1

(D + 1) (D − 1)=

1

2

1

D − 1− 1

2

1

D + 1.

Hence the integral problem to be solved is

p (x) = 12e

x

∫

(2 + 5x) e−xdx − 12e

−x

∫

(2 + 5x) exdx,

which, after your expending a lot more work than you would using the first method,yields the same p (x).

Page 312 (§ 130. Particular solutions) case 4, clarification of notation: Thefirst line of (5) on page 312 would be clearer as

Dn (eaxX) = eax (D + a)n(X) ,

so as to show the extent of application of the differential operators. The final result of(5) would be clearer as

D−n (eaxX) = eax (D + a)−n

(X) .

This is simply assertion that the first formula works even when n is negative.

Page 312 (§ 130. Particular solutions) case 4, example 2: By (5) on page 312,

(D − 1)−1 (ex log x) = eax (D + 1− 1)−1 (log x) = eaxD−1 (log x) .

The D−1 operator is simply the taking of the antiderivative. The antiderivative of log xis x (log (x) − 1) = x log (x/e). The book’s answer follows from this.

Page 313 (§ 130. Particular solutions) case 4, example 1: By (5) on page 312,

(D + 1)−3 (

e−x)

= e−x (D − 1 + 1)−3 · 1 = e−xD−3 · 1.

The D−3 operator tells us to take the antiderivative of 1 three times. The first timegives x, the second, 1

2x2, and the third, 1

6x3. The book’s answer follows from that.

Page 313 (§ 130. Particular solutions) case 4, example 2: First factor theoperator:

(

D3 − 1)

= (D − 1)(

D2 +D + 1)

. By (5) on page 312,

(D − 1)−1 (

D2 +D + 1)−1

xex = exD−1(

D2 + 3D + 3)−1

x.

Applying D−1 to x gives 12x

2. We then apply(

D2 + 3D + 3)−1

to 12x

2. If p (x) is thefunction that solves this, then

(

D2 + 3D + 3)

(p) = 12x

2. (a)

We can apply the method of undetermined coefficients to this. Let p = λ2x2+λ1x+λ0.

Applying the differential operator in (a) to p,

2λ2 + 6λ2x+ 3λ1 + 3λ2x2 + 3λ1x+ λ0 = 1

2x2.

Sorting terms by powers of x gives the linear system,

3λ2x2 = 1

2x2,

6λ2x + 3λ1x = 0,2λ2 + 3λ1 + 3λ0 = 0,

yielding λ2 = 16 , λ1 = − 1

3 , and λ0 = 29 . Hence p (x) =

(

16x

2 − 13x+ 2

9

)

ex. The book’sanswer is missing the constant term inside the parentheses. You are invited, though, todemonstrate for yourself that

d3p

dx3=(

16x

2 + 23x+ 2

9

)

ex,

which means that(

D3 − 1)

p = xex, as the problem requires. The book’s answer doesnot, however, solve this equation.

S-25

Page 313 (§ 130. Particular solutions) case 5, example 1: This one factors

into p (x) =(

D2 + 1)−1

(D + 1)−1

sin 2x. Since n = 2, we have −n2 + 1 = −3.So according to (9) on page 313, we have p (x) = − 1

3 (D + 1)−1

sin 2x, or equiva-lently, (D + 1) p (x) = − 1

3 sin 2x. When you multiply both sides by (D − 1), youget

(

D2 − 1)

p (x) = − 13 (D − 1) sin 2x. On the left we replace D2 with −n2 to get

−5p (x) = − 13 (D − 1) sin 2x. Now apply the remaining differential operator to sin 2x

and divide both sides by −5 to get the book’s answer.

Page 314 (§ 130. Particular solutions) case 5, example 2: In the D-form, thisequation is

(

D2 − k2)

y = cosmx. Clearly the roots for the formation of the comple-mentary solution are ±k. We set up

p (x) =(

D2 − k2)−1

cosmx.

Here n = m, so we replace D2 with −m2, so according to (10) on page 313

p (x) =1

−m2 − k2 cosmx.

The book’s answer follows from this.

Page 314 (§ 130. Particular solutions) case 5, example 3: Setting this up inD-form, (

D2 +mD + n2)

y = a sinnt.So for the particular solution, p (x), we have

p (x) =(

D2 +mD + n2)−1

a sinnt.

Replacing D2 with −n2 (in accordance with (9) on page 313) results in a cancellationinside the parentheses to give

p (x) = 1mD

−1a sinnt.

Again the D−1 operator is simply taking the antiderivative. Taking the antiderivativeof sinnt gives − 1

n cosnt. The book’s answer follows from that and from the giveninformation that the roots for forming the complementary solution are α and β.

In many engineering books, this equation is given in the form of

d2y

dt2+ 2ζω

dy

dt+ ω2 = a sinωt.

Here ω is known as the resonant frequency of the system and ζ is known as the damping See pageS-61 forfurtherdevelop-ment ofresonantsystems.

factor. The reciprocal of 2ζ is known as the Q-factor. You are encouraged to solve thisproblem in this form as well. What happens to the roots of the auxillary equation when0 < ζ < 1? How does that affect the complementary solution? As ζ approaches zero,how does this affect the particular solution? This equation represents what engineerscall a resonant system.

Page 314 (§ 130. Particular solutions) case 5, example 1: That − 12x cosx

results from(

D2 + 1)−1

sinx can be proved in the same way as is shown on page 314for the identical operator on cosx. But with a little a priori knowledge, we can applythe method of undetermined coefficients to this too. The a priori knowledge is that if(

D2 + n2)

is a factor of the operator, and a sinnx+b cosnx is the forcing function, thenthe particular integral will be in the form of p (x) = λsx sinnx+λcx cosnx. In this case,a = 1, b = 0 and n = 1.

(

D2 + 1)

(λsx sinx+ λcx cosx) = sinx.

Hence

λs (2 cosx−���x sinx+���x sinx) + λc (−2 sinx−����x cos x+����x cosx) = sinx.

Clearly λs = 0 and λc = − 12 . The book’s answer follows from that.

S-26

Similarly you can find that for arbitrary n, where(

D2 + n2)

is a factor and sinnx isthe forcing function,

λs(

2n cosnx−(((((n2x sinnx+(((((

n2x sinnx)

+λc

(

−2n sinx−�����n2x cosx+�����

n2x cosx)

= sinnx.

In this case, λs = 0 and λc = − 12n .

Page 314 (§ 130. Particular solutions) case 5, example 4: In the D-form, this is(

D3 − 1)

y = x sinx, which factors into (D − 1)(

D2 +D + 1)

y = x sinx. The roots of

the auxillary equation are 1 and − 12±ι12

√3. These roots account for the complementary

solution in accordance with case 3 on page 309.

The book offers no satisfactory strategy for solving the particular integral on this oneuntil case 6, so this may be regarded as a challenge problem. Here are two approachesyou might take. The first is to observe that according to Euler’s formula

p (x) =(

D3 − 1)−1

x sinx =(

D3 − 3)−1

x

(

eιx − e−ιx

2ι

)

,

where ι =√−1. By the principle of superposition

p (x) =

{

(

D3 − 1)−1

xeιx

2ι

}

−{

(

D3 − 1)−1

xe−ιx

2ι

}

.

By case 4 on page 312, this is the same as

p (x) ={

eιx(

D3 + ι3D2 − 3D − ι− 1)−1 x

2ι

}

−{

e−ιx(

D3 − ι3D2 − 3D + ι− 1)−1 x

2ι

}

.

So if p (x) = p1 (x)− p2 (x) where

p1 (x) =eιx(

D3 + ι3D2 − 3D − ι− 1)−1 x

2ιand

p2 (x) =e−ιx

(

D3 − ι3D2 − 3D + ι− 1)−1 x

2ι,

then we can solve for p1 and p2 separately using the method of undetermined coefficients,and then recombine the solutions to form p. If p1 = eιx (λ0 + λ1x), then we need onlyconcern ourselves with the (−3D− ι− 1) part of the differential operator, as higherderivatives of (λ0 + λ1x) are zero. So for p1 we must solve

(−3D− ι− 1) (λ0 + λ1x) =x

2ιfor λ0 and λ1. Expanding this,

(−ι− 1)λ1x = x2ι ,

−3λ1 + (−ι− 1)λ0 = 0.

Hence λ1 = 14 (1 + ι) and λ0 = − 3

4 , from which we can write p1 (x). Using the samemethodology we also solve

(−3D + ι− 1) (λ2 + λ3x) =x

2ι.

The result is λ3 = 14 (−1 + ι) and λ2 = 3

4 . This allows us to write p2 (x) as wellas p1 (x).

p (x) = p1 (x)− p2 (x) = 14e

ιx {−3 + (1 + ι)x} − 14e

−ιx {3 + (−1 + ι)x} ,p (x) = − 3

4

(

eιx + e−ιx)

+ 14

(

eιx + e−ιx)

x+ 14

(

eιx − e−ιx)

ιx,

and by Euler’s formula

p (x) = − 32 cosx+ 1

2x cos x− 12x sinx,

which is the same as the book’s answer.

S-27

The second way to solve this is to use the method of undetermined coefficients frombeginning to end. When the forcing function is x times sine or cosine of x (and D2 + 1is not a factor of the operator), then the particular integral will be in the form of

p (x) = λc0 cosx+ λc1x cosx+ λs0 sinx+ λs1x sinx. (a)

By applying Leibniz’s rule we find that

d3p

dx3= λc0 sinx+ λc1x sinx− 3λc1 cosx− λs0 cosx− λs1x cos x− 3λs1 sinx.

When you take the difference, d3pdx3 − p = x sinx, then pick the result apart into separate

equations according to whether the coefficient is multiplied by cosx or x cosx or sinxor x sinx, it gives the linear system,

λc1 − λs1 = 1,− λc1 − λs1 = 0,

− λc0 − 3λc1 − λs0 = 0,λc0 − λs0 − 3λs1 = 0.

Solving the above yields λc0 = − 32 , λc1 = 1

2 , λs0 = 0, and λs1 = − 12 . Using these

coefficients in (a) gives the book’s answer.

Page 315 (§ 130. Particular solutions) case 6, example 2: If you tried themethod given on the top half of page 315 directly, you probably ran into difficulty. Thisis because

(

D2 + 1)

is a factor of(

D4 − 1)

, making sinx a part of the complementarysolution. A way to do this is to solve it in two phases.

(

D4 − 1)−1

x sinx =(

D2 + 1)−1 (

D2 − 1)−1

x sinx.

Applying (11) on page 315 to the(

D2 − 1)

factor,(

D2 + 1)−1 (

D2 − 1)−1

x sinx =(

D2 + 1)−1

{

x−(

D2 − 1)−1

(2D)}

(

D2 − 1)−1

sinx.

By (9) on page 313 this is(

D2 + 1)−1 (

D2 − 1)−1

x sinx = − 12

(

D2 + 1)−1

{

x−(

D2 − 1)−1

(2D)}

sinx,

= − 12

(

D2 + 1)−1

{

x sinx− 2(

D2 − 1)−1

cosx}

.

By (10) on page 313(

D2 + 1)−1 (

D2 − 1)−1

x sinx = − 12

(

D2 + 1)−1 {x sin x+ cosx} .

By superposition(

D2 + 1)−1 (

D2 − 1)−1

x sinx = − 12

{

(

D2 + 1)−1

x sinx}

− 12

{

(

D2 + 1)−1

cosx}

.

By the development on page 314(

D2 + 1)−1 (

D2 − 1)−1

x sinx = − 12

{

(

D2 + 1)−1

x sinx}

− 14x sinx. (a)

For the second phase, it remains to find(

D2 + 1)−1

x sinx. Unfortunately this is theexceptional case of (11) on page 315 that does not immediately yield to this method. You

end up with an expression in which you must, once again, evaluate(

D2 + 1)−1

x sinx.To continue you need to apply a trick. Here then is that trick.

(

D2 + 1)−1

x sin x ={

x−(

D2 + 1)−1

(2D)}

(

D2 + 1)−1

sinx,

={

x−(

D2 + 1)−1

(2D)}

(

− 12

)

x cosx,

= − 12x

2 cosx+(

D2 + 1)−1

D (x cosx) ,

= − 12x

2 cosx+(

D2 + 1)−1

(−x sinx+ cosx) .

S-28

Now add(

D2 + 1)−1

x sinx to both sides of the above:

2(

D2 + 1)−1

x sinx = − 12x

2 cosx+(

D2 + 1)−1

cosx,

= − 12x

2 cosx+ 12x sinx,

from which we see that(

D2 + 1)−1

x sin x = − 14x

2 cosx+ 14x sinx.

We can also use the method of undetermined coefficients to solve this. In this case,rather than set it up formally, I will use the inspection method (that is making educated

guesses and refining them). Because − 12x cosx results when you apply

(

D2 + 1)−1

tosinx, we should suspect that when we apply the same to x sinx, one of the terms of theresult ought to be a multiple of x2 cosx. So we apply the operator to that expression.

(

D2 + 1)

x2 cosx =�����−x2 cosx− 4x sinx+ 2 cosx+����x2 cosx.

We need to add another term that will cancel the 2 cosx term in the above. Based uponexperience, we suspect that term will be a multiple of x sinx.

(

D2 + 1) (

x2 cosx+ λx sin x)

=

�����−x2 cosx− 4x sinx+ 2 cosx−����λx sinx+ 2λ cosx+����x2 cosx+����λx sin x.

Clearly if λ = −1, then(

D2 + 1) (

x2 cosx− x sinx)

= −4x sinx. From this we concludeagain that (

D2 + 1)−1

x sinx = − 14x

2 cosx+ 14x sinx.

We substitute this into (a) on the previous page to get(

D2 + 1)−1 (

D2 − 1)−1

x sinx = 18x

2 cosx− 18x sinx− 1

4x sinx.

Gathering terms in the above yields the book’s answer.

Page 315 (§ 130. Particular solutions) case 6, example 3: To solve thisone it is useful first to evaluate

(

D2 − 1)

ex sinx and(

D2 − 1)

ex cosx. This is doneeasily using the method of undetermined coefficients. In each case we assume thatp (x) = λce

x cosx+ λsex sinx. We find

(

D2 − 1)

p (x) = (2λs − λc) ex cosx+ (−2λc − λs) ex sinx.

Separating terms according multiples of ex cosx and ex sinx and putting it all intomatrix form we have

λc λs(

−1 2−2 −1

)(

10

)

row for ex cosxrow for ex sinx

}

for finding(

D2 − 1)−1

ex cosx.

By Cramer’s rule, λc = − 15 and λs =

25 . Hence

(

D2 − 1)−1

ex cosx = − 15e

x cosx+ 25e

x sinx. (a)

Also

λc λs(

−1 2−2 −1

)(

01

)

row for ex cosxrow for ex sinx

}

for finding(

D2 − 1)−1

ex sinx.

By Cramer’s rule, λc = − 25 and λs = − 1

5 . Hence(

D2 − 1)−1

ex sinx = − 25e

x cosx− 15e

x sinx. (b)

The problem is to solve(

D2 − 1)

y = xex sinx. Clearly the complementary solution isyc = C1e

x + C2e−x. To find the particular solution, p (x), we apply (11) on page 315.

p (x) =(

D2 − 1)−1

ex sinx ={

x−(

D2 − 1)−1

(2D)}

(

D2 − 1)−1

ex sinx.

S-29

By (b) above we have

p (x) ={

x−(

D2 − 1)−1

(2D)}

(

− 25e

x cosx− 15e

x sinx)

= − 25xe

x cosx− 15xe

x sinx− 2(

D2 − 1)−1 (− 3

5ex cosx+ 1

5ex sinx

)

.

Finally by superposition and by (a) and (b) above we have

p (x) = − 25xe

x cosx− 15xe

x sinx− 225e

x cosx+ 1425e

x sinx.

Factoring out − 125e

x and gathering terms from the above yields the book’s answer.

Page 315 (§ 130. Particular solutions) case 6, example 4: To solve this one

it is useful first to evaluate(

D2 − 1)−1

x cosx and(

D2 − 1)−1

x sin x. Each of themsuccumb easily to combining (11) on page 315 with (9) and (10) on page 313.

(

D2 − 1)−1

x cosx ={

x−(

D2 − 1)−1

(2D)}

(

D2 − 1)−1

cosx,

={

x−(

D2 − 1)−1

(2D)}

(

− 12 cosx

)

,

= − 12x cosx+

(

D2 − 1)−1

D cosx,

= − 12x cosx−

(

D2 − 1)−1

sinx,

= − 12x cosx+ 1

2 sinx. (a)

Likewise(

D2 − 1)−1

x sinx ={

x−(

D2 − 1)−1

(2D)}

(

D2 − 1)−1

sinx,

={

x−(

D2 − 1)−1

(2D)}

(

− 12 sinx

)

,

= − 12x sinx+

(

D2 − 1)−1

D sinx,

= − 12x sinx+

(

D2 − 1)−1

cosx,

= − 12x sinx− 1

2 cosx. (b)

Now apply (11) on page 315 then (a) above and finally (b) above to the problem athand.

(

D2 − 1)−1

x2 cosx ={

x−(

D2 − 1)−1

(2D)}

(

D2 − 1)−1

x cosx,

={

x−(

D2 − 1)−1

(2D)}

(

− 12x cosx+ 1

2 sinx)

,

= − 12x

2 cosx+ 12x sinx+

(

D2 − 1)−1

D (x cosx− sinx) ,

= − 12x

2 cosx+ 12x sinx−

(

D2 − 1)−1

x sinx,

= − 12x

2 cosx+ 12x sinx−

(

− 12x sinx− 1

2 cosx)

,

= − 12x

2 cosx+ x sinx+ 12 cosx,

which is the book’s answer.

S-30

The Method of Laplace. Section § 130 details a variety of artifices you can employto determine the particular integral when you have a linear differential equation withconstant coefficients together with a forcing function. I have also described, here in thesupplement, the additional method of undetermined coefficients, which Mellor does notcover, as well as stating the principle of superposition. By applying these methods, youcan find the particular integral for any such equation whenever the forcing function is anysum and/or product of exponentials, sines, cosines, and polynomials. The only difficultyis that the method you need to bring to bear depends on the type of forcing function.Here I will describe the method of Laplace, with which you can find the particularintegral in any problem using only this single method. The method of Laplace hasthe additional advantage in that it can deal with initial conditions and it can easily beapplied to simultaneous equations. It is, to quote Tolkein, the one ring to rule them all.

You may have noticed that there is a correspondence between functions of the equation’sindependent variable (usually x or t), and the functions that result from the differentialoperator, D. For example, if you have

dy

dx− ay = ebx hence y =

1

D − aebx,

You know that the complementary function will have an eax term in it. If we were tofind the particular integral by integrating, the integral expression to be solved would be

p (x) = eax∫

ebx

eaxdx.

We see that there is a correspondence between seeing the (D − a)−1 as an operator andthe function, eax. This correspondence works in both directions. So we can observe thatsince the forcing function is ebx in the example, we should be able to replace it with(D − b)−1

. We end up solving for y as a function of D instead of as a function of x.

Y (D) =1

(D − a) (D − b) =1

a− b

(

1

D − a −1

D − b

)

.

Notice that when you go from a function of x (or t) to a function of D, as we did with y,it is customary notation to use the upper case symbol for the function of D (e.g., Y (D)).Once the D-function has been expanded into partial fractions, it is immediately evidentwhat the final solution to this problem is. We simply translate from the D-function

back to a function of x. We know that (D − a)−1corresponds with eax, which is a part

of the complementary function. So that part of the D-function becomes Ceax. The(D − b)−1

portion translates back to ebx. Since it is a part of the particular integral,we must pay attention to its constant multiplier and carry that into the final solution.Hence the solution to the example is

y = Ceax − 1

a− bebx.

The table on the following page shows the pairings of various functions of x with theircorresponding functions of D. It also shows some rules by which you can infer thecorresponding D-function of more complicated functions of x. If you have a functionof x, then its corresponding D-function is known as its Laplace transform. Many ifnot most modern textbooks use the symbol, s (or sometimes p), in place of D, as theindependent variable of a function’s Laplace transform. This is a matter of nomenclatureonly, and it does not change the essence of the method one bit. Because Mellor usesthe symbol, D, throughout, I will stick with it here in the further explanation of theLaplace method.

What the Laplace method does is to convert the problem of finding a particular integralinto nothing more than a partial fractions problem. Take, for example, the problem

d2y

dx2+ 9y = x2, or equivalently,

(

D2 + 9)

y = x2.

(text continues on page S-33).

S-31

f (x) F (D)

11

D

eax1

D − a

sin bxb

D2 + b2

cos bxD

D2 + b2

xn, where n ≥ 0n!

Dn+1

xneax, where n ≥ 0n!

(D − a)n+1

Rule: eaxg (x) G (D − a)

Hence eax sin bxb

(D − a)2 + b2

and eax cos bxD − a

(D − a)2 + b2

Rule: xng (x) (−1)n dnG

dDn

Hence xeax sin bx2b (D − a)

{

(D − a)2 + b2}2

and xeax cos bx(D − a)2 − b2

{

(D − a)2 + b2}2

Rule: g (ax)1

aG

(

D

a

)

Rule:dg

dx(D)G (D)

Rule:

∫ x

0

g (u)du1

DG (D)

Rule: µg (x) + νh (x) µG (D) + νH (D)

sinh bxb

D2 − b2

cosh bxD

D2 − b2

Short table of Laplace Transforms.

S-32

Using the table to look up the forcing function and solving for Y (D), it becomes

Y (D) =2

(D2 + 9)D3=

2D

81 (D2 + 9)− 2

81D+

2

9D3.

We know that the complementary function for the original equation isC1 sin 3x + C2 cos 3x. Once expanded into partial fractions, the first term of Y (D)translates back to a multiple of cos 3x, so it is a part of the complementary function.The remaining terms are a part of the particular integral. Using the table to translatethem back to a function of x, we find the full solution is

y = C1 sin 3x+ C2 cos 3x− 281 + 1

9x2.

A type of example that tends to trip up beginners is where the forcing function is linearlydependent upon a term of the complementary function. The Laplace method handlesthese seamlessly. For example

d2y

dx2+ b2y = cos bx, or equivalently,

(

D2 + b2)

y = cos bx.

The complementary function is C1 sin bx + C2 cos bx, so clearly the forcing function islinearly dependent upon a term of the complementary function. Using the table to solvethis one for Y (D),

Y (D) =D

(D2 + b2)2 .

Using the table to translate back we see that the particular integral in this case isp (x) = 1

2bx sin bx, which is consistent with the result on page 314 of the text (note thatthe table entry for xeax sin bx gives x sin bx when a = 0). If we had started with a forcingfunction of sin bx, which is also linearly dependent upon the complementary function,the solution for Y (D) would have been

Y (D) =b

(D2 + b2)2 ,

=1

b

b2

(D2 + b2)2,

=1

b

12b

2 − 12D

2 + 12D

2 + 12b

2

(D2 + b2)2 ,

= − 1

2b

{

D2 − b2

(D2 + b2)2− 1

D2 + b2

}

.

The second term inside the brackets is a part of the complementary function. The firstterm corresponds in the table to x cos bx, so the particular solution is p (x) = − 1

2bx cos bx,which is consistent with example 1 on page 314.

As stated before, the method of Laplace can also determine the undetermined con-stants in the complementary function given initial conditions at x (or t) equal tozero. If y is a function of x (or t), then the Laplace transform of the first deriva-tive of y is DY (D) − y (0). The Laplace transform of the second derivative of y isD2Y (D) − Dy (0) − y′ (0), where y (0) and y′ (0) are initial conditions at x = 0 (ort = 0). Higher derivatives follow the same pattern. As an example, we solve the follow-ing equation with a stipultated initial condition:

dy

dt+ ay = sinωt, where y (0) = 1.

In the Laplace transform version, this equation is

(D + a)Y (D) − 1 =ω

D2 + ω2.

Make sure you understand how the initial condition results in the −1 just to the left ofthe equal.

S-33

Solving for Y ,

Y (D) =ω

(D2 + ω2) (D + a)+

1

D + a,

=ω

a2 + ω2

{

a−DD2 + ω2

+1

D + a

}

+1

D + a. (a)

Letting A = ωa2+ω2 , we find that

y (t) = Aa

ωsinωt − A cosωt + (A+ 1) e−at.

The solution above derives from looking up the terms of (a) on the table. As an exercise,you are encouraged to confirm that y (0) = 1 and to verify that this solution satisfies theoriginal equation. Observe that there are no undetermined constants in this solution.This is because we stipulated the initial condition, and doing so fixes the multiplier ofthe complementary function.

Applying the Laplace method to simultaneous equations. Here is an example.

dy

dx+ z = 0, where y (0) = 0,

dz

dx− 4y = x, where z (0) = −3.

Translating this into Laplace form,

DY + Z = 0,

DZ − 4Y =1

D2− 3.

In matrix form this is

Y Z(

D 1

−4 D

)(

01

D2− 3

)

The determinant of this matrix is D2 + 4. Cramer’s rule yields

Y (D) =3

D2 + 4− 1

D2 (D2 + 4)=

3

D2 + 4− 1

4

(

1

D2− 1

D2 + 4

)

,

Z (D) =1

D (D2 + 4)− 3D

D2 + 4=

1

4

(

1

D− D

D2 + 4

)

− 3D

D2 + 4.

Gathering terms and then translating back using the table gives

y = 138 sin 2x− 1

4x,

z = 14 − 13

4 cos 2x.

Again it is left as an exercise for you to verify that this solution satisfies both the originalsystem of equations and the initial conditions.

In engineering, the problem of simultaneous linear equations occurs over and over againin analog electronics and in feedback and control theory. In those fields, Laplace is thepreferred method for solving such problems. Indeed manufacturers of analog electronicparts and electro-mechanical transducers frequently characterize their products by citingproduct responses in terms of Laplace functions. The power of the Laplace method isthat it converts calculus problems into algebra problems. You are urged to review theexample above to see how solving for Y and Z became, in the Laplace domain, is simplya problem of solving a system of linear algebra equations. Decomposing that solutioninto partial fractions is again just an algebra problem. The final step of translating backis nothing more than a matter of looking it up on the table.

S-34

Page 315 (Linear equations with variable coefficients) case 1, info: The funny-looking symbol, ϑ, that the text uses throughout this case is called vartheta, and is acursive form of the Greek letter, θ.

Page 316 (§ 131. Linear equations with variable coefficients) case 1, firstexample 1: Page 315 establishes that z = log x. Hence

dz

dx=

1

x, so by the chain rule,

1

x

d

dz=dz

dx

d

dz=

d

dx.

Multiplying through by x completes the proof.

Page 316 (§ 131. Linear equations with variable coefficients) case 1, firstexample 2: The ϑ operation consists of first taking the derivative, then multiplyingby x. d

dx(xm) = mxm−1; Now multiplying by x, x

d

dx(xm) = mxm.

Alternatively, from page 315, x = ez, so xm = emz. By example 1,

ϑ (xm) =d

dz(emz) = memz = mxm.

Page 316 (§ 131. Linear equations with variable coefficients) case 1, secondexample 2: Using the ϑ operator according to identities at the bottom of page 315,

ϑ(

ϑ− �1)

y +��ϑy + q2y = 0, ∴ ϑ2y + q2y = 0.

Per page 315, we make the substitution, z = log x, and apply the identity from ex-ample 1. So with the substitution and D = d

dz , you have(

D2 + q2)

y = 0. Hencey = C1 sin qz +C2 cos qz = C1 sin (q log x) +C2 cos (q log x), which is the book’s answer.

Page 316 (Linear equations with variable coefficients) case 1, thirdexample 2: Substituting xm for y,

m (m− 1)xm + 4mxm + 2xm = 0.

Dividing out xm and solving for m:

m2 + 3m+ 2 = 0, hence m = −1 or m = −2.The solution, then, is y = C1x

−1 + C2x−2, which is is the book’s answer.

Observe that we could also have applied this method to the second example 2 (that is theprevious example shown on this page), but m would have been imaginary in that case.For that example we would have gotten m = ±ιq, for a solution of y = A1x

ιq +A2x−ιq.

As an exercise, apply Euler’s formula to show that this solution is equivalent to thesolution already shown of y = C1 sin (q log x)+C2 cos (q log x). What is the relationshipbetween the undetermined constant sets, A1, A2 and C1, C2?

Page 316 (§ 131. Linear equations with variable coefficients) case 1, example3: Replacing x with ez, we have:

{D (D − 1)− 3D + 4} y = e3z,

(D − 2)2y = e3z. (a)

The complementary function is y = C1e2z + C2ze

2z. Replacing ez with x and z withlog x yields y = C1x

2+C2x2 log x, which is equivalent to the book’s answer. In addition,

we can apply the method of undetermined coefficients to (a) to establish the particularintegral. Let p (x) = λe3z. Then

(D − 2)2p (x) =

(

D2 − 4D + 4)

p (x) = (9− 12 + 4)λe3x = e3x,

hence λ = 1. So p (x) = e3x = x3.

S-35

Page 316 (§ 131. Linear equations with variable coefficients) case 1, example4: The auxillary equation for xm is m2 −m− 2 = 0. So m = 2 or m = −1. That leadsto y = C1x

2 + C2x−1, which is the book’s answer.

Page 317 (§ 131. Linear equations with variable coefficients) case 1, example1: The equation is equivalent to {ϑ (ϑ− 1)− 2ϑ− 4} y = x4. This factors into

(ϑ+ 1) (ϑ− 4) y = x4.

The roots are 4 and −1, which establishes the complementary solution as C1x4+C2x

−1.Substituting x = ez and ϑ with D, we can find the particular integral by solving

p (z) = (D + 1)−1

(D − 4)−1e4z.

According to (5) on page 312, this is the same as

p (z) = e4z (D + 5)−1D−1 · 1.

Applying the D−1 first,p (z) = e4z (D + 5)

−1z.

From this you get p (z) = e4z(

15z − 1

25

)

. But e4z = x4 is a part of the complementarysolution. So we can drop the 1

25e4z term. Substituting z = log x gives p (x) = 1

5x4 log x.

Page 317 (§ 131. Linear equations with variable coefficients) case 1, example4: This one is equivalent to

{ϑ (ϑ− 1) + 4ϑ+ 2} y = ex,(

ϑ2 + 3ϑ+ 2)

y = ex,

(ϑ+ 1) (ϑ+ 2) y = ex.

Hence the complementary function is C1x−1+C2x

−2. The book offers no direct methodfor finding the particular integral of this one. But we can cobble together a procedureout of various methods we’ve learned so far. First we replace x with ez and ϑ with D,as we have done in other examples.

p (z) = (D + 2)−1

(D + 1)−1e(e

z).

Then we apply integral in (4) from page 311 to the (D + 1)−1

operator.

p (z) = (D + 2)−1e−z

∫

eze(ez)dz = (D + 2)

−1e−ze(e

z).

Now do the same with the (D + 2)−1

operator.

p (z) = (D + 2)−1e−ze(e

z) = e−2z

∫

e2ze−ze(ez)dz = e−2ze(e

z).

Back-substitute x for ez, and you get the book’s answer.

Taking a more general view of the procedure we used on this example, we solvep (x) = (ϑ+ a)−1R (x) by solving p (z) = (D + a)R (ez). We do this by applyingthe integral in (4) on page 311.

p (z) = e−az

∫

eazR (ez) dz.

Since log x = z, we also have dx/x = dz. Back-substituting this integral gives theformula,

p (x) = (ϑ+ a)−1R (x) = x−a

∫

xa−1R (x) dx, (a)

which is a general formula for such problems. If you apply (ϑ+ 1)−1

to R (x) = ex

using (a), then apply (ϑ+ 2)−1

to the result, you will end up with the same particularintegral for example 4 as we arrived at above.

S-36

Suppose we do the same example, but this time we first apply (ϑ+ 2)−1

to R (x) = ex,

then apply (ϑ+ 1)−1

to the result. We expect to end up with the same function asbefore.

(ϑ+ 2)−1ex = x−2

∫

x exdx = x−2 (x− 1) ex =ex

x− ex

x2.

Before we apply the second operator to this result, we must deal with the fact that wewill end up with integrals that cannot be expressed in closed form. Let

E1 (x) =

∫

ex

xdx and E2 (x) =

∫

ex

x2dx.

Using integration by parts (letting u = ex, du = exdx, dv = dx/x2, and v = −1/x),

E2 (x) =

∫

u dv = uv −∫

v du = −ex

x+

∫

ex

xdx = E1 (x)−

ex

x.

We now apply (ϑ+ 1)−1

to ex/x − ex/x2, using (a) from the previous page and theabove identity to substitute for E2 (x).

(ϑ+ 1)−1

(

ex

x− ex

x2

)

= x−1

∫

x0(

ex

x− ex

x2

)

dx = x−1

{

���E1 (x)−���E1 (x) +ex

x

}

,

demonstrating that we get the same result no matter in which order we apply theoperators.

The more common class of forcing function for problems like this is a polynomial. If weapply (a) from the previous page to R = xm, we get

p (x) = (ϑ+ a)−1xm = x−a

∫

xa+m−1dx =xm

a+mfor a+m 6= 0. (b)

When a +m = 0, you have p (x) = x−a log x. The formula (b) also works for negativeand noninteger exponents. This formula is applicable to the remaining examples in thiscase (5 through 8).

Page 317 (§ 131. Linear equations with variable coefficients) case 1, example5: Equation becomes

{ϑ(ϑ− 1)(ϑ− 2) + 2ϑ(ϑ− 1) − ϑ + 1} y = x+ x3,{

ϑ3 − 3ϑ2 +��2ϑ+ 2ϑ2 −��2ϑ− ϑ+ 1}

y = x+ x3,{

ϑ3 − ϑ2 − ϑ+ 1}

y = x+ x3,

(ϑ− 1)2 (ϑ+ 1) y = x+ x3.

Observe that in general, a repeated root, (ϑ− r)n, produces terms in the complementary

function of Ck xr (log x)

kfor k = 0 to k = n−1. So in this example, the complementary

function is C0x + C1x log x + C2x−1, which is equivalent to the book’s complementary

function. To find the particular integral, we apply (b) on this page, and where necessary,(a) on the previous page to

p (x) = (ϑ− 1)−2

(ϑ+ 1)−1 (

x+ x3)

,

= (ϑ− 1)−2

(

x

2+x3

4

)

by (b),

= (ϑ− 1)−1

(

x

2log x+

x3

8

)

by (b),

=x

2

∫

log x

xdx +

x3

16by (a) on the first term and (b) on the second,

=x

4(log x)

2+x3

16, which is the book’s answer.

S-37

Page 317 (§ 131. Linear equations with variable coefficients) case 1, example6: Equation becomes

{ϑ(ϑ− 1)(ϑ− 2) + 2ϑ(ϑ− 1) + 2} y = 10x+10

x,

{

ϑ3 − 3ϑ2 +��2ϑ+ 2ϑ2 −��2ϑ+ 2}

y = 10x+10

x,

{

ϑ3 − ϑ2 + 2}

y = 10x+10

x,

(ϑ+ 1)(

ϑ2 − 2ϑ+ 2)

y = 10x+10

x.

The second factor above has no real roots. Its nonreal roots are ϑ = 1 ± ι. So thecomplementary function is C1x

−1 + C2x1+ι + C3x

1−ι. To turn the second and thirdterms into real form,

C2x1+ι + C3x

1−ι = C2xeι log x + C3xe

−ι log x, and then by Euler’s formula

= C2x {cos (log x) + ι sin (log x)}+ C3x {cos (log x)− ι sin (log x)} .If you stipulate that C2 and C3 be complex complements of each other, then the aboveis equal to C′

2x cos (log x)+C′3x sin (log x), where both C

′2 and C′

3 are both real. Puttingthe pieces together we find that we now have the same complementary function as thebook. To find the particular integral, the setup is

p (x) = (ϑ+ 1)−1 (ϑ− 1− ι)−1 (ϑ− 1 + ι)−1

(

10x+10

x

)

.

Formula (b) on the previous page works even when the ϑ-factor is nonreal. Applyingthe second and third factor to each of the forcing function terms using (b) yields

p (x) = (ϑ+ 1)−1

(

10x

(−ι) (ι) +10x−1

(−2− ι) (−2 + ι)

)

= (ϑ+ 1)−1 (

10x+ 2x−1)

.

Finally using (b) to apply (ϑ+ 1)−1

to each of the terms that result,

p (x) = (ϑ+ 1)−1 (

10x+ 2x−1)

= 5x+ 2x−1 log x.

Putting this together with the complementary function yields the book’s answer.

Page 317 (§ 131. Linear equations with variable coefficients) case 1, example8: The third example 2 on page 316 uses the operator,

(

ϑ2 + 3ϑ+ 2)

= (ϑ+ 1) (ϑ+ 2).So the setup for finding the particular integral when R = x−1 is

p (x) = (ϑ+ 1)−1

(ϑ+ 2)−1x−1, and by (b),

= (ϑ+ 1)−1 x−1

2− 1= (ϑ+ 1)

−1x−1. by (b) again,

= x−1 log x, which is the book’s answer.

Page 317 (Linear equations with variable coefficients) case 2, example 1: Abetter substitution than the one suggested in the book for this is z = a/b + x. This isbecause all derivatives of y with respect to z are equal to the corresponding derivativeswith respect to x. Equation becomes

b2z2d2y

dz2+ b2z

dy

dz+ c2y = 0 ≡

{

b2ϑ (ϑ− 1) + b2ϑ+ c2}

y =(

b2ϑ2 − c2)

y = 0.

By the methods given for case 1, the solution to this is C′1 sin (c/b log z)+C

′2 cos (c/b log z).

See page 316, second example 2, which is worked on page S-35. Back-substitute for z toobtain at C′

1 sin (c/b log (a/b+ x)) + C′2 cos (c/b log (a/b+ x)), which is the same as

C′1 sin

(c

blog (a+ bx)− c

blog b

)

+ C′2 cos

(c

blog (a+ bx)− c

blog b

)

.

Use trig identities for sin (u− v) and for cos (u− v) to arrive at the book’s answer.

S-38

Page 317 (§ 131. Linear equations with variable coefficients) case 2, example2: Substitute z = x+ a. Equation becomes

z2d2y

dz2− 4z

dy

dz+ 6y = x ≡ {ϑ (ϑ− 1)− 4ϑ+ 6} y = (ϑ− 2) (ϑ− 3) y = z − a.

Clearly the complementary function for this is C1z2+C2z

3 = C1 (x+ a)2+C2 (x+ a)

3.

The setup for finding the particular integral is

p (z) = (ϑ− 2)−1

(ϑ− 3)−1

(z − a) , which by (b) on S-37 is

= (ϑ− 2)−1

(

z

(−3 + 1)− a

(−3 + 0)

)

= (ϑ− 2)−1(

−z2+a

3

)

; by (b) again,

= − z

2 (−2 + 1)+

a

3 (−2 + 0)=

z

2− a

6=

x+ a

2− a

6=

x

2+a

3,

which, when combined with the complementary function, is the book’s answer.

Page 318 (§ 132. Exact equations) clarification: You might have found book’sdevelopment of how to deal with higher order exact equations to be confusing. The easyway to understand it is to work it backwards from the first integral. If you begin with

X0d2y

dx2+ (X1 −X ′

0)dy

dx+ (X2 −X ′

1 +X ′′0 ) y =

∫

Rdx+ C,

then take its derivative using the product rule, you get

X0d3y

dx3+ X ′

0

d2y

dx2+

(X1 −X ′0)d2y

dx2+ (X ′

1 −X ′′0 )dy

dx+

(X2 −X ′1 +X ′′

0 )dy

dx+(X ′

2 −X ′′1 +X ′′′

0 ) y = R,

X0d3y

dx3+ X1

d2y

dx2+ X2

dy

dx+ X3y = R,

which holds provided that X3 = X ′2 −X ′′

1 +X ′′′0 . So you can see that by differentiating

(5) on page 318, we have recovered (1) on page 317. At the same time we have alsoevolved the exactness criterion shown in (4) of page 318. This method can easily beextended to determine exactness and first integrals of equations of order higher than 3.

Page 318 (§ 132. Exact equations) example 2: We have X3 = 2, X ′2 = 4, X ′′

1 = 2,and X ′′′

0 = 0. Hence X3 −X ′2 +X ′′

1 −X ′′′0 = 2− 4 + 2− 0 = 0, so the equation is exact.

By (5) on page 318, the first integral is

xd2y

dx2+(

x2 − 3− 1) dy

dx+ (4x− 2x) y = C1.

If you allow the coefficient functions of the above to be Y0 = x, Y1 = x2−4, and Y2 = 2x,then Y2 − Y ′

1 + Y ′′0 = 2x − 2x = 0. So this new equation is also exact. We apply (5)

again to get

xdy

dx+(

x2 − 4− 1)

y = C1x+ C2 ≡ dy

dx+

(

x− 5

x

)

y = C1 +C2

x.

Using the method on page 294, the solution to the above is

y = x5e−x2/2

{

C3 +

∫

(

C1x−5 + C2x

−6)

ex2/2dx

}

.

The integral in the expression above cannot be expressed in elementary functions, butcan be described using something called the incomplete gamma function. You can findmore about incomplete gamma in Handbook of Mathematical Functions by Abramowitzand Stegun (U.S. Department of Commerce, National Bureau of Standards, AppliedMathematics Series · 55).

S-39

Page 318 (§ 132. Exact equations) example 3: Example 1 has already demon-strated that this equation is exact. So we apply (5) on page 318 to find the first integral.

x5d2y

dx2+ (15− 5)x4

dy

dx+ (60− 60 + 20)x3y = x5

d2y

dx2+ 10x4

dy

dx+ 20x3y = ex + C.

If we let Y0 = x5, Y1 = 15x4, and Y2 = 20x3, then Y2 = Y ′1 − Y ′′

0 , hence this equation isalso exact. So we apply (5) again to get the second integral.

x5dy

dx+ (10− 5)x4y = x5

dy

dx+ 5x4y = ex + Cx+ C2.

This equation is also exact. Its left side is a perfect differential.

d(

x5y)

dx= ex + Cx + C2 ∴ x5y = ex + C1x

2 + C2x+ C3,

with C1 = 12C, which is the book’s answer.

Observe that if you divide the original equation by x2, the resulting equation is homo-geneous in powers,

x3d3y

dx3+ 15x2

d2y

dx2+ 60x

dy

dx+ 60y =

ex

x2,

and can be solved using the method from § 131.

{ϑ (ϑ− 1) (ϑ− 2) + 15ϑ (ϑ− 1) + 60ϑ+ 60} y =ex

x2,

(

ϑ3 + 12ϑ2 + 47ϑ+ 60)

y =ex

x2,

(ϑ+ 5) (ϑ+ 4) (ϑ+ 3) y =ex

x2.

The complementary function is C1x−3+C2x

−4+C3x−5. To find the particular integral,

we need to solve the following by repeated application of (a) on page S-36:

(ϑ+ 5) (ϑ+ 4) (ϑ+ 3) p (x) =ex

x2,

(ϑ+ 5) (ϑ+ 4) p (x) = x−3

∫

x2ex

x2dx =

ex

x3,

(ϑ+ 5) p (x) = x−4

∫

x3ex

x3dx =

ex

x4,

p (x) = x−5

∫

x4ex

x4dx =

ex

x5.

Combining this result with the complementary function, we get a solution for y that isequivalent to the book’s answer.

Page 319 (§ 132 Exact equations) Comments on Forsyth’s elimination ofterms: That xmdny/dxn is a perfect differential whenever m < n is proved by takingthe integral of this expression by parts:

∫

xmdny

dxndx = xm

dn−1y

dxn−1− m

∫

xm−1 dn−1y

dxn−1dx.

Note that to the right of the equal both the exponent and the order of the derivative ofthe expression under the integral have been reduced by unity. So by repeating the processover and over, we can see that the exponent reaches zero before the order of the derivativedoes, provided m < n. At that point only one more integration is required, and sincewe would be integrating only a derivative of y (not multiplied by any polynomial of x)we would simply get the next lower derivative (or y itself if m = n− 1). An easy way tojustify eliminating such terms from the exactness criterion, (4) on page 318, is that theorder of the derivative we’d have to take of Xk would exceed its highest exponenent ofx, and so that derivative would be zero and contribute nothing to the exactness criterionformula.

S-40

Page 320 (§ 133. Equations with missing terms) example 1: Back-substitutingp with dy/dx into the text’s result, p2y2 = y4 + C4,

y2(

dy

dx

)2

= y4 + C4 hence ydy

dx=√

y4 + C4.

Now let u = y2, and 12du = y dy, the above becomes

1

2

du

dx=√

u2 + C4, which by separation of variables isdu√

u2 + C4= 2dx.

Integrating,

sinh−1( u

C2

)

= 2x+ C2 henceu

C2= sinh (2x+ C2) .

Back-substituting y2 for u, multiplying by C, and changing the symbol, C, to C1 resultsin the book’s answer.

Page 320 (§ 133. Equations with missing terms) example 2: This equation islinear with constant coefficients and constant forcing function, and as such it can besolved by methods developed in § 129 and § 130. Doing it using the method suggestedin the example, we divide through by µ and make the substitution, observing thatd2x1/dt

2 = d2x/dt2. Equation becomes

1

µ

d2x1dt2

+ x1 = 0 henced2x1dt2

+ µx1 = 0.

You can solve this by inspection: x1 = C1 cos(

tõ)

+ C2 sin(

tõ)

. Back-substitutex1 = x− ν/µ to arrive at the book’s answer.

Page 320 (§ 133. Equations with missing terms) example 4: Let p = dy/dx.Equation becomes

dp

dx− ap2 = 0, which by separation of variables is

dp

p2= a dx.

Integrating then back-substituting,

−1

p= ax+ C hence

dy

dx= − 1

ax+ C.

Integrating again gives y = − (1/a) log (ax+ C) + C′. Multiply both sides by −a, thentake the exponential function of both sides to get,

e−ay = e−aC′

(ax+ C) = C1x+ C2 where C1 = ae−aC′

and C2 = e−aC′

C,

which is the book’s answer.

Page 320 (§ 133. Equations with missing terms) example 5: Substitute p fordy/dx and p dp/dy for d2y/dx2, and the equation becomes

1 + p2 = ypdp

dy, which by separating variables is

dy

y=

p dp

1 + p2.

Integrating,

log y + C = 12 log

(

1 + p2)

hence C1y =√

1 + p2 where C1 = eC .

Back-substitute dy/dx for p,

C1y =

√

1 +

(

dy

dx

)2

hence√

C21y

2 − 1 =dy

dx.

Separating variables and integrating,∫

dy√

C21y

2 − 1=

∫

dx hencecosh−1 (C1y)

C1= x+ C2.

Letting a = C1 and b = C1C2, and multiplying by a, then taking cosh of both sides,then dividing both sides by a, the above results in the book’s answer.

S-41

Page 320 (§ 133. Equations with missing terms) example 6: The book’s answeris evident by inspection using methods from earlier sections. But doing it using themethod from this section, we substitute p for dV/dx and p dp/dV for d2V/dx2. Equationbecomes

pdp

dV− β2V = 0 or equivalently p dp = β2V dV.

Integrating and cancelling the common factor of 12 ,

p2 = β2V 2 + C hence p = β√

V 2 + C′ where C′ =C

β.

Back-substituting and separating variables,

dV

dx= β

√

V 2 + C′ hencedV√

V 2 + C′= β dx.

Integrating,

sinh−1

(

V√C′

)

= βx + C′′ hence V =√C′ sinh (βx+ C′′) .

Using the exponential identity to expand sinh,

V = 12

√C′(

eβx+C′′ − e−βx−C′′

)

.

Let C1 = 12

√C′eC

′′

and C2 = − 12

√C′e−C′′

to obtain the book’s answer. The questionthen arises, can you choose C′ and C′′ in such a way as to arrive at an arbitrary C1 andC2? This would be true provided that the ratio of eC

′′

to e−C′′

can be made to be anydesired value, positive or negative, by proper choice of C′′. So if for arbitrary A,

eC′′

e−C′′= e2C

′′

= A, then C′′ =logA

2.

If we allow nonreal values for C′′ (such as C′′ = 12 log 2 +

12 ιπ for A = −2 for example)

then by Euler’s formula, we can solve this given any A. Also by allowing C′ to beimaginary when desired, we can satisfy the cases where we would want

√C′ to be

negative.

Page 322 (§ 133. Equations with missing terms) example 5: Letting p = dy/dx,the equation becomes

xdp

dx= 1, or equivalently

∫

dp =

∫

dx

x.

Integrating yields p = log x+ C. Back-subsituting for p,

dy

dx= log x+ C, or equivalently

∫

dy =

∫

log x+ C dx.

By integrating the above and letting C1 = C − 1 you get the book’s answer.

Page 335 (§ 135. Chemical Reaction Rates) solving (16): The equation to besolved is

dz

dy=a− zz − y , or equivalently

dy

dz=z − ya− z or

dy

dz+

y

a− z =z

a− z .

The last version is in the form given by (1) of § 122 (page 296). Hence the integratingfactor for this is µ = e− log(a−z) = 1

a−z . The exact equation is

dy

a− z +y − z

(a− z)2dz = 0.

Integrating the second term dz yields the general solution,y

a− z −a

a− z − log (a− z) = C.

Substituting the conditions on page 335, a = 1, y = 0, and z = 0, into the above forcesC = −1. The solution, (17) on page 335, follows from that.

S-42

Page 335 (§ 135. Chemical Reaction Rates) solving (18): The equation to besolved is

dz

dx=a− zy − x , or equivalently

dx

dz=y − xa− z or

dy

dz+

x

a− z =y

a− z ,

where y = z + (a− z) log (a− z). Like the last equation, this is first-order linear, withthe integrating factor being µ = e− log(a−z) = 1

a−z . Multiplying by the integrating factorand substituting for y,

dx

a− z +x dz

(a− z)2=z + (a− z) log (a− z)

(a− z)2dz,

=

{

z

(a− z)2+

log (a− z)a− z

}

dz,

which is exact. Integrating yieldsx

a− z = log (a− z) + a

a− z −12 {log (a− z)}

2+ C.

Using the analogous initial conditions as were stipulated for (16), that is a = 1, x = 0,and z = 0, we find again that C = −1. Substitute −1 for C, then multiply the abovethrough by a− z and take the cancellation to arrive at the book’s answer.

Page 336 (§ 135. Chemical Reaction Rates) example: Taking the quotient of(14) and (15) on page 335 gives

dy

dz=

k2k3

z − ya− z = ρ

z − ya− z where ρ =

k2k3.

Once again this equation is first-order linear. The integrating factor isµ = e−ρ log(a−z) = 1

(a−z)ρ . So the exact equation is

dy

(a− z)ρ +ρy dz

(a− z)ρ+1 =ρz dz

(a− z)ρ+1 .

Integrating, y

(a− z)ρ =

∫

ρz dz

(a− z)ρ+1 .

To integrate the expression to the right of the equal, let u = a − z, a − u = z, and−du = dz. Integral becomes

−ρ∫

a− uuρ+1

du = −ρ∫

a

uρ+1du + ρ

∫

du

uρ=

a

uρ− ρ

ρ− 1

1

uρ−1+ C.

Back-substituting and multiplying through by (a− z)ρ,

y = a− ρ

ρ− 1(a− z) + C (a− z)ρ = − a

ρ− 1+

ρ

ρ− 1z + C (a− z)ρ .

Applying the same conditions as indicated on page 335 of a = 1, z = 0, and y = 0, wefind C = − 1

ρ−1 .

To find the equation for x in terms of z, we let σ = k1/k3 and solve it the same way aswe did for y, except that σ replaces ρ.

dx

dz= σ

y − xa− z .

Replace y with the solution we got above for y. The integrating factor will be µ = 1(a−z)σ .

You getx

(a− z)σ = σ

∫ − aρ−1 + ρ

ρ−1z + C (a− z)ρ

(a− z)σ+1 dz + C′.

Taking of the integral above is left to the reader.

S-43

Page 336 (§ 136. Simultaneous equations) example 1: This one can also be doneusing the Laplace method (see supplement pages S-31 through S-34). The equations are

Dx+ ay = 0, and Dy + bx = 0.In matrix form we have

x y(

D ab D

)

.

The determinant of this matrix is D2 − ab =(

D −√ab)(

D +√ab)

. Because there

are no initial conditions given and forcing functions are zero, we can determine only thecomplementary function from this system of equations. That function is determined bythe roots of the determinant. If we apply these roots to x, the complementary function

is x = C1e√abt + C2e

−√abt. Substitute the derivative of this for dx/dt into the first

equation to solve for y’s complementary function.

Page 337 (§ 136. Simultaneous equations) example 2: Doing it using the book’smethod, we take the derivative of the first equation,

d2x

dt2+dy

dt= 3

dx

dt.

Subtracting the first equation from this gives

d2x

dt2− dx

dt+dy

dt− y = 3

dx

dt− 3x.

Subtracting the second from the above gives

d2x

dt2− dx

dt= 3

dx

dt− 4x.

As indicated in the text, this eliminates y and its derivative and we are left with(

D2 − 4D + 4)

x = (D − 2)2x = 0. This gives a complementary function for x of

C1e2t + C2te

2t. Taking the derivative of this gives dx/dt = (2C1 + C2) e2t + 2C2te

2t.Substituting both of these into the first equation,

(2C1 + C2) e2t + 2C2te

2t + y = 3(

C1e2t + C2te

2t)

.

Solve for y to get the book’s complementary function for y.

Using the Laplace method we subtract 3x from both sides of the first equation andx from both sides of the second. This puts the equation in standard form with alldependent variables to the left of the equal. Forming the matrix from those equationsyields

x y(

D − 3 1−1 D − 1

)

.

The determinant of this matrix is (D − 3) (D − 1) + 1 = D2 − 4D + 4. You can seehow the same differential operator results from this method as from the book’s method.Apply this operator to x to get its complementary function, then use the substitutionmethod shown above to find y’s complementary function from that of x.

Page 337 (§ 136. Simultaneous equations) example 3: First using the book’smethod.

d2x

dt2= µ

dy

dt, derivative of 1st equation

−µ2x = µdy

dt2nd equation times µ, but reversed

d2x

dt2+ µ2x = 0 difference eliminates y.

S-44

Which is the same as(

D2 + µ2)

x = 0. Observe the same operator evolves from usingthe matrix method. We rewrite the equations as dx/dt − µx = 0 and dy/dt+ µx = 0.The matrix is

x y(

D −µµ D

)

,

whose determinant is D2 + µ2. If we ascribe to x the complementary solution,x = C1 cosµt + C2 sinµt, then dx/dt = −µC1 sinµt + µC2 cosµt. Substituting intothe first equation, −µC1 sinµt+ µC2 cosµt = µy,hence y = −C1 sinµt + C2 cosµt, and the relationships between the x-coefficients andthe y-coefficients are as the book indicates.

Page 337 (§ 136. Simultaneous equations) example 5: In the case where thereare more than two dependent variables, the matrix method shows its superiority to thedifferentiate-and-substitute method. All three equations, as given, are in standard form– that is all of the dependent variables are to the left of the equal. The matrix, then, is

x y z

D b ca1 D c1a2 b2 D

,

whose determinant is D3 − (a1b+ a2c+ b2c1)D+ a1b2c+ a2bc1. This cubic is the sameas the one shown in the book for this example, except using a different independentvariable. Note that the book carelessly reused the symbol, z, as an independent variablefor the cubic after it had already used it as one of the dependent variables in the originalequation set. The two usages are not equivalent, and the book should have chosen adifferent symbol for the cubic to avoid confusion.

Page 337 (§ 136. Simultaneous equations) example 6: In a modern electrical en-gineering class, this problem would almost certainly be solved using the Laplace method.We shall do it that way here. The matrix is

i1 i2

MD L2D +R2

L1D +R1 MD

,

the determinant of which is

M2D2 − (L1D +R1) (L2D +R2)

=(

M2 − L1L2

)

D2 − (L1R2 + L2R1)D −R1R2.

Observe that the above quadratic is equivalent to the one given on page 338. Theremainder of the problem – that is finding the particular solutions given that E1 andE2 are constants – is shown in the book.

Page 339 (§ 136. Simultaneous equations, variable coefficients) example 3: Tomake the second solution work out, you have to multiply the equations by an integratingfactor of x. x dx

x2 − y2 − z2 =dy

2y=dz

2z.

Hence P =(

x2 − y2 − z2)

/x, Q = 2y, and R = 2z. Multiplying these respectively byx, y, and z, and then taking the sum yields x2 + y2 + z2. Hence by (3) on page 339,

x dx

x2 − y2 − z2 =dy

2y=dz

2z=x dx + y dy + z dz

x2 + y2 + z2.

S-45

The last expression is a perfect differential of 12du/u, where u = x2 + y2 + z2. Setting

this equal to either of latter two expressions in the modified original equations (we’llchoose dz/2z) and integrating yields

12 log

(

x2 + y2 + z2)

= 12 log z + C,

from which the book’s answer follows. Observe that you could have chosen dy/2y andhad 1

2 log(

x2 + y2 + z2)

= 12 log y + C,

which would also be correct.

Supplemental problem: If you simplify example 3 of page 339 by dropping z, youhave the differential equation,

dx

x2 − y2 −dy

2xy= 0. (a)

Based upon the solution for example 3 of page 339, we can infer that the solution to(a) is x2 + y2 = Cy. Note that (a) is not exact. The problem then is, based upon thesolution given here, to infer an integrating factor, µ (x, y), that makes (a) become exact.

First we work the solution to (a) into a form in which taking the derivative eliminatesthe undetermined constant, C.

log(

x2 + y2)

= log y + logC, whose derivative is2x dx+ 2y dy

x2 + y2=dy

y,

Rearranging the derivative,

2x dx

x2 + y2+

(

2y

x2 + y2− 1

y

)

dy = 0,

which is exact because we arrived at it by taking the derivative of (a)’s solution. Sum-ming the dy-fractions over a common denominator,

2x dx

x2 + y2+

y2 − x2y (x2 + y2)

dy = 0,

which remains exact. In this form, it becomes clear how to form the integrating factor.To go from (a) to the above, you would have to divide by x2 + y2, and multiply by2x(

x2 − y2)

. Hence

µ (x, y) =2x(

x2 − y2)

x2 + y2.

S-46

Page 340, apparent error in book: If u = f(

ax3 + by3)

, then applying the chainrule to take the partial derivatives of u with respect to x and y results in

∂u

∂x= 3ax2f ′ (ax3 + by3

)

and∂u

∂y= 3by2f ′ (ax3 + by3

)

,

which differs from the partial derivatives shown in the book for this function. The book’sderivatives appear to be incorrect. Multiplying the partial (above) with respect to x byby2 and the partial with respect to y by ax2, we find that the right-hand side of bothequations become equal. Hence

by2∂u

∂x− ax2

∂u

∂y= 0,

which also differs from the book’s conclusion on page 340.

Page 341 (partial differential equations), example 1: Taking the partial deriva-tives with respect to x and y respectively,

−b ∂u∂x

= b f ′ (bx− ay) and 1− b ∂u∂y

= −af ′ (bx− ay) .

Substituting for f ′ (bx− ay) from the first equation into the second,

1− b ∂u∂y

= a∂u

∂x, hence a

∂u

∂x+ b

∂u

∂y= 1.

Page 341 (partial differential equations), example 2: Taking the partial deriva-tives with respect to x and y respectively,

− 1

z2∂z

∂x+

1

x2=

1

x2f ′(

1

y− 1

x

)

and − 1

z2∂z

∂y= − 1

y2f ′(

1

y− 1

x

)

.

Using the second equation to substitute into the first for f ′(

1y − 1

x

)

,

− 1

z2∂z

∂x+

1

x2=

y2

x2z2∂z

∂y, hence − x2 ∂z

∂x+ z2 = y2

∂z

∂y,

from which the book’s answer follows.

Page 341 (partial differential equations), example 3: This one is trivial comparedwith the preceding two examples. Both partial derivatives (with respect to x and withrespect to y) are identically equal to a′ (x+ y), and therefore equal to each other. Soclearly the difference between the two partials must be zero, as asserted by the book’sanswer.

Page 344 (partial differential equations), example 3: The equation to solve is

a

(

∂z

∂x+∂z

∂y

)

= z or equivalentlya

z

∂z

∂x+a

z

∂z

∂y= 1.

Make the substitutions, Z = log z, X = x/a, and Y = y/a. Then

a

z

∂z

∂x=∂Z

∂Xand

a

z

∂z

∂x=∂Z

∂Y; equation becomes

∂Z

∂X+∂Z

∂Y= 1.

In this case we have f (α, β) = α + β − 1 based upon the modified equation. Hence

for f (α, β) = 0, we have β = 1 − α. The solution to the modified equation then is

Z = αX + (1− α)Y + C. Converting back to the original variables,

Z = log z = αx

a+ (1− α)

y

a+ C, hence z = Ce

1a{αx+(1−α)y}.

S-47

Page 345 (partial differential equations), example 4: The equation to be solvedis

x2(

∂z

∂x

)2

+ y2(

∂z

∂y

)

= z2 or equivalently

(

x

z

∂z

∂x

)2

+

(

y

z

∂z

∂y

)2

= 1.

If Z = log z, then dZ = dz/z, hence that substitution yields(

x∂Z

∂x

)2

+

(

y∂Z

∂y

)2

= 1.

Similarly substituting X = log x and Y = log y, the equation becomes(

∂Z

∂X

)2

+

(

∂Z

∂Y

)2

= 1.

The auxillary function for this modified equation is f (a, b) = a2 + b2 − 1 = 0. Sob =√1− a2. By (1) on page 344, the solution is

Z = aX +√

1− a2 Y + C hence log z = a logx +√

1− a2 log y + C.

Taking the antilog of the latter equation results in the book’s answer.

Page 345 (partial differential equations, type II), example 1: The book hasdefinite errors on this problem. The equation,

√a2 + z · dz/dx = 2, is correct. From (2)

on page 345, the solution is

x+ ay + C = 12

∫ √a2 + z dz = 1

3

(

a2 + z)

32 .

Squaring both sides results in the solution,

(x+ ay + C)2 = 19

(

a2 + z)3, (b)

which differs from the book’s answer (because the book misplaced a scalar). To showthat this solution is correct, we substitute it back into the equation. Taking ∂/∂xof (b),

2 (x+ ay + C) = 13

(

a2 + z)2 ∂z

∂x, hence 6

x+ ay + C

(a2 + z)2 =

∂z

∂x.

By design, ∂z/∂y = a ∂z/∂x, which is easily confirmed by taking ∂/∂y of (b). Hence

(

∂z

∂x

)2

z +

(

∂z

∂y

)2

=

(

∂z

∂x

)2(

z + a2)

= 36

(

x+ ay + C

(a2 + z)2

)2(

z + a2)

.

Substituting from (b) for (x+ ay + C)2,

3619

(

a2 + z)3

(a2 + z)4

(

z + a2)

= 4,

which verifies that the solution is correct. Indeed if you replace 19 in (b) with any other

value, the above test fails.

Page 345 (partial differential equations, type II), example 2: Replacing q withbp, the equation becomes p

(

1 + b2p2)

= bp (z − a), which simplifies to

1 + b2p2 = b (z − a). Solving for p we have p = φ (z) = 1b

√bz − ab− 1. By (2) on

page 345, the solution is

x+ by + C =

∫

dz

φ (z)= b

∫

dz√bz − ab− 1

= 2√bz − ab− 1.

The book’s answer follows from this. From that answer it’s easy to isolate z as a functionof x and y.

z =1

4b(x+ by + C)

2+ a+

1

b, hence

∂z

∂x=

1

2b(x+ by + C) .

S-48

Substituting these results into 1 + b2p2 = b (z − a), we have

1 + b2(

∂z

∂x

)2

= 1 + 14 (x+ by + C)

2and b (z − a) = 1

4 (x+ by + C)2+ 1,

which verifies that the solution is correct.

Page 345 (partial differential equations, type III), example 2: Rearranging thiswe have p2 − x = y − q2 = a. Now separating, it becomes p2 = x + a and q2 = y − a.By (3) on page 345, the integral is

z =

∫ √x+ a dx+

∫ √y − a dy.

The book’s answer follows from taking the integrals. Observe that you could also haveseparated it into p2 = x − a and q2 = y + a and obtained a slightly different solution.Explain how it is that these two solutions are, in fact, equivalent.

Page 345 (partial differential equations, type III), example 3: Rearranging wehave q/y = 2p2 = 2a2. Notice that we set this to 2a2 rather than a in order to avoida radical in the solution. You could set it to a and arrive at an equivalent solution.Separating it becomes q = 2a2y and p = a. By (3) on page 345,

z =

∫

a dx+

∫

2a2y dy = ax+ a2y2 + C,

which is the book’s answer.

Page 346 (equations analogous to Clairault’s), how to find singular solutions:The book offers no clue on how to do this. In this supplement beginning on page S-17,we developed the method that can be found in Ordinary Differential Equations and theirSolutions by George M. Murphy (van Nostrand, 1960) for finding singular solutions toordinary differential equations. That same method can be extended to apply to partialdifferential equations such as the ones found at the top of page 346. Murphy suggeststhat if p = dy/dx, then we take ∂/∂p of the entire equation, solve for p, then substitutethat back into the original equation to obtain the singular solution. We now have thesituation where p = ∂z/∂x and q = ∂z/∂y. The extended method is to take both ∂/∂pand ∂/∂q of the original equation, solve for p and q simultaneously, then substitutethose solutions back into the original equations. Here is that method applied to the firstexample on page 346:

∂

∂p(z = px+ qy + pq) gives 0 = x+ q, and

∂

∂q(z = px+ qy + pq) gives 0 = y + p.

Clearly p = −y and q = −x. Substituting for p and q into the original equation, wehave z = −xy − xy + xy = −xy, which is the singular solution given in the book.

Applying the same method to the third example,

∂

∂p

{

z = px+ qy − n (pq)1n

}

gives 0 = x− q (pq)1n−1

, and

∂

∂q

{

z = px+ qy − n (pq)1n

}

gives 0 = y − p (pq)1n−1

.

Solving for q in the first equation, q = xnpn−1. Substituting q into the second equation,

y = p{

p(

xnpn−1)}

1n−1

= p (xnpn)1n−1

= p (xp)1−n

, hence p = y1

2−nxn−12−n .

By symmetry we also have q = x1

2−n yn−12−n , and therefore pq = (xy)

n2−n . The book’s

singular solution follows by substituting these into the original equation.

S-49

The second example on page 346 is by far the most difficult on which to apply thismethod, but its singular solution also emerges in the same way.

∂

∂p

(

z = px+ qy + r√

1 + p2 + q2)

gives 0 = x+rp

√

1 + p2 + q2, (a)

∂

∂q

(

z = px+ qy + r√

1 + p2 + q2)

gives 0 = y +rq

√

1 + p2 + q2. (b)

From (a) we have

x2 =r2p2

1 + p2 + q2, hence x2 + p2x2 + q2x2 = r2p2 and x2 + q2x2 =

(

r2 − x2)

p2.

Solving for p2,

p2 =x2(

1 + q2)

r2 − x2 ; substituting p2 into (b), y2 =r2q2

1 + x2(1+q2)r2−x2 + q2

.

Now multiply top and bottom of the nasty fraction by r2−x2, expand the denominator,and then take the cancellations:

y2 =r2q2

(

r2 − x2)

r2 −��x2 +��x2 +���x2q2 + q2r2 −���q2x2=q2(

r2 − x2)

1 + q2,

from which

y2(

1 + q2)

= q2(

r2 − x2)

, hence y2 = q2(

r2 − x2 − y2)

.

We easily find q from this, and by the symmetry of (a) and (b) we can find p as well:

q = ± y√

r2 − x2 − y2and p = ± x

√

r2 − x2 − y2.

When these expressions are substituted for p and q in the original equation, we have(choosing the minus of the ±):

z = − x2 + y2√

r2 − x2 − y2+ r

√

1 +x2 + y2

r2 − x2 − y2 .

Multiplying by√

r2 − x2 − y2 and taking the cancellations,

z√

r2 − x2 − y2 = −x2 − y2 + r2,

from which the book’s singular solutions follows.

Page 348 (§ 141. Linear Partial Equations with Constant Coefficients) exam-ple 2: The book’s solution is in error. The operator equation is

(

D2 − 4DD′ + 4D′2) z = 0.

This factors into (D − 2D′)2z = 0. If you substitute f (mx+ y) into the equation, the

auxillary equation is m2 − 4m+ 4 = (m− 2)2= 0. Hence m = 2. By (9) on page 348,

the solution is f (2x+ y). Substituting this into the original equation verifies it as asolution, whereas the book’s answer fails to verify. Furthermore, the auxillary equationhas a double root. We can’t just say that z = f1 (2x+ y) + f2 (2x+ y) is the generalsolution, since the f1 and f2 terms are equivalent. Equation (11) on page 349 showshow to deal with repeated roots. Accordingly the complete solution is

z = f1 (2x+ y) + xf2 (2x+ y) .

If you think about it, there is no reason that yf3 (2x+ y) shouldn’t also be a solution.And indeed, if you substitute it into the original equation, you will see that it is. Sowhy don’t we include it as a third term in the general solution? Because it is implicitin the existing two terms of that general solution. Consider that (2x+ y) f (2x+ y) isa single function of 2x + y, and is therefore a solution by being equivalent to the firstterm of the general solution. But the 2xf (2x+ y) part of this is also a solution by beingequivalent to the second term of the general solution. As such it becomes zero when

S-50

the equation’s operator is applied to it. That means that the yf (2x+ y) part must alsobecome zero when the operator is applied to it, so it must also be a solution. Hence it’sbeing a solution is implied by the general solution given above. So yf (2x+ y), thougha solution, is not independent of the general solution given above.

Page 349 (§ 141. Linear Partial Equations with Constant Coefficients) ex-ample 1: Again the book gets the sign wrong. The auxillary equation factors into(m+ 1)

2. The double root is m = −1. The solution, z = f1 (x− y)+xf2 (x− y) follows

from (11) on page 349. Observe that z = f1 (x− y) + yf2 (x− y) would also have beena suitable general solution.

Page 349 (§ 141. Linear Partial Equations with Constant Coefficients) exam-

ple 2: The answer given in the book implies an auxillary equation of (m+ 1)2(m− 1) = 0.

This expands into m3 + m2 − m − 1 = 0. The partial differential equation that goeswith this auxillary equation is

(

D3 +D2 −D − 1)

z = 0. You will find that the book’ssolution solves this equation and fails to solve the equation given in the example (al-though the f (x+ y) term by itself does solve the book’s equation, as m = 1 is a root ofm3 − 3m2 +m+ 1 = 0). The book’s equation factors into

(

m− 1)(

m− 1 +√2)(

m− 1−√2)

= 0.

Hence the general solution to the book’s equation (which has no repeated roots) is

z = f1 (x+ y) + f2

(

(

1−√2)

x+ y)

+ f3

(

(

1 +√2)

x+ y)

.

Page 351 (§ 141. Non-homogeneous linear partial equations) example 3:Using the trial solution, z = eαx+βy results in the auxillary equation,

αβ + aα+ bβ + ab = 0, which factors into (α+ b) (β + a) = 0.

The book’s solution follows from the paragraph following (12) on page 349.

Page 351 (§ 141. Non-homogeneous linear partial equations) example 4:The equation factors into

(D +D′ − 1) (D −D′ + 2) z = 0.

Again the book’s solution follows from the paragraph following (12) on page 349.

Page 351 (§ 141. Non-homogeneous non-factorable linear partial equa-tions) supplemental example: Some equations cannot be factored at all. This ex-ample is not in the book:

∂2z

∂x2− ∂2z

∂y2+ z = 0.

The trial solution, z = eαx+βy yields the auxillary equation, α2 − β2 + 1 = 0, orequivalently, α = ±

√

β2 + 1. Substituting for α and inserting the undetermined scalarsfor each possible β, we have a solution of

z =∑

β

Cβ e±√

β2+1 x+βy

according to (15) on page 350. We can modify this solution by allowing β to assumepurely imaginary values. So if β = ιω, then for ω ≥ 1,

z =∑

ω

Cω sin(

±√

ω2 − 1 x+ ωy + φω

)

is also a solution in accordance with Euler’s formula.

Example 1 on page 351 is also not factorable. Here the prototype solution wasz = eαx+α2y. This one becomes of special interest when you allow α to assume thepurely imaginary value of ιω.

S-51

Its general solution becomes

z =∑

ω

Cω e−ω2y sin (ωx+ φω) .

If you replace y with kt, where k is a heat conductivity rate, then this equation represents SeeS-67 formore onFourier’sheatequationand itssolution.

the temperature over time of a long uniform bar. Observe that if the temperature att = 0 varies sinusoidally over the length of the bar, then the magnitude of that sinusoiddecays in time at a rate in proportion to the square of its spacial frequency, ω. Diffusionof fluids in a long narrow tube complies with this same equation.

Page 352 (Particular integrals of homogeneous partial equations) first exam-ple 2: The problem is to solve

(

D2 + 3DD′ −D′2)−1(x+ y) = (D + 2D′)

−1(D +D′)

−1(x+ y) . (a)

Applying the right-hand factor first, we follow the procedure and let m = −1. Subtract-ing mx from y of the forcing function gives x+ y+x = 2x+ y. Integrating that dx givesx2 + xy. Now adding mx = −x to y of this result gives x2 + x (y − x) = xy. So at thispoint we’ve reduced the problem to finding

(D + 2D′)−1

(xy) .

We repeat the same procedure, but this time withm = −2. Subtractingmx = −2x from

y gives x (y + 2x) = xy + 2x2. Integrating that gives 12x

2y + 23x

3. Now add mx = −2xto y of this result to get 1

2x2 (y − 2x) + 2

3x3 = 1

2x2y − x3 + 2

3x3, which is equal to the

answer in the book.

Alternatively we could have integrated (a) starting with the left-hand factor first. Thenwe have m = −2. Subtracting mx = −2x from y in (a) gives x + y + 2x = 3x + y.Integrating dx gives 3

2x2+xy. Now addingmx to y we have 3

2x3+x (y − 2x) = xy− 1

2x2.

At this point we’ve reduced the problem to finding

(D +D′)−1 (

xy − 12x

2)

.

We repeat the procedure, this time with m = −1. So subtract mx = −x from y to get

x (y + x) − 12x

2 = xy + 12x

2. Integrating this dx gives 12x

2y + 16x

3. Add mx = −x to

y in this result gives 12x

2 (y − x) + 16x

3 = 12x

2y − 12x

3 + 16x

3, which is also equal to the

book’s answer.

Page 352 (Particular integrals of homogeneous partial equations) second ex-ample 2: Here F (D,D′) = D2 + 5DD′ + 6D′2, which has two factors, hence we willintegrate twice. Dividing by D2 gives φ (D′/D) = 1 + 5D′/D+ 6D′2/D2. We also haveb/a = 1

2 . So φ (b/a) = 1 + 52 + 6

4 = 5. So the particular integral will be

p (x, y) =1

5

∫∫

dx2

y + 2x,

=1

10

∫

log (y + 2x) dx,

= 120 {(y + 2x) log (y + 2x)− y − 2x} .

S-52

The 1902 edition of the book originally gave this example as(

D2 + 5DD′ + 6D′2) z =1

(y − 2x).

In this case b/a is a root of φ (b/a). So the problem falls into the category explained atthe top of page 353. Factoring this one, it becomes

(D + 2D′) (D + 3D′) z =1

y − 2x,

hence p (x, y) = (D + 2D′)−1

(D + 3D′)−1(

1

y − 2x

)

.

The factor that has the problem with b/a = − 12 is the left-hand factor. So we integrate

just once using the right-hand factor. For that factor alone we have φ (b/a) = 1 + 3ba

hence φ (b/a) = 1− 32 = − 1

2 . So we take

p (x, y) = (D + 2DD′)−1{

−2∫

dx

y − 2x

}

,

= (D + 2DD′)−1

log (y − 2x) .

Now applying (18) on page 353 to the vanishing factor, we have p (x, y) = x log (y − 2x).You are encouraged to substitute this expression for p into

(

D2 + 5DD′ + 6D′2) p andapply the operator to confirm that you get back 1/ (y − 2x).

Page 353 (Particular integrals, case 3) first example 2: As the book suggests, wesolve (D +DD′ − 2D′) z = sin (x− y) and (D +DD′ − 2D′) z = sin (x+ y) separately,and then by the principle of superposition (see page S-24) we add the results to getthe composite solution. Instead of using the method given on page 353, which, in thisexample, is prone to mistakes, we apply the method of undetermined coefficients (alsointroduced on S-24). In the case of the forcing function of sin (x− y), the particularintegral must be a linear combination of sin (x− y) and cos (x− y). Hence

(D +DD′ − 2D′) {λc cos (x− y) + λs sin (x− y)} = sin (x− y) .Applying the operator to the expression in the curly-brackets,

λc {− sin (x− y) + cos (x− y))− 2 sin (x− y)} +λs {cos (x− y) + sin (x− y) + 2 cos (x− y)} = sin (x− y) .

Separating terms and dividing out the sines and cosines yields

−3λc + λs = 1,

λc + 3λs = 0,

hence λc = − 310 and λs =

110 .

Likewise we solve for the particular integral of sin (x+ y) by solving

(D +DD′ − 2D′) {λc cos (x+ y) + λs sin (x+ y)} = sin (x+ y) .

Applying the operator to the expression in the curly-brackets,

λc {− sin (x+ y)− cos (x+ y)) + 2 sin (x+ y)} +λs {cos (x+ y)− sin (x+ y)− 2 cos (x+ y)} = sin (x− y) .

Separating terms and dividing out the sines and cosines yields

λc − λs = 1,

−λc − λs = 0.

Hence λc =12 and λs = − 1

2 . So the complete particular integral is

p (x, y) = − 310 cos (x− y) + 1

10 sin (x− y) + 12 cos (x+ y)− 1

2 sin (x+ y) ,

which differs from the book’s answer. And by the way, did you try to find the complemen-

S-53

tary solution to this one? If z = eαx+βy, then the auxillary equation is α+αβ− 2β = 0.A few more algebraic steps gets you to

z =∑

β

Cβ

{

e2x+(1+β)y}

β1+β

.

Now that we have solved the more difficult problem, I’ll reveal that I believe that thetypographical error in the book was not in the answer given, but in the original equation,which should have been

(

D2 +DD′ − 2D′2) z = sin (x− y)+sin (x+ y). The particularintegral for this one is easily solved using the method in case 3 of page 353.

1

D2 +DD′ − 2D′2 sin (x± y) = 1

−1∓ 1 + 2sin (x± y) .

Observe that by using the ± notation, we solve both cases at the same time. In the− case of the ±, we immediately see that we get 1

2 sin (x− y). In the + case, thedenominator vanishes. But the operator factors into

D +DD′ − 2D2 = (D + 2D′) (D −D′) .

So by case 2 and (18) on page 353, with a = 1 and b = 1, we have

(D + 2D′)−1

(D −D′)−1

sin (x+ y) =1

1 + 2x

∫

sin (x+ y) dx = − 13x cos (x+ y) .

Combining the two results still differs from the book’s answer by the sign of the co-sine term. You are encouraged to apply the original operator to the expression aboveto see which is the right answer. The complementary solution to this equation isf1 (2x− y)+ f2 (x+ y) in accordance with (9) on page 348. So the complete solution is

z = f1 (2x− y) + f2 (x+ y) + 12 sin (x− y)− 1

3x cos (x+ y) .

Page 353 (Particular integrals, case 4) second example 1: We havea = 2 and b = 3. Accordingly if F (D,D′) = D2 − DD′ − 2D′2 + 2D + 2D, thenF (a, b) = a2−ab− 2b2+2a+2b = 4− 6− 18+4+6 = −10. The book’s answer follows.Page 353 (Particular integrals, case 4) second example 2: We haveF (n,m) = nm+ an+ bm+ ab = (n+ b) (m+ a). The book’s answer follows.

Page 354 (Particular integrals, case 5) second example 2: First divide out 3Dfrom the operator:

1

3D

(

D

3− D′2

3D− 1 +

D′

D

)−1

xy = − 1

3D

{

1−(

D

3− D′2

3D+D′

D

)}−1

xy.

No sooner does the book state, “The expansion is not usually carried higher than thehighest power . . . in f (x, y),” than it gives an example that requires carrying out theexpansion higher than that. In this case we need to take the expansion out to the cubedterm. The reason is the D that appears in the denominator of terms that will be in theexpansion. You have to use your careful judgement on this. Look at how the powersexpand. Any D or D′ to the squared or higher power will cause xy to become zero.Hence the D′2/3D term in the above contributes nothing to the solution and can beignored throughout your analysis. There is another very sneaky aspect to this problemthat I will show as we analyze the expansion. The expansion out to the cubed term isthen

− 1

3D

{

1 +

(

D

3− D′2

3D+D′

D

)

+

(

D

3− D′2

3D+D′

D

)2

+

(

D

3− D′2

3D+D′

D

)3

+ . . .

}

xy.

Going through term by term, we see that

− 1

3Dxy contributes − 1

6x2y.

S-54

Next, recalling that the D′2/D contributes nothing, we have,

− 1

3D

(

D

3− D′2

3D+D′

D

)

xy which contributes − 19xy − 1

18x3.

Next, again ignoring the D′2/D term,(

D

3− D′2

3D+D′

D

)2

expands toD2

9+

2D′

3+D′2

D2,

Only the middle term has powers of D or D′ less than squared, so it is the only termthat contributes. Hence

− 1

3D

2D′

3xy contributes − 1

9x2.

Finally, again ignoring D′2/D,(

D

3− D′2

3D+D′

D

)3

expands toD3

27+DD′

3+D′2

D+D′3

D3,

of which only the DD′/3 term will make any contribution. This expansion is in accor-dance with the binomial theorem, but here is where there is a trap for you. The binomialtheorem assumes multiplication to be commutative. But in this case that assumptionleads to a wrong answer. Letting a = D/3 and b = D′/D, we have,

(a+ b) (a+ b) (a+ b) = a3 + aab+ abb+ aba+ baa+ bab+ bba+ b3.

But notice that if you apply aa to xy before applying b to it, you get zero. The orderof application here does matter. Hence only two out of the three terms containing twoa’s and one b contribute anything. The one that applies both a’s before applying the bcontributes nothing. So the real contribution of the cubed expansion is

− 1

3D

2DD′

9xy which contributes − 2

27x.

You should confirm for yourself that expanding powers higher than cubed leads to allof the terms containing a factor of Dn or D′n where n ≥ 2. So no higher powers areneeded. Gathering all of the contributions shown above and summing them yields thebook’s answer. You are encouraged to substitute that answer back into the originalequation to confirm that it does work.

Note that you could have started by dividing by 3D′ rather than 3D. Using thesame procedure as above, you would have arrived at 1

6xy2+ 1

18y3+ 1

9xy+19y

2+ 227y. This

solution also works. If p is the book’s answer and q is this answer, then Cp+ (1− C) qalso solves the particular integral.

Page 355 (nonconstant coefficients) first example 2: Making the substitution,the equation becomes

∂2z

∂u2− ∂z

∂u+ 2

∂2z

∂u∂v+∂2z

∂v2− ∂z

∂v= 0.

Letting D and D′ be partial derivatives with respect to u and v respectively, we have(

D2 −D + 2DD′ +D′2 −D′) z = 0.

The auxillary equation is

α2 − α+ 2αβ + β2 − β = (α+ β)2 − (α+ β) = 0.

Solving we find that either α + β = 0 or α + β = −1. The first root leads to thesolution, z = g1 (v − u), in accordance with (9) on page 348. The second root leadsto the solution, z = eug2 (v − u) in accordance with B on page 349. So the completesolution is

z = g1 (v − u) + eug2 (v − u) .Back-substituting x = eu and y = ev, and allowing fk (s) = gk (e

s) for both functionsyields the book’s answer.

S-55

Page 355 (nonconstant coefficients) example 3: Making the suggested substitu-tion, we have

v =∂z

∂xand

∂2z

∂x∂y=dv

dy.

Equation becomes(x+ y)

dv

dy− av = 0.

Separating variablesady

x+ y=dv

v.

Integrating gives

a log (x+ y) + g (x) = log v or equivalently f1 (x) (x+ y)a

= v,

where f1 = eg. Back-substitute for v and we have

f1 (x) (x+ y)a

=∂z

∂x.

Integrating dx yields the book’s answer.

Page 355 (nonconstant coefficients) second example 2: Expanding the book’sformulation in ϑ,

ϑ2 − (1 + n)ϑ+ 2ϑϑ′ + ϑ′2 − (1 + n)ϑ′ + n = 0,

or equivalently(ϑ+ ϑ′)

2 − (1 + n) (ϑ+ ϑ′) + n = 0.This is a quadratic in ϑ+ ϑ′, and is easily factored into

(ϑ+ ϑ′ − 1) (ϑ+ ϑ′ − n) = 0.

Allowing u = log x and v = log y, then by B on page 349,

z = eug1 (v − u) + enug2 (v − u) ,from which the book’s answer follows by back-substituting x = eu, y = ev, andfk (s) = gk (e

s).

Page 356 (Solutions in power series) Jumbled equation: In the uniquenessproof, it should go like this: If y = φ (x) is the power series solution to dy/dx = y, andy = vφ (x) were a different solution, then

dy

dx− y = v′φ (x) + vφ′ (x)− vφ (x) = 0.

But since φ (x) solves the differential equation, that means φ′ (x) = φ (x). Hence theabove equation collapses into

dy

dx− y = v′φ (x) = 0,

requiring either v′ or φ (x) to be identically zero. So if φ is not zero, v must be aconstant.

This same strategy can demonstrate uniqueness of the general solution to any first orderlinear equation that has no forcing function. If y = φ (x) solves the equation,

dy

dx+ p (x) y = 0. (a)

We suppose that y = vφ (x) also solves (a). Then

vdφ

dx+ v′φ (x) + p (x) vφ (x) = 0. (b)

But since φ (x) solves the original equation, we have

dφ

dx= −p (x)φ (x) .

Substituting the above into (b) yields v′φ (x) = 0, which is the same result as before.

S-56

Page 357 (solutions in power series) example 4: The method given in the bookfor determining series solutions is also known as the method of Frobenius, after math-ematician, Ferdinand Georg Frobenius (1849-1917). The modern book, MathematicalMethods in the Physical Sciences, 2nd edition, by Mary L. Boas, (1983, Wiley and Sons)provides an especially lucid presentation of this method in chapter 12. The problem athand is to develop a series for the solution to

d2y

dx2+ xy = 0.

The table below breaks the series solution into its elements according to the terms ofthe above equation.

xm xm+1 xm+n

d2y

dx2(m+ 2) (m+ 1) a2 (m+ 3) (m+ 2) a3 (n+m+ 2) (n+m+ 1) an+2

xy a0 an−1

Each row of the above table is the contribution of one of the terms of the differentialequation. Each column shows the contributions attributed to each power of x by variousterms of the equation. The quantity, m, is the exponent of the lowest nonzero term ofthe series solution (note that in general m is not necessarily positive and not necessarilyan integer).

We use the right-most column of the table to determine the recursion relationship amongthe an’s. The sum of that column must be zero in order to satisfy the original equation.Hence

an−1+(n+m+2) (n+m+1)an+2 = 0 consequently an+2 =−an−1

(n+m+ 2) (n+m+ 1).

Replacing n with n+ 1, we have

an = − (n+m+ 3) (n+m+ 2)an+3 consequently an+3 =−an

(n+m+ 3) (n+m+ 2).

To determine legitimate values for m, we use the left-hand equation above to run thethe recursion backward. At the point where n < 0, we need to get a zero result in orderto prevent the series from continuing indefinitely into negative indices. Note that therecursion relationship requires that n − m be a multiple of 3. This means that whenn−m = −3, we have 0 = m (m− 1). Hence m = 0 or m = 1.

Running the recursion forward with m = 0 and a0 = 1, we see that

a0 = 1; a3 = − 1

2 · 3; a6 =1

(2 · 3) (5 · 6) ; a9 = − 1

(2 · 3) (5 · 6) (8 · 9) ; . . .

Doing the same with m = 1 and a1 = 1, we see that

a1 = 1; a4 = − 1

3 · 4 ; a7 =1

(3 · 4) (6 · 7) ; a10 = − 1

(3 · 4) (6 · 7) (9 · 10); . . .

You should be able to see that these coefficients are equal to the ones given by the book’sanswer. The two series (where m = 0 and where m = 1) are independent of each other,so the general solution is a linear combination of those two series.

It is useful to see what happens when we allow m = 2. That means thata2, a5, a8, . . . would all be nonzero. Hence there is a nonzero term in the series, a2x

2.The second derivative of this term is the constant, 2a2. That means that some otherterm of the series times x must cancel 2a2. But that term would have to come froma−1x

−1, which, by assumption, is zero. So if we were to allow nonzero a2, then theseries would continue with nonzero terms indefinitely into negative indices. You areencouraged to experiment for yourself with applying the differential equation to a seriesconsisting of

∑

a3n+2x3n+2 to see for yourself why it can never work unless all of these

terms are zero.

S-57

It is fairly easy to prove that both the m = 0 and the m = 1 series converge for all x.Each nonzero term of either series is arrived at by multiplying the last nonzero term byx3 and dividing it by (n− 1)n or by n (n+ 1), where n is the index of the new term.Clearly you can always find large enough n that the divisor will exceed x3, no matterlarge x is. So for large enough n, the series will always pass the ratio test and thereforemust converge. This convergence analysis applies to this example only. There are otherdifferential equations for which the series solutions have finite radii of convergence.

Returning now to the problem of solving for possible values ofm, there is an abbreviatedmethod for doing this once you have established the table. Usually all non-blank rowsthe left-most column of the table will have only a single index for a. If so, take the sumof all rows in that column. In the case of this example, only the first row of the left-mostcolumn is non-blank. It shows an expression involving a2, namely (m+ 2) (m+ 1)a2.Let k be the index of a, in this case, k = 2. The procedure is to replace m with m− k,drop the a, and set the resulting expression to zero. In this case you would have

(m+ 2− k) (m+ 1− k) = m (m− 1) = 0. (c)

Equation (c) is known as the indicial equation. Simply solve it for m. In this case itis clear that m = 0 or m = 1. For this example the solution is easy and results in mhaving two nonnegative integer values. More complicated examples may have more thanone row of the left-most column of the table containing nonzero entries. Solving for mwill then involve finding the roots of an indicial polynomial whose degree is equal to theorder of the original equation. Those roots will not necessarily be integer or positive.Indeed they could even be non-real complex complements. Whatever they be, each suchsolution for m will be the basis for an independent series solution for the differentialequation in question, and will appear as a summand in each exponent. As such, it willalways be possible to factor xm from the series solution to arrive at a solution in theform of

y =

∞∑

n=0

anxm+n = xm

∞∑

n=0

anxn.

Page 358 (Expansion of J0 into a power series): Substituting n = 0 into theBessel equation given on page 358 gives

d2y

dx2+

1

x

dy

dx+ y = 0 or equivalently x

d2y

dx2+dy

dx+ xy = 0. (d)

The equation on the left is a canonical form, d2y/dx2 + P (x) dy/dx+Q (x) y = 0, thatis useful in the analysis of second order linear equations. We see that in the case inwhich we are interested, P (x) = 1/x and Q (x) = 1. We shall come back to that later,but for the purposes of finding a series solution, J0 (x), to this, we shall start with theright-hand form of (d). We could just go ahead and assemble the table from that form,which would yield the correct series, but doing it that way would hide an importantpiece of information. Instead we observe that

xd2y

dx2+dy

dx=

d

dx

(

xdy

dx

)

.

Making that substitution, (d) becomes

d

dx

(

xdy

dx

)

+ xy = 0. (e)

This is the form out of which we build the table. If yn (x) = anxn, then

x yn (x) = anxn+1, and

d

dx

(

xdyndx

)

= ann2xn−1 hence

d

dx

(

xdyn+1

dx

)

= an+1 (n+ 1)2xn.

S-58

Based upon those facts, we fill out the table as follows:

xm xm+1 xm+n

xy a0 an−1

d

dx

(

xdy

dx

)

a1 (m+ 1)2

a2 (m+ 2)2

an+1 (m+ n+ 1)2

Using the method we used on the previous example, the indicial polynomial ism2 = 0. Observe that this has a double root at m = 0. So we cannot come up withtwo independent series arising from the two solutions for m. Instead we will develop theseries for just one of them, using m = 0, and make some comments on the other. Theright-most column of the table shows the recursion to be

an−1 + an+1 (m+ n+ 1)2 = 0 consequently an+2 = − an

(m+ n+ 2)2

Because m = 0 and the recursion goes in steps of two, we get a series with only evenpowers of x. Allowing a0 = 1, we have

J0 (x) = 1− x2

(2)2 +

x4

(2 · 4)2− x6

(2 · 4 · 6)2+ · · · =

∞∑

n=0

(−1)n x2n

22n n!2.

Given that Γ (1) = 1, this is the same as the series given in on page 358 with n = 0.Using the same type of argument as in the previous example, we see that this seriesconverges for all x.

We return to the canonical form of the original equation. Recall that the canonical formis d2y/dx2 + P (x) dy/dx + Q (x) y = 0. In the case of J0, we have P (x) = 1/x andQ (x) = 1. The series development we just did results in a Taylor series taken aroundx = 0, which is precisely the point at which P (x) is discontinuous. Compare that withexample 4 on page 357, which is already in canonical form with P (x) = 0 and Q (x) = x.In that example, both coefficient functions are continuous and differentiable everywhere.The distinction has a bearing on how we use the roots of the indicial equation. Thistopic is quite involved, and we can only touch on it here. Ordinary Differential Equationsand Their Solutions by George M. Murphy (van Nostrand, 1960) goes into full detailin part I, unit B1. Other advanced books on differential equation also deal with theseissues. When P and Q are continuous and differentiable at the point around which youare taking the series, then you can use both roots of the indicial equation as we did inexample 4 on page 357, provided the roots are distinct. If, however, there is a pole ineither P or Q at the center-point of the series, as we have in the equation for J0, thesituation becomes more complicated. In that case, if the two roots differ by an integer,then the two series developed from those two roots will not be independent of eachother. This happens when you develop the series for Jn, where n is any integer otherthan zero. The roots of the indicial equation are m = ±n. The series from the root,m = −n, has negative powers of x. The denominators of those terms contain a factorof Γ (k), where k is a nonnegative integer. But the Γ function of nonnegative integers isinfinite, so those terms of the series are all zero. The terms that remain are all a scalarmultiple of those of the other series developed from the root, m = n. Hence the twoseries are dependent. In addition when you have a double root in the indicial equation,as we do for J0, we also have no way of developing two independent series.

We can use the formula on page 307 to establish a second independent solution whenwe already know a particular solution. In the of the zeroth order Bessel function, weknow that J0 (x) solves equation (d). Using P (x) from the canonical form, we have

Y0 (x) = J0 (x)

∫

1

J0 (x)2 e

∫−P (x)dx dx,

S-59

where Y0 (x) is the second independent solution to (d). Since P (x) = 1/x, the integralbecomes

Y0 (x) = J0 (x)

∫

dx

xJ0 (x)2 .

Observe that J0 (x) is nonzero at x = 0. Based upon the series solution for J0 (x), it

is clear that 1/J0 (x)2 and all its derivatives exist at x = 0. This means that 1/J0 (x)

2

can be expanded as a Taylor series around x = 0, and that series will have a radius ofconvergence equal to the lowest magnitude x that solves J0 (x) = 0. Carrying out thatexpansion is beyond the scope of this tutorial. It is clear, however, that the constantterm of that expansion will be equal to 1. That term divided by the x that appearsinside the integral is responsible for J0 (x) log x term that appears in the expansion forY0 (x). The first few terms of this expansion is shown on page 358, although the bookuses the old symbol, K0, for this function rather than the modern symbol, Y0.

Developing a series solution around points other than x = 0: That is, if we seeka series around x = b, for example, we are looking to establish the ak’s in

y (x) = a0 + a1 (x− b) + a2 (x− b)2 + a3 (x− b)3 + . . . =

∞∑

n=0

an (x− b)n .

The method here is a straightforward variation on the method of Frobenius alreadyintroduced on page S-57. Simply make the substitution, z = x − b, into the originaldifferential equation. As an example, the zeroth order Bessel equation, (d), becomes

d2y

dz2+

1

z + b

dy

dz+ y = 0 or equivalently b

d2y

dz2+

d

dz

(

zdy

dz

)

+ (z + b) y = 0.

Develop the series solution for z around z = 0 using the methods already presented,then back-substitute at the end. In the case of the zeroth order Bessel function, theseries is more complicated than any we have attempted. We will only go as far as to fillout the table for this example.

zm zm+1 zm+n

zy a0 an−1

by a0b a1b anb

d

dz

(

zdy

dz

)

a1 (m+ 1)2

a2 (m+ 2)2

an+1 (m+ n+ 1)2

bd2y

dz2a2b (m+2) (m+1) a3b (m+3) (m+2) an+2b (m+n+2)(m+n+1)

To form the indicial equation, sum up the left-most column. Observe that we havemixed indices of a. Let k be the highest of those indices – in this case k = 2. Replacem with m− k. The resulting equation is

a−2b+ a−1 (m− 1)2+ a0bm (m− 1) = 0.

By assumption, a0 is the lowest index coefficient that is nonzero. It follows that a−2 anda−1 are both zero. So the indicial equation is m (m− 1) = 0. Summing the right-mostcolumn, though, we see that the recursion is nasty and involves four different indicesof a. In this example there is no clear path to a closed form for an. Were we able toestablish an, however, we could not expect that either the m = 0 or the m = 1 series tobe equal to either J0 or Y0. Instead each would be some linear combination of the two.And because Y0 has a pole at z = −b, we would expect that the radius of convergenceof either series would be b.

S-60

Resonant systems revisited: On page S-26 is a paragraph introducing the canonicalform of the equation for resonant systems. Resonant systems occur again and again inengineering. Here are three physical examples.

i. An RLC-circuit. The di-agram shows such a circuit. Avoltage, vin, which varies sinu-soidally in time, is applied on theleft and results in an output volt-age, vout, on the right. Here vinis the forcing function and voutis the particular integral derivingfrom the differential equation implicit in the circuit acting upon the forcing function.We let vin = cosωf t, where we shall call ωf , the forcing frequency. When ωf = 0, theimpedance of the capacitor, C, is infinite and the impedance of the inductor, L, is zero.So we expect then that vout = vin. As ωf goes to infinity, the impedance of the inductorgoes to infinity as well while the impedence of the capacitor goes to zero. Under thoseconditions we should expect that vout should approach zero. Analysis of this circuitnetwork yields

vout =1

LC

D2 + RL D + 1

LC

vin.

The analysis is based upon the impedance of the inductor being LD, the impedence ofthe capacitor being 1/ (CD), and the impedence of the resistor being R.

Letting ω =

√

1

LCand ζ =

1

2R

√

C

L, we find that

vout =ω2

D2 + 2ζωD + ω2vin,

which is similar to the resonant equation on page S-26, where ω is the resonant frequencyand ζ is the damping factor. Substituting vin = cosωf t, this becomes

vout =ω2

D2 + 2ζωD + ω2cosωf t. (a)

The goal is to see how vout varies with the forcing frequency, ωf . We can solve for theparticular integral, vout, using the method detailed in case 5 on page 313. Replacing D2

with −ω2f ,

vout =ω2

2ζω D −(

ω2f − ω2

) cosωf t.

Now multiply top and bottom by 2ζω +(

ω2f − ω2

)

to get

vout =

{

2ζωD +(

ω2f − ω2

)}

ω2

4ζ2ω2D2 −(

ω2f − ω2

)2 cosωf t.

Again replacing D2 with −ω2f ,

vout = −

{

2ζωD +(

ω2f − ω2

)}

ω2

4ζ2ω2 ω2f +

(

ω2f − ω2

)2 cosωf t.

To simplify, divide top and bottom by ω4 and replace ωf/ω with the frequency ratio, σ.

vout = − 2 ζωD +

(

σ2 − 1)

4ζ2σ2 + (σ2 − 1)2 cosωf t =

(

1− σ2)

cosωf t+ 2ζσ sinωf t

(1− σ2)2+ 4ζ2σ2

. (b)

It is possible to resolve this function into two components. One is the amplitude of the

S-61

resulting sinusoid. To find that we take the square root of the sum of the squares of thecoefficients of cosωf and sinωf .

‖vout‖‖vin‖

=1

√

(σ2 − 1)2 + 4ζ2σ2

. (c)

A plot of this functionis shown to the right forvarious values of ζ. Thistype of plot showing themagnitude of output ofa system as a functionof sinusoidal frequencyof input is known asa Bode plot, after Hen-drik Wade Bode (1905-1982), who invented it.Note that the axes ofthe plot are scaled bythe log base 10 of fre-quency and the log base10 of amplitude.

Based upon thisplot we can seethat, as expected, forωf � ω (that is for0 < σ � 1), we have‖vout‖/‖vin‖ ≈ 1, whichis consistent with ourearly observation that vout = vin when ωf = 0. When ωf � ω, (that is when σ � 1),we see that on the log/log scale, the plot decreases with a slope of −2. This is equivalentto ‖vout‖ varying inversely as σ2, which is consistent with our early observation that asωf goes to infinity, we expect vout to go to zero. Of course the region of greatest interestis near the resonant frequency – that is the region around σ = 1, or equivalently theregion around ωf = ω. When ζ < 1, we see that the plot reaches a maximum just to theleft of σ = 1. As ζ gets closer and closer to zero, the maximum becomes sharper andsharper, as well as becoming higher. The position of the maximum approaches σ = 1.Observe that at ζ = 0.1, we have the amplitude of vout being five times the input voltageat the point of resonance. Indeed, when σ = 1, equation (c) predicts that

‖vout‖ =‖vin‖2ζ

.

So as ζ approaches zero, at resonance the ratio of the output amplitude to the inputamplitude goes to infinity. Circuits such as this one designed to have a sharp resonance(that is a very small value for ζ) are useful in radio receivers, among other applications,where it is necessary to select a narrow band of frequencies out of a broader spectruminput.

By contrast, look at what happens when ζ is much greater than 1. The trace for ζ = 5shows that for ωf up to about 0.1ω, we continue to have ‖vout‖ ≈ ‖vin‖. From about0.1ω to about 10ω, the trace decreases with a slope of −1. Only for ωf > 10ω do we seethe decrease rate become a slope of −2. What is happening here is that when ζ > 1, thedenominator of equation (a) will have two real roots, one of them with |D| < ω, the otherwith |D| > ω. In the case of ζ = 5, we see D =

(

−5 +√24)

ω and D =(

−5 +√24)

ω,or D ≈ −0.1ω and D ≈ −10ω. When −ωf falls between those two roots, the factor

S-62

(D + 0.1ω) is growing proportionally larger, whereas the factor (D + 10ω) is changingvery little. So between the two roots, the amplitude varies in inverse proportion tothe forcing frequency. But when the forcing frequency is larger than the magnitude ofboth the roots, the amplitude varies in inverse proportion to the square of the forcingfrequency. Also observe that as ζ becomes larger and larger, the two roots will spreadfarther and farther apart (although both roots will always be negative).

Recalling that the method of finding the particular integral given on page 313 replacesD2 with −ω2

f , this means that it also replaces D with ι ωf , where ι =√−1. If the two

roots are at −0.1ω and −10ω, then the contribution to the denominator of (a) of thetwo roots will be ι ωf +0.1ω and ι ωf +10ω respectively. The use of complex quantitiesin the analysis of linear circuits is commonplace in the world of electrical engineering.Indeed if all you had done to equation (a) was replace D with ι ωf and evaluated theresulting complex function, the real part of the result would give you the correct cosinecoefficient and the imaginary part the correct sine coefficient. From those coefficientsthe right-hand version of equation (b) would follow immediately. And just so you know,the customary symbol in electrical engineering for

√−1 is j,which distinguishes it from

i, which electrical engineers use as a variable name for electrical current. Electricalengineers also use the symbol, s, instead of D.

The discussion of equa-tion (b) already statedthat it has a secondcomponent besides itsmagnitude. That sec-ond component is itsphase angle. When youadd the sine and co-sine terms of equation(b), you get a sinu-soid that lags in phasefrom vin. The plot hereshows that phase differ-ence as a function ofσ = ωf/ω for variousvalues of ζ. We cansee that when the forc-ing frequency is substantially less than ω, there is very little phase lag between theoutput and the input. At the resonant frequency, the output always lags the input byexactly 90◦. And as the forcing frequency becomes much greater than the resonant fre-quency, the phase lag approaches 180◦, which means that the vout is almost completelyopposite to vin. The other trend to notice is that as ζ approaches zero, the phase lagtransition from nearly 0◦ to nearly 180◦ occurs more and more sharply right at theresonant frequency.

The function used to construct the above plot is:

∠vout|vin = tan−1 2ζσ

1− σ2,

then summing in 180◦ where the arctangent was negative to account for quadrant.

If you had processed equation (a) by replacing D with ι ωf , then you would have had apoint on the complex plane for each value of ωf . Drawing a ray from the origin throughthat point would make an angle with the real axis that is exactly equal to the phase lagof vout to vin corresponding to that ωf . In addition, the distance of that point from theorigin would be exactly equal to the corresponding magnitude, ‖vout‖/‖vin‖.

S-63

As promised earlier, there are other examples of physical systems that follow the reso-nance equation, (a).

ii. An automobile’s suspension: The axle of an automobile is connected to thechassis by a spring and a dashpot (which mechanics call a shock-absorber) in parallel.Let the chassis have mass, m, and let xin be the vertical position of the axle and xoutbe the position of the chassis. The spring applies a force, k (xin − xout) to the chassis.The dashpot applies a force, c

(

dxin

dt − dxout

dt

)

to the chassis. Be aware that this is a verysimplified description of an automobile suspension. Other factors in real suspensionsare couplings between the four wheels, that the suspension is responsive not only inthe vertical axis but the lateral and longitudinal axes as well, and a number of otherengineering details. But the single-wheel, single-axis system described here will showhow resonant systems occur in mechanics.

By Newton’s laws of motion, we have

ftotal = k (xin − xout) + c

(

dxindt− dxout

dt

)

= md2xoutdt2

, (d)

or equivalently

k (xin − xout) + c (xin − xout)D = mxoutD2.

Solving for xout in terms of xin,

xout =

k

m+

c

mD

D2 +c

mD +

k

m

xin.

If we let ω =

√

k

mand ζ =

1

2

c√km

, the equation becomes

xout =ω2 + 2ζωD

D2 + 2ζωD + ω2xin.

Finally we let xin = cosωf t:

xout =ω2 + 2ζωD

D2 + 2ζωD + ω2cosωf t, (e)

which is similar to equation (a), in that it has the same denominator, but is different inthat it has an extra term in the numerator. We will solve this for xout by substitutingD with ιωf , as suggested on the previous page.

xout =ω2 + 2ζιωfω

2ζιωfω − (ωf − ω2)cosωf t,

Multiply top and bottom by 2ζιωfω +(

ωf − ω2)

, then divide top and bottom by ω4

and replace ωf/ω with the frequency ratio, σ, take the cancellation, rearrange a bit, andyou have

xout =4ζ2σ2 −

(

σ2 − 1)

− 2ζισ3

4ζ2σ2 + (σ2 − 1)2cosωf t. (f)

S-64

A plot of the magnitude of‖xout‖/‖xin‖ for various val-ues of ζ is shown to theright. Observe that likeBode plot of equation (c),for ωf � ω, the outputmagnitude is nearly equalto the input magnitude.The region around the reso-nance point is nearly identi-cal to that of (c). But forωf > ω, the behavior isdifferent. For low valuesof ζ the trace slopes downsteeply after resonance. Forhigh values of ζ, the tracedoesn’t even start to slopedown appreciably until wellto the right of resonance.For all values of ζ, the slopefar to the right of resonanceeventually approaches −1.This of course begs the ques-tion of what value of ζshould be chosen for an au-tomobile suspension. The goal is that the suspension will isolate the chassis from irreg-ularities in the road. So for high values of ωf , which correspond to rapid changes inthe road surface, we would like to see them well below the 1.0 line in the plot. For lowvalues of ωf we would like the chassis to follow the axle. A low value for ζ does thatthe best except for the region right around resonance. In that region a low ζ amplifiesthose frequencies, and the effect will be that the car bounces at the resonant rate eachtime it goes over a bump. So the actual value of ζ chosen by automotive designers is acompromise. Typically the suspension is designed for ω ≈ 2π radians/second and withζ between 0.5 and 1.0.

iii. A servo-system: A potentiometer is arranged in a circuit so that its wiper-pinproduces a voltage that is in proportion to how far the potentiometer shaft is turned.That voltage goes to an amplifier that delivers a current to a servomotor in proportionto the input voltage. The servomotor delivers a torque to a flywheel in proportion toits input current. The flywheel also has a mechanical damper that provides a resistivetorque in proportion to how fast the flywheel spins. Finally, the shaft of the motor iscoupled to the shaft of a second potentiometer arranged in the same way as the first.The wiper-pin voltage of that potentiometer is subtracted from the wiper-pin voltageof the first, and it is actually the difference of the two voltages that is the input of theamplifier.

The desired effect is that if an operator turns the shaft of the first potentiometer bysome angle, the servo-system will turn the flywheel by the same angle, after which theflywheel will come to rest. Or more broadly, we desire the flywheel’s angular positionto follow that of the first potentiometer shaft.

Let k be conversion factor that relates servomotor torque to potentiometer positiondifference. Let c be the resistive factor that relates resistive torque on the flywheel tothe angular velocity of the flywheel. Let m be the moment of inertia of the flywheel.Then the equation descibing the whole servosystem is identical to (d) in the previous

S-65

example, where xin is the angular position of the first potentiometer shaft, xout is theangular position of the flywheel, and ftotal is replaced by total torque. Observe thatfor low values of ζ, the flywheel will overshoot, then oscillate around the control pointbefore settling, each time the first potentiometer is moved. For high values of ζ, theflywheel response will be sluggish, undershooting the first potentiometer’s change ofposition and settling into the control point only over time. Again the ideal choice for ζis a compromise value between 0.5 and 1.0.

The the galvanometer on page 323 is yet another example forcing an input on a resonantsystem.

Differential Equation Solution to Keplerian Orbits: Here we solve the problemof two bodies attracted to each other by gravity. In particular we solve the problemwhere one of the two bodies outweighs the other by a large factor (for example, thesun’s mass is over 300,000 times the mass of the earth). This way we can consider thelarger mass as fixed and solve only for the orbit of the smaller mass, m.

We set the problem up in polar coordinates. In the radial axis we have

md2r

dt2−mr

(

dθ

dt

)2

= −f (r) . (a)

Here f (r) is the attractive gravitational force felt by the smaller object as a function ofhow far it is from the larger object. The first term on the left is simply Newton’s lawof motion. The second term is the so-called centrifugal force. That is it is the apparentforce a rotating object feels away from the center of rotation.

The gravitational force acts only in the radial direction. Hence in the tangential axis wehave conservation of angular momentum. Angular momentum, L, is given by

L = mr2dθ

dt. (b)

By saying that this quantity is conserved, we are saying that it is constant and thereforeits time derivative is zero.

md

dt

(

r2dθ

dt

)

= 0, or equivalently rd2θ

dt2+ 2

dr

dt

dθ

dt= 0.

Observe that geometrically, angular momentum, L, is the mass, m, times twice the rateof radial area swept by the smaller object as it orbits the larger object. Johanas Keplerobserved that planets orbiting the sun sweep equal area in equal time (Kepler’s secondlaw). So we see that conservation of angular momentum is equivalent to Kepler’s secondlaw.

Solving (b) for dθ/dt and substituting the result into (a), we have

md2r

dt2−mr

(

L

mr2

)2

= −f (r) , or equivalently md2r

dt2=

L2

mr3− f (r) .

Newton’s universal law of gravitation postulates that

f (r) =GMm

r2,

where M and m are the masses of the two bodies, G is a universal physical constant,and r is the distance between the two objects. So the equation we must solve is

md2r

dt2=

L2

mr3− GMm

r2. (c)

The trick to solving this is to make the substitution, r = 1/u. Then differentiating ronce yields

mdr

dt= −m 1

u2du

dt= −m 1

u2du

dθ

dθ

dt= −��m

1

��u2du

dθ

L��u2

��m.

S-66

Differentiating a second time yields

md2r

dt2= −L d

dt

du

dθ= −L d

2u

dθ2dθ

dt= −L

2u2

m

d2u

dθ2.

Substituting this result into (c) we get

−L2u2

m

d2u

dθ2=

L2u3

m−GMmu2, or equivalently

d2u

dθ2+ u =

GMm2

L2. (d)

The general solution to (d) (see case 3 on page 309) is

u = u0 cos (θ + φ) +GMm2

L2; back-substuting, r =

L2

u0L2 cos (θ + φ) +GMm2, (e)

where u0 and φ are arbitrary constants. Equation (e) does not tell us anything about thetime-behavior of the orbit, but it does give the shape of the orbit. Setting φ = 0 alignsthe symmetry of the orbit with the x-axis. By substituting r2 = x2+y2 and x = r cos θ,we find that the resulting Cartesian formula is the equation of a conic section, of whichan ellipse with one of the foci at the origin is one of the possibilities fitting the equation.The latter is mathematical confirmation of Kepler’s first law. Replacing φ with nonzerovalues simply rotates the major axis of the ellipse away from the x-axis.

The time-behavior of the orbit can only be determined by numeric methods. Suchmethods, as well as more details on the derivation above, can be found in An Introductionto Celestial Mechanics by Forest Ray Moulton (1914 MacMillan. Available since 1970from Dover Publications).

Introduction to Fourier’s Heat Equation and its Solution: This topic is coveredhere only as the barest of outlines. Imagine a long thin strand of heat conductingmaterial, such as a length of wire. We also imagine that it is completely insulatedfrom the outside world except maybe at its endpoints. So heat conduction is restrictedto heat traveling along the length of the wire. A flow of heat along the wire resultsin a temperature gradient according to the constant, σ. That is, for example, if σ =1 calorie/C◦cm second, then if the temperature gradient is 1 C◦per cm, heat flow pastany point of that gradient will be 1 calorie per second. We also let the thermal density ofthe wire be represented by ρ. Then, for example, if ρ = 1 calorie/cm C◦ and a point onthe wire is at temperature 100 C◦, then that point on the wire contains 100 calories/cmof heat.

If we take the case where one end of the wire of length, L, is held at temperature, T1 andthe other at T2, and the temperature gradient is constant throughout the length at arate of (T2 − T1) /L, then heat flow will be σ (T2 − T1) /L. If the initial heat gradient isconstant at this value, then as long as the ends are held at T1 and T2, the heat gradientalong the entire length of the wire will remain unchanged over time. That is, a constantheat gradient of (T2 − T1) /L is the steady state condition for the boundary conditionsof T1 and T2 held at the respective ends of the wire.

Having constant heat gradient implies that ∂2T/∂x2 = 0, where x represents positionalong the wire, and whenever this condition is true, the wire is in steady state, whichis to say that the temperature profile will remain unchanged through time. But when∂2T/∂x2 6= 0, we find that the temperature profile changes over time even if the bound-ary conditions at either end of the wire are held constant. Finding the time function ofthe temperature profile under this condition is the problem Jean Baptist Joseph Fourier(1768-1830) sought to solve in 1822 (see God Created the Integers, a collection of math-ematical breakthrough papers edited and commented by Stephen Hawking, RunningPress, 2005 for more on Fourier and his work).

S-67

We can see intuitively that wherever there is a “hump” in temperature, that is where∂2T/∂x2 < 0, there is a net outward flow of heat from that region to dissipate the hump,and wherever there is a “dip” in temperature, that is where ∂2T/∂x2 > 0, there is a netinward flow of heat to that region to “fill up” the dip. Fourier’s heat equation,

∂2T

∂x2=

1

k

∂T

∂t, (a)

where k = σ/ρ quantifies this. This equation is similar to example 1 on page 351, andpage S-52 offers a solution to this equation in terms of decaying sinusoids,

T (x, t) =∑

ω

Cωe−kω2t sin (ωx+ φω) ,

or equivalently

T (x, t) =∑

ω

e−kω2t {Aω cos (ωx) +Bω sin (ωx)} , (b)

where Aω and Bω are sets of constants, one A and one B for each value of ω used inthe solution.

What Fourier recognized and is remembered for is the observation that you can alwaysconvert any initial bounded function, T (x, 0), into a weighted sum of sines and cosines,and likewise, you can always convert any weighted sum of sines and cosines into afunction of x. So the procedure is to first convert the initial T (x, 0) into a set ofAω’s and Bω’s for some set of values of ω. Second, to find what has happened aftersome elapsed time, t, we apply the decay function, e−kω2t, for each value of ω to thecorresponding sines and cosines. Third and finally, we sum up the decayed sines andcosines each weighted by the Aω’s and Bω’s that we found in the first step.

If we restrict our consideration of a finite length, L, of the wire, then the set of ω’s weneed to consider is ωk = 2πk/L, for each nonnegative integer, k. To demonstrate howwe break an arbitrary function T (x, 0), defined over 0 ≤ x ≤ L, into sines and cosines,we first demonstrate the principle of orthogonality of sines and cosines. Observe that

∫ L

0

cos (ωjx) sin (ωkx) dx = 0 for any j and k

and∫ L

0

cos (ωjx) cos (ωkx) dx = 0 and

∫ L

0

sin (ωjx) sin (ωkx) dx = 0 whenever j 6= k

In words, we have a collection of functions, the product of any two of them whenintegrated from 0 to L yields zero, provided the two functions are distinct. Such functionsare said to be orthogonal to one another under this operation. The operation itself isfundamental to this problem and we give it the name, scalar product. We define thescalar product of two function, f (x) and g (x) as

〈f, g〉 =

∫ L

0

f (x) g (x) dx,

and we say that nonzero f and g are orthogonal whenever 〈f, g〉 = 0.

If we assume that the function, T (x, 0), is the sum,

T (x, 0) =∑

k

Ak cos (ωkx) +Bk sin (ωkx) ,

for some set of Ak and Bk, then taking the scalar product of T with any any cos (ωjx)yields 〈T (x, 0) , cos (ωjx)〉 = 〈Aj cos (ωjx) , cos (ωjx)〉 .Observe that by orthogonality, all the terms in the original sum except the one aboveto the right of the equal are zero. Only for the cosine term where k = j do we get a

S-68

nonzero term. Likewise with

〈T (x, 0) , sin (ωjx)〉 = 〈Bj sin (ωjx) , sin (ωjx)〉 .It is easily verified using our definition of scalar product that

〈Af (x) , g (x)〉 = A 〈f (x) , g (x)〉 and 〈f (x) , Bg (x)〉 = B 〈f (x) , g (x)〉 ,for any functions, f and g, and for any constants, A, and B. This makes it easy to solvefor the value of any Ak or Bk. Simply take the product of T (x, 0) with correspondingcos (ωkx) or sin (ωkx) and integrate the product from 0 to L, then divide by L/2. Thedividing by L/2 is because

〈cos (ωkx) , cos (ωkx)〉 = 〈sin (ωkx) , sin (ωkx)〉 =L

2for all k.

Proving this last fact is left as an exercise for the reader.

To prove that the desired set of Ak’s and Bk’s always exists, we sample the interval,[0, L] at 2n+ 1 equally spaced points.

xi =iL

2nfor i from 0 to 2n.

We also isolate 2n+ 1 sinusoid functions, cos (ωkx) for k from 0 to n and sin (ωkx) fork from 1 to n. Note that we omit consideration of sin (ω0x) because it is zero for all x.We allow rows k = 0 through n of a matrix, M , to be cos (ωkxi), where i goes from 0 to2n. We allow rows k = n+ 1 through 2n to be sin (ωk−nxi). This gives us an invertiblesquare matrix. Then taking a vector of n + 1 Ak’s and n Bk’s times the matrix, M ,gives a weighted sum of the sine and cosine functions over the various sample points,xi. It follows that M−1T (xi, 0)must give back a vector of the weights, Ak and Bk. As we let n go to infinity, we see thatwe can reproduce the original function, T (x, 0), on a denser and denser set of samplepoints.

There is much more can be said about solving Fourier’s heat equation. Indeed thereare entire textbooks devoted to this one equation. The point here is to introduce youto the notion of decomposing a function into weighted sinusoids. This procedure isknown as Fourier analysis, and is the topic of chapter VIII of J.W. Mellor’s book. Itis also applicable to a wide variety of problems other than the solving Fourier’s heatequation. In an earlier section, for example, where we made Bode plots of the responsesof resonant systems to sinusoidal inputs, we were laying the groundwork for applyingFourier analysis to this problem as well. In real life the inputs to such systems are notsinusoids, but could be any function. But the fact that any legitimate input function canbe decomposed into sinusoids allows us to solve the action of the system on each sinusoid,then sum the output sinusoids to determine the system’s response to the original input.Even the action of the human ear is connected with Fourier’s method. The human ear isan instrument that resolves an input function, namely sound acting upon the eardrum,into a set of sinusoids of various frequencies and transmits the magnitudes of the resolvedfrequencies to the brain.

Also be aware that Fourier analysis is just a single example of a more general method-ology of decomposing an arbitrary function into a weighted sum of some set of basisfunctions. In the more general setting, the basis set need not be sinusoids. Dependingupon the type of problem, resolution into other basis functions is often convenient. Asmentioned on page 358 of the text, Bessel functions provide a useful basis for systemsthat have cylindrical symmetry. And there are others for other symmetries.

S-69

Selected referenced sections of original text begin

§ 20. Leibnitz’ Theorem. To find the nth differential coefficient of the productof two funnctions of x in terms of the differential coefficients of each function.

On page 26 the differential coefficient of the product of two variables was shown tobe dy

dx=d (uv)

dx= v

du

dx+ u

dv

dx,

where u and v are fuuctions of x. By successive differentiation and analogy with thebinomial theorem it may be shown that The co-

efficient,n, is abinomialcoeffi-cient,as arethe re-spectiveunshowncoeffi-cientsin thisequation.KH

dn (uv)

dxn= v

dnu

dxn+ n

dv

dx

dn−1u

dxn−1+ · · ·+ u

dnv

dxn. (1)

This formula, due to Leibnitz, will be found very convenient in Chapter VII “How toSolve Differential Equations”. The reader must himself prove the formula, as an exerciseon (§ 19, by comparing the values of d2(uv)/dx2, d3(uv)/dz3,..., with the developments

of (x+ h)2, (x+ h)

3,..., of page 22.

49

§ 22. Euler’s Theorem on Homogeneous Functions. The followingdiscussion is convenient for reference:

To show that if u is an homogeneous function∗ of the nth degree, say u = Σaxαyβ,†whereα+ β = n, then

x∂u

∂x+ y

∂u

∂y= nu. (14)

By differentiation of the homogeneous function,

u = axαyβ + bxα1

yβ1

+ · · · =∑ axαyβ ,

where α+ β = α1 + β1 = · · · = n, we obtain

∂u

∂x=∑

aαxα−1yβ and∂u

∂y=∑

aβxαyβ−1.

Hencex∂u

∂x+ y

∂u

∂y=∑

a (α+ β)xαyβ = n∑

axαyβ.

The theorem may be extended to include any number of variables (see footnote, page340).

EXAMPLES.– (1) If u = x2y+ y2x+3xyz, then x ∂u∂x + y ∂u

∂y + z ∂u∂z = 3u. Prove this

result by actual differentiation. It of course follows directly from Euler’s theorem, sincethe equation. is homogeneous and of the third degree.

(2) If u = x3+x2y+y3

x2+xy+y2 , x∂u∂x + y ∂u

∂y = u, since the equation is of the first degree andhomogeneous.

(3) Put Euler’s theorem into words. Ansr. In any homogeneous function, the termsof the products of each variable with the partial differential coefficients of the originalfunction with respect to that variable is equal to the product of the original function withits degree.

∗An homogeneous function is one in which all the terms containing the variables havethe same degree. Examples: x2 + bxy+ z2; x4 + xyz2 + x3y+ x2 + z2 are homogeneousfunctions of the second and fourth degrees respectively.†The sign “Σ” is to be read “the sum of all terms of the same type as. . . ” or here

the sum of all terms containing x, y and a constant”. The symbol “Π” is sometimesused in the same way for “the product of all terms of the type”.

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§ 23. Successive Partial Differentiation. We can get the higher par-tial derivatives by successive differntiation, using processes analogous tothose used on page 47. Thus when

u = x2 + y2 + x2y2,

∂u

∂x= 2x+ 2y3x;

∂u

∂y= 2y + 3x2y2; (1)

repeating the differentiation,

∂2u

∂x2= 2

(

1 + y3)

;∂2u

∂y2= 2

(

1 + 3x2y)

; (2)

If we had differentiated ∂u/∂x with respect to y, and ∂u/∂y with respectto x, we should have obtained two identical results, viz.:–

∂2u

∂y∂x= 6y2x, and

∂2u

∂x∂y= 6y2x. (3)

This rule is general.The higher partial derivatives are independent of the order of differ-

entiation. By differentiation of ∂u/∂x with respect to y, assuming x to

be constant, we get ∂

(

∂u

∂x

)/

∂y, which is written∂2u

∂y∂x; on the other

hand, by the differentiation of ∂u/∂y with respect to x, assuming y to be

constant, we obtain∂2u

∂x∂y. That is to say

∂2u

∂y∂x=

∂2u

∂x∂y. (4)

This was only proved in (3) for a special case. As soon as the reader. hasgot familiar with the idea of differentiation, he will no doubt be able todeduce the general proof for himself, although it is given in the regulartext books. The result stated in (4) is of great importance.

§ 24. Exact Differentials. To find the condition that u may be a func-tion of x and y in the equation

du = M dx+N dy, (5)

where M and N are functions of x and y.We have just seen that if u is a function of x and y

du =∂u

∂xdx+

∂u

∂ydy, (6)

that is to say, by comparing (5) and (6)

M =∂u

∂x; N =

∂u

∂y.

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Differentiating the first with respect to y, and the second with respectto x, we have, from (4)

∂M

∂y=

∂N

∂x. (7)

In the chapter on differential equations this condition is shown to be nec-essary and sufficient in order that certain equations may be solved, or“integrated” as it is called. Equation (7) is therefore called the criterionof integrability. An equation that satisfies this condition is said to be acomplete or an exact differential. For examples, see page 290.

§ 25. Integrating Factors. The equation

M dx+N dy = 0 (8)

can always be made exact by multiplying through with some function of x,called an integrating factor. (M and N are functions of x and y.)

Since M and N are functions of x and y, (8) may be written

dy

dx= − M

N.(9)

or the variation of y with respect to x is as −M is to N ; that is to say, xis some function of y, say

f (x, y) = a,

then from (5), page 52,

∂f (x, y)

∂xdx+

∂f (x, y)

∂ydy = −. (10)

By a transformation of (10), and a comparison of the result with (9), wefind that

dy

dx= − ∂f (x, y)

∂xdx

/

∂f (x, y)

∂ydy = −M

N. (11)

Hence∂f (x, y)

∂x= µM ; and

∂f (x, y)

∂y= µN, (12)

where µ, is either a function of x and y, or else a constant. Multiplying theoriginal equation by the integrating factor µ, and substituting the valuesof µM , µ,N so obtained in (12), we obtain

∂f (x, y)

∂xdx+

∂f (x, y)

∂ydy = 0,

which fulfils the condition of exactness.EXAMPLE.– Show that the equation y x dy = 0 becomes exact when multiplied by

1/y2. ∂M

∂y= − 1

y2;∂N

∂x= − 1

y2.

Hence ∂M/∂y = ∂N/∂x, the condition required by (7). In the same way show that1/xy and 1/x2 are also integrating factors.

58

§ 67. Envelopes.

The equationm

a+ ax,

represents a family of curves, since for each value of a, we get a distinctcurve. If a varies continuously it will determine a succession of curves, eachof which is a member of the family denoted by the above equation. a issaid to be the variable parameter of the family, since the different membersof the family are obtained by assigning arbitrary values for a. Let theequations

y1 =m

a+ ax (1)

y2 =m

a+ δa+ (a + δa)x (2)

y3 =m

a+ 2δa+ (a + 2δa)x (3)

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be three successive members of the family. As a general rule two distinctcurves in the same family will have a point of intersection.

Fig. 78. – Envelope

Let P (Fig. 78) be the point of intersectionof curves (1) and (2); P1 the point of inter-section of curves (2) and (3), then, since P1,and P2, are both situated on the curve (2),PP1, is part of the locus of a curve whoseare PP1, coincides with an equal part of thecurve (2). It can be proved, in fact, thatthe curve P P ,. . . touches the whole familyof curves represented by the original equa-tion. Such a curve is said to be an envelopeof the family.

To find the equation to the envelope,bring all the terms of the original equationto one side,

y =m

a− ax = 0.

Then differentiate with respect to the variable parameter, and putm

a2− x = 0.

Eliminate a, between these equations,

y −√m · x− x

√

m

x= 0, or, y − 2

√m · x = 0.

∴ y2 = 4mx.

EXAMPLES.– Find the envelope of the family of circles

(x− a)2 + y2 = r2,

where a is the variable parameter. Differentiate with respect a and x − a = 0;

Fig. 79. – Double Envelope

eliminating a, we get y = ±a, whichis the required envelope. The enve-lope y = ±a represents two straightlines parallel to the x-axis and at adistance +a, and −a from it. ShownFig. 79.

(2) Show that the envelope of thefamily of curves (x−m− a)2 + y2 =4ma, is a parabola y2 = 4mx. See§§ 126 and 138.

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§ 72. How to find a Value for the Integration Constant. It isperhaps unnecessary to remind the reader that integration constants mustnot be confused with the constants belonging to the original equation. Forinstance, in the law of descent of a falling body

dv/dt = g;∫

dv = g∫

dt, or, v = gt+ C. (1)

Here g is a constant representing the increase of velocity due to the earth’sattraction, C is the constant of integration. The student will find someinstructive remarks in § 118.

There are two methods in general use for the evaluation of the integra-tion constant.

First Method. Returning to the falling body and to its equation ofmotion,

v = gt+ C.

On attempting to apply this equation to an actual experiment, weshould find that, at the moment we began to calculate the velocity, thebody might be moving upwards or downwards, or starting from a positionof rest. All these possibilities are included in the integration constant C.Let v0, denote the initial velocity of the body. The computation beginswhen t = 0, hence

v0 = g × 0 + C, or, C = v0.

If the body starts to fall from a position of rest, v0 = C = 0, and∫

dv = gt, or, v = gt.

This suggests a method for evaluating the constant whenever ’the natureof the problem permits us to deduce the value of the function for particularvalues of the variable.

If possible, therefore, substitute particular values of the variables inthe equation containing the integration constant and solve She resultingexpression for C.

EXAMPLE.– Find the value of C in the equation

t =1

klog

1

a− x + C, (2)

which is a standard “velocity equation” of physical chemistry. t represents the timerequired for the formation of an amount of substance x. When the reaction is justbeginning, x = 0 and t = 0. Substitute these values of x and t in (2).

1

klog

1

a+ C = 0, or, C = − 1

klog

1

a.

Substitute this value of C in the given equation and we get

t =1

k

(

log1

a− x − log1

a

)

=1

klog

a

a− x.

Second Method. Another way is to find the values of x corresponding totwo different values of t. Substitute the two

162

sets of results in the given equation. The constant can then be made todisappear by subtraction.

EXAMPLE.– In the above equation, (2), assume that when t = t1, x = x1 and whent = t2, x = x2; where x1, x2, t1, and t2, are numerical measurements. Substitute theseresults in (2).

t1 =1

klog

1

a− x1+ C; t2 =

1

klog

1

a− x2+ C.

By subtraction and rearrangement of terms

t2 − t1 =1

klog

a− x1a− x2

.

The result of this method is to eliminate, not evaluate the constant.

Numerous examples of both methods will occur in the course of thiswork. Some have already been given in the discussion on the “CompoundInterest Law in Nature”.

163