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Differential Equation INTRODUCTION Differential equations are fundamental in engineering mathematics since many of the physical laws and relationships between physical quantities appear mathematically in the form of such equations. The transition from a given physical problem to its mathematical representation is called modeling. This is of great practical interest to engineer, physicist or computer scientist. Very often, mathematical models consist of a differential equations or system of simultaneous differential equations, which needs to be solved. In this chapter we shall look at classifying differential equations and solving them by various methods. DIFFERENTIAL EQUATIONS OF FIRST ORDER Definitions A differential equation is an equation which involves derivatives or differential coefficients or differentials. Thus the following are all examples of differential equations. (a) x2dx + y2 dy = 0

(b) d2x

dt2 + a2x = 0

(c) y = xdy

dx+

x2

dy/dx

(d) [1 + (𝑑𝑦

𝑑𝑥)

2]

−5/3

= a d2y

dx2

(e) dx

dy wy = a cos pt,

dy

dt + wx = a sin pt

(f) x2 z

x+ y

z

y = 3z

(g)

2y

t2 = a2

2y

x2

An ordinary differential equation is an equation which all the differential coefficients al with respect to a single independent variable. Thus the equations (a) (c) are all ordinary differential variables and partial differential coefficients with respect to any of them. The equations (f) and (g) are partial differential equations. The order of a differential equation is the order of the highest derivative appearing in it. The degree of a differential equation is the order of the highest derivative appearing in it. The degree of a differential equation is the degree of the highest derivative occurring in its, after the equation has been expressed in a form free from radicals and fractions as far as the derivatives are concerned. Thus from the examples above, (a) is of the first order and first degree (b) is of the second order and first degree

(c) written as y𝑑𝑦

𝑑𝑥 = x(

dy

dx)

2+ x2

is of the first order but of

second degree; (d) After removing radicals is written as

[1 + (𝑑𝑦

𝑑𝑥)

2]

−5

= a3 (𝑑2𝑦

𝑑𝑥2)3

and is of the second order and third degree. Solution of a differential equation A solution (or integral) of a differential equation is a relation between the variable which satisfies the given

differential equation.

For example, y = c 𝑒𝑥3

3 …(i)

is a solution of dy

dx = x2 y …(ii)

The general (or complete) solution of a differential equation is that in which the number of arbitrary constants is equal to the order of the differential equation. Thus (i) is a general solution of (i) is a general solution of (ii) as the number of arbitrary constants (one constant c) is the same as the order of the equations (ii) (first order). Similarly, in the general solution of a second order differential equation, there will be two arbitrary constants. A particular solution is that which can be obtained from the general solution by giving particular values to the arbitrary constants.

For example, y = 4ex3

3 Equations of the first order and first degree It is not possible to analytically solve such equations in general, we shall, however, discuss some special methods of solution which are applied to the following types of equations: 1. Equations where variables are separable. 2. Homogeneous equations 3. Linear equations 4. Exact equations Variables Separable If in an equation it is possible to collect all functions of x and dx on one side and all the functions of y and dy on the other side, then the variables are said to be separate. Thus the general form of such an equation is f(y) dy =

(x)dx + c as solution. Homogeneous Equations

Are of the form 𝑑𝑦

𝑑𝑥=

f(x,y)

(x,y )

Where f(x, y) and (x) (x, y) homogeneous functions of the same degree in x and y. Homogenous function: An expression of the form a0xn +

a1xn1 y + a2xn-2y2 + …+ anyn in which every term is of the nth degree, is called a homogenous function of degree n. This can be rewritten as xn [a0 + a1(y/x) + a2(y/x)2 + …+ an(y/x)n]. Thus any functions f(x, y) which can be expressed in the form xnf(y/x), is called a homogenous function of degree n in x and y. For instance x3 cos (y/x) is a homogenous function of degree 3 in x and y. To solve a homogenous equation

1. Put y = vx, then dy

dx = v + x

dy

dx

2. Separate the variables v and x, and integrate. Linear Equations of first order A different equation is said by linear if the dependent variable and its differential coefficients occur only in the first degree and not multiplied together. Thus the following differential equations are linear

1. dy

dx + 4y = 2

2. x2 d2y

dx2 + 3x

dy

dx + 4y = 2

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equation (i) is linear first order differential equation while equation(ii) in linear second order differential equation. The following equations are not linear

1. (𝑑𝑦

𝑑𝑥)

2+ y = 5

2. dy

dx + y1/2 = 2

3. ydy

dx = 5

Leibnitze linear equation The standard form of a linear equation of the first order, commonly known as leibnitz’s linear equation, is dy

dx + Py = Q where P, Q are arbitrary functions of x.

To solve the equation, multiply both sides by 𝑒∫ Pdx + y

(𝑒∫ Pdx P) = Q𝑒∫ Pdx i.e. d

dx (𝑦𝑒∫ Pdx) = 𝑄𝑒∫ Pdx

Integrating both sides, we et 𝑦𝑒∫ Pdx = ∫ Qe∫ Pdx dx + c as the required solution.

Note. The factor 𝑒∫ Pdx on multiplying by which the left-hand side of (1) becomes the differential coefficient of a single function, is called the integrating factor(l.F.) of the linear equation (i) So remember the following:

l.F. = 𝑒∫ Pdx and the solution is

y(l.F.) = ∫ Q (l.F.)dx + c

Bernoulli’s Equation

The equation dy

dx+ Py = Qyn (i)

where P, Q are functions of x, is reducible to the Leibnitz’s linear and is usually called the Bernoulli’s equation.

To solve (i), divide both sides by yn, so that yn dy

dx + Py1 n

= Q …(ii)

Put y1 n = z so that (1 n) yn dy

dx = dz

dx

Eq. (ii) becomes 1

1−𝑛

dz

dx + Pz = Q

or dz

dx + P(1 n) z = Q(1 n),

Which is Leibnitz’s linear in z and can be solved easily. Exact differential equations Def. A differential equation of the form M(x, y) dx + N(x, y) = 0 is said to be exact if its left hand member is the exact differential of some function u(x, y) i.e. du = Mdx + Ndy = 0. Its solution, therefore, is u(x, y) c. Theorem. The necessary and sufficient condition for the differential equations Mdx + Ndy = 0 to be exact is

M

y =

N

x

Method of solution. It can be shown that, the equation Mdx + Ndy = 0 becomes

d[u + ∫ f(y)dy] = 0 Integrating

u + ∫ f(y)dy = 0. But u = ∫ Mdx and f(y) = terms of N not containing x.

The solution of Mdx + Ndy = 0 is ∫ Mdx + ∫(terms of N not containing x ) dy = c (Porvides of course that the equation is exact.

i.e. M

y =

N

x )

Note: While finding ∫ Mdx, y is treated as constant since we are integrating with respect to x. Equations Reducible To Exact Equations Sometimes a differential equation which is not exact, can be made so on multiplication by a suitable factor called an integrating factor. The rules for finding integrating factors of the equation Mdx + Ndy= 0 are as given in theorem 1 and 2 below: In the equation Mdx + Ndy = 0

Theorem 1: if

M

y−N

x

N be a function of x only =

f(x) say, then 𝑒∫ f(y)dy is an integrating factor. Clairaut’s Equation* An equation of the form = y = px + f(p), where

p = dy

dx , is known as clairaut’s equation

…..(i) Differentiating with respect to x, we have p = p

+ xdp

dx+ f(p)

dp

dx

[x + f(p)] dp

dx = 0

dp

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

dp

dx = 0, gives p = c ….(ii)

Linear Differential Equations (of nTH order) Definitions Linear differential equations are those in which the dependent variable its derivatives occur only in the first degree and are not multiplied together. The general linear differential equation of the nth order is of the form

𝑑𝑛𝑦

𝑑𝑥𝑛 + p1

dn−1y

dxn−1 + p2 dn−2y

dxn−2 + … + pny = X

where p1, p2, pn and X are functions of x only. Linear differential equations with constant coefficients are of the form

dny

dxn + k1 dn−1y

dxn−1 + k2dn−2y

dxn−2 + … + kny = X

where k1, k2….., kn are constants and X is a function of x only. Such equations are most important in the study of electromechanical vibrations and other engineering problems.

1. Theorem: If y1, y2 are only two solutions of the equations

𝑑𝑛𝑦

𝑑𝑥𝑛 +dn−1y

dxn−1 + k2 dn−2y

dxn−2 + … + kny = X Then

c1y1 + c2y2 + …+ kny = X since it can be easily shown by differentiating is that dnu

dxn+ k1dn−1u

dxn−1 + …. + knu = 0….(ii)

2. Since the general solution of a differential equation of the nth order contains n arbitrary constants, if follows, from above, that if y1, y2, y3,…, yn, are n independent solutions of (1), then c1y1 + c2y2 + ….+ cnyn(= u) is its complete solution.

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3. If y = v be any particular solution of

dny

dxn + k1 dn−1v

dxn−1 + … + kny = X ….(iii)

then dnv

dxn + k1

dn−1v

dxn−1 + … + knv = X ….(iv)

Adding (ii) and (iv), we have

dn(u+v)

dxn + k1

dn(u+v)

dxn + k1

dn−1(u + v)

dxn−1 + …+ kn (u +

v) = X This shows that y = u + v is the complete solution of (iii). The part u is called the complementary function (C. F) and the part v is called the particular integral (P.l.) of (iii)

The complete solution (C.S.) of (iii) is y = C.F. + P.l. Thus in order to solve the equation (iii), we have to first the C.F. i.e., the complementary function of (i), and then the P.l., i.e. a particular solution of (iii).

Operator D denoting d

dx,

d2

dx2,

d3

dx3 = D3 etc. , the

equation (iii) above can be written in the symbolic form

(Dn + k1Dn1 + ….+ kn) y = X,

i.e. f(D)y = X,

where f(D) = Dn + k1Dn1+… + kn, i.e. a polynomial in D. Thus the symbol D stands for the operation of differential and can be treated much the same as an algebraic quantity i.e. f(D) can be factorised by ordinary rules of algebra and the factors may be taken in any other. For instance

d2y

dx2 + 2dy

dx 3y = (D2 + 2D 3) y

= (D + 3)(D 1)y or (D 1) (D + 3) y Rules for finding the complementary function

To solve the equation dny

dxn + k1 dn−1y

dxn−1 + k2 dn−2

dxn−2 + …+ kny =

0 ….(i) where k's are constants. The equation (i) in symbolic form is

(Dn + k1Dn1 + k2 Dn2 +….+ kn)y= 0 ….(ii) Its symbolic co-efficient equated to zero i.e.

Dn + k1Dn1 + k2Dn2 + …+ kn = 0 is called the auxiliary equation (A. E.). Let m1, m2,…., mn be its roots. Now 4 cases arise. Case I. If all the roots be real and different. then (ii) is equivalent to

(D m1) (D m2) …. (D mn)y = 0 ….(iii)

Now (iii) will be satisfied by the solution of (D mn)y = 0,

i.e. by dy

dx mny = 0.

This is a Leibnitz’s linear and l.F. = e−mnx

Its solution is ye−mnx = cn, i.e. y = cnemnx Similarly, since the factors in (iii) can be taken in any order, it will be satisfied by the solutions of

(D m1)y = 0, (D m2) = 0 etc., i.e. by y = c1emnx etc. Thus the complete solution of (i) is y = c1em1x + c2em2x+ … + cnemnx …(iv)

Case II. If two roots are equal (i.e. m1 = m2), then (iv) becomes y = (c1 + c2) em2x + c3em1x+ …. + cnemnx y = Cem1x + c3em3x + … + cnemnx

[ c1 + c2 = one arbitrary constant C] It has only n - 1 arbitrary constants and is, therefore, not the complete solution of (i). In this case, we proceed as follows: The part of the complete solution corresponding to the repeated root is the complete solution of

(D m1) (D m1) y = 0

Putting (D – rn1) y = z. it becomes (D m1) z = 0 or dz

dx

m1z = 0 This is Leibnitz's linear in z and I.F. = e−m1x

its solution is ze−m1x = c1 or z = c1em1x

Thus (D m1) y = z = c1em1x or dy

dx m1y = c1em1x ….(v)

Its l.F. being e−m1x, the solution of (v) is

ye−m1x = ∫ 𝑐1em1xdx + c2

y = (c1x + c2) em1x Thus the complete solution of (i) is y = (c1x + c2) em1x + c3em3x + …+ cnemnx If, however, the A.E. has three equal roots (i.e. m1 = m2 = m3), then the complete solution is y = (c1x2 + c2x + c3) em1x + c4em4x + … + Cnemnx Case III. If one pair of roots be imaginary , i.e.

m1 = + i,

m2 = i, then the complete solution is

y = c1e( + ) + c2e( )x+ c3em3x + … + cnemnx

= ex (c1eix + c2e−ix

) + c3em3x + … + cnemnx

= ex [c1 (cos x + i sin

x) + c2(cos x i sin x)] + c3em3x + … + cnemnx

[ by Euler’s

Theorem, ei = cos + i sin ]

= ex [c1 cos x + c2 sin x_ + c3em3x + …+ cnemnx where C1 = c1 + c2

and C2 = i(c1 c2) Case IV. If two pair of imaginary roots be equal i.e.

m1 = m2 = + i,

m3 = m4 = i, then by case II, the complete solution is

y = ex [(c1x + c2)cos x +

(c3x + c4) sin x] + … + cnemnx. Inverse Operator

Definition, 1

f(D) X is that function of x, not containing

arbitrary constants, which when operated upon by f(D) gives X.

i.e. f(D){1

f(D)} = X

Thus 1

f(D)X satisfies the equation f(D)y = X and is,

therefore, its particular integral. Obviously f(D) and 1/f(D) are inverse operators.

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1

D X = ∫ Xdx

Let 1

Dx = y

i.e. X = dy

dx

integrating w.r..t. x, y = ∫ Xdx

Thus 1

DX = ∫ Xdx

1

D−aX = eax ∫ Xe−axdx

Let 1

D aX = y.

Operating by D a,

(D a). 1

D a X = (D a) y.

or X = dy

dx ay, i.e.

dy

dx ay = X

which is a leibnitz’s linear equations.

l.F. being eax, its solution is

yeax = ∫ Xe−axdx, no constant being added as (ii) doesn’t contain any constant.

Thus, 1

D−aX = y = eax ∫ 𝑋e−ax dx.

TWO OTHER METHODS OF FINDDING P.I. Method of Variation of Parameters This method is quite general and applies to equations of the from

yn + py + qy = X ….(i) where p, q, and X are functions of x. It gives

P.l. = y1 ∫y2X

Wdx +

y2∫y1X

Wdx ….(ii)

where y1 and y2 are the solutions of y” + py + qy = 0

and W = |y1 y2

y1′ y2

′ | is called the wronskian of y1, y2.

Questions: 1. What is the derivative of 𝑓(𝑥) = |𝑥| at = 0 ?

(A) 1 (B) −1 (C) 0 (D) does not exist

2. The minimum point of the function 𝑓(𝑥) = (𝑥3

3) − 𝑥

is at (A) 𝑥 = 1 (B) 𝑥 = −1

(C) 𝑥 = 0 (D) 𝑥 =1

√3

3. Which of the following functions is not differentiable in the domain [−1, 1]? (A) 𝑓(𝑥) = 𝑥2

(B) 𝑓(𝑥) = 𝑥 − 1

(C) 𝑓(𝑥) = 2

(D) 𝑓(𝑥) = Maximum (𝑥, −𝑥)

4. The solution of the differential equation 𝑑𝑦

𝑑𝑥+ 𝑦2 =

0 is

(A) 𝑦 =1

𝑥+𝑐

(B) 𝑦 =−𝑥3

3+ 𝑐

(C) 𝑐𝑒𝑥 (D) Unsolvable as equation is non‐ linear

5. If 𝑥 = 𝑎(𝜃 + sin 𝜃) and 𝑦 = 𝑎(1 − cos 𝜃) , then 𝑑𝑦

𝑑𝑥

will be equal to

(A) sin (𝜃

2) (B) cos (

𝜃

2)

(C) tan (𝜃

2) (D) cot (

𝜃

2)

Data question 6 and 7 The complete solution of the ordinary differential equation

𝑑2𝑦

𝑑𝑥2+ 𝑝

𝑑𝑦

𝑑𝑥+ 𝑞𝑦 = 0 is 𝑦 = 𝑐1𝑒−𝑥 + 𝑐2𝑒−3𝑥

6. Then 𝑝 and 𝑞 are (A) 𝑝 = 3, 𝑞 = 3

(B)𝑝 = 3, 𝑞 = 4

(C) 𝑝 = 4, 𝑞 = 3

(𝐷) 𝑝 = 4, 𝑞 = 4

7. Which of the following is a solution of the differential equation

𝑑2𝑦

𝑑𝑥2+ 𝑝

𝑑𝑦

𝑑𝑥+ (𝑞 + 1)𝑦 = 0

(A) 𝑒−3𝑥 (B) 𝑥𝑒−𝑥

(C) 𝑥𝑒−2𝑥 (D) 𝑥2𝑒−2𝑥

8. If 𝑥2 𝑑𝑦

𝑑𝑥+ 2𝑥𝑦 =

2 ln (𝑥)

𝑥 and 𝑦(1) = 0, then what is

(𝑒) ?

(A) 𝑒 (B) 1 (C) 1/𝑒 (D) 1/ 𝑒2

9. By a change of variable 𝑥(𝑢, 𝑣) = 𝑢𝑣, 𝑦(𝑢, 𝑣) = 𝑣𝑙𝑢

is double integral, the integrand 𝑓(𝑥, 𝑦) changes to (𝑢𝑣, 𝑣𝑙𝑢)𝜑(𝑢, 𝑣) . Then, 𝜑(𝑢, 𝑣) is (A) 2𝑣/𝑢 (B) 2uv (C) 𝑣2 (D) 1

10. For 𝑑2𝑦

𝑑𝑥2 + 4𝑑𝑦

𝑑𝑥+ 3𝑦 = 3𝑒2𝑥, the particular integral

is

(A) 1

15𝑒2𝑥 (B)

1

5𝑒2𝑥

(C) 3𝑒2𝑥 (D) 𝐶1𝑒−𝑥 + 𝐶2𝑒−3𝑥

11. The solution of the differential equation 𝑑𝑦

𝑑𝑥+ 2𝑥𝑦 =

𝑒−𝑥2 with 𝑦(𝑂) = 1 is

(A) (1 + 𝑥)𝑒+𝑥2

(B) (1 + 𝑥)𝑒−𝑥2

(C) (1 − 𝑥)𝑒+𝑥2

(D) (1 − 𝑥)𝑒−𝑥2

12. The solution of 𝑑𝑦

𝑑𝑥= 𝑦2 with initial value 𝑦(𝑂) = 1

bounded in the interval (A) −∞ ≤ 𝑥 ≤ ∞

(B) −∞ ≤ 𝑥 ≤ 1

(C) 𝑥 < 1, 𝑥 > 1

(D) −2 ≤ 𝑥 ≤ 2

13. The partial differential equation 𝜕2𝜙

𝜕𝑥2 +𝜕2𝜙

𝜕𝑦2 +𝜕𝜙

𝜕𝑥+

𝜕𝜙

𝜕𝑦= 0 has

(A) Degree 1 order 2

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(B) degree 1 order 1 (C) degree 2 order 1 (D) degree 2 order 2

14. If 𝜙(𝑥, 𝑦) and 𝜓(𝑥, 𝑦) are functions with continuous

second derivatives, then 𝜙(𝑥, 𝑦) + 𝑖𝜓(𝑥, 𝑦) can be expressed as an analytic ffinction 𝑜𝑓𝑥 + 𝑖𝜓(𝑖 =

√−1) , when

(A) 𝜕𝜙

𝜕𝑥= −

𝜕𝜓

𝜕𝑥,

𝜕𝜙

𝜕𝑦=

𝜕𝜓

𝜕𝑦

(B) 𝜕𝜙

𝜕𝑦= −

𝜕𝜓

𝜕𝑥,

𝜕𝜙

𝜕𝑥=

𝜕𝜓

𝜕𝑦

(C) 𝜕2𝜙

𝜕𝑥2+

𝜕2𝜙

𝜕𝑦2=

𝜕2𝜓

𝜕𝑥2+

𝜕2𝜓

𝜕𝑦2= 1

(D) 𝜕𝜙

𝜕𝑥+

𝜕𝜙

𝜕𝑦=

𝜕𝜓

𝜕𝑥+

𝜕𝜓

𝜕𝑦= 0

15. It is given that 𝑦′′ + 2𝑦′ + 𝑦 = 0, 𝑦(𝑂) = 0, 𝑦(1) =

0. What is (𝑂. 5) ?

(A) 0 (B) 0.37 (C) 0.62 (D) 1.13

16. Let 𝑓 = 𝑦𝑥 What is 𝜕2𝑓

𝜕𝑥𝜕𝑦 at 𝑥 = 2, 𝑦 = 1 ?

(A) 0 (B) ln2

(C) 1 (D) 1

ln 2

17. Given that �̈� + 3𝑥 = 0, and 𝑥(𝑂) = 1, �̇�(0) = 0,

what is 𝑥(1) ?

(A) −0.99 (B) −0.16 (C) 0.16 (D) 0.99

18. The solution of 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 = 𝑥4 with the condition

𝑦(1) =6

5 is

(A) 𝑦 =𝑥4

5+

1

𝑥

(B) 𝑦 =𝑥4

5+ 1

(C) 𝑦 =4𝑥4

5+

4

5𝑥

(D) 𝑦 =𝑥5

5+ 1

19. An analytic function of a complex variable 𝑧 = 𝑥 +

𝑖𝑦 is expressed as 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) where

𝑖 = √−1. If 𝑢 = 𝑥𝑦, the expression for 𝑣 should be

(A) (𝑥+𝑦)2

2+ 𝑘 (B)

𝑥2−𝑦2

2+ 𝑘

(C) 𝑦2−𝑥2

2+ 𝑘 (𝐷)

(𝑥−𝑦)2

2+ 𝑘

20. The Blasius equation, 𝑑3𝑓

𝑑𝜂3 +𝑓

2

𝑑2𝑓

𝑑𝜂2 = 0, is a

(A) second order nonlinear ordinary differential equation (B) third order nonlinear ordinary differential equation (C) third order linear ordinary differential equation (D) mixed order nonlinear ordinary differential equation

21. Consider the differential equation 𝑑𝑦

𝑑𝑥= (1 + 𝑦2)𝑥.

The general solution with constant 𝑐 is

(A) 𝑦 = tan 𝑥2

2+ tan 𝑐

(B) 𝑦 = tan2(𝑥

2+ 𝑐)

(C) 𝑦 = tan2(𝑥

2) + 𝑐

(D) 𝑦 = tan (𝑥2

2+ 𝑐)

22. A series expansion for the function sin 𝜃 is

(A) 1 −𝜃2

2!+

𝜃4

4!− ⋯

(B) 1 + 𝜃 +𝜃2

2!+

𝜃3

3!+ ⋯

(C) 𝜃 −𝜃3

3!+

𝜃5

5!− ⋯

(D) 𝜃 +𝜃3

3!+

𝜃5

5!+ ⋯

23. Consider the differential equation 𝑥2(𝑑2𝑦/𝑑𝑥2) +

𝑥(𝑑𝑦/𝑑𝑥) − 4𝑦 = 0 with the boundary conditions of 𝑦(𝑂) = 0 and 𝑦(1) = 1. The complete solution of the differential equation is (A) 𝑥2

(B) sin (𝜋𝑥

2)

(C) 𝑒𝑥 sin (𝜋𝑥

2)

(D) 𝑒−𝑥 sin (𝜋𝑥

2)

Answers

1 D

2 A

3 D

4 A

5 C

6 C

7 C

8 D

9 A

10 B

11 B

12 C

13 A

14 B

15 A

16 C

17 D

18 A

19 C

20 B

21 D

22 C

23 A