Ordinary DEQ's Notes

October 31, 2010

Copyright © 2010 by A. R. Saleem.

ALI RAZA SALEEM2008-EE-17Section: A

Email: [email protected]

Univeristy of Engineering & Technology Lahore1, 54890

Punjab, Pakistan

1www.uet.edu.pk

2

mailto:[email protected]

http://www.uet.edu.pk

Preface

These notes are my gratitude toMadam Samina, one of the best teacher,i have in my prestigious institution. Without whom i may not be able towrite and compile these notes. These are Dierential Equation notes, themost important tool in the eld of science and engineering. It is very use-ful particularly for engineering students because they are mostly in searchof tricks or techniques to solve problems. In these notes, the material isresearched, and the cited one.

Enjoy it.................

3

In nature, many processess are characterized in terms of derivatives and dier-entials. The mathematical equations containing these derivatives and dierentialsare called Dierenitial Equations (DEQ's). There are some methods to solvesthese equations so that you can nd the original function. Dierential equationshave many applications in engineering, physics, economics, medicines and in manyother discipline.There are two main types of dierential equations as followed:

Ordinary Dierential Equations:

An ordinary dierential equation ODE2 is a dierential equation inwhich the unknown function (also known as the dependent variable) isa function of a single independent variable i.e. y = f (x).

For example:

1. dydx = 2x +c

2. d2ydx2 + cy = 0.

where c is a constant.

Partial Dierential Equations:

A partial dierential equation (PDE) is a dierential equation inwhich the unknown function is a function of multiple independent vari-ables and the equation involves its partial derivatives, i.e. y = f (x, t)oru = f(x,y).

For example:

1. σ2uσx2 + σ2u

σy2 = 0

2. σ2yσt2 = α2 σ

2yσx2

Everything in this world has some purpose and identication. Similarly, ODEhas degree and order .

Order: The highest order derivative in the equation.

Degree:The power of the highest order derivative in the equation.

2We 'll discuss only ordinary dierential equations (ODE).

4

DEQ'S Degree Order(d3ydx3

)2+ 5

(d2ydx2

)+ ex = x 2 3

y1(

d3ydx3

)2+ 5

(dydx

)+ x = x 3 3

x1(

d3ydx3

)1+ 5dy

dx + x = ex 1 3

3ODE's are classied according to their order and they are further classiedaccording to their method of solution as following:

Ordinary Dierential Equations (ODE's)

1st Order 2nd Order Higher OrderVariable Separable Constant Coecient Variable Coecient By Synthetic DivisionHomogeneous Homogeneous Homogeneous By Factorization

Linear Non-homogeneous Non-homogeneousExact

Part I

1st Order DEQ's

1 Variable Separable

If DEQ is of the type:

f(x)dx + f(y)dy = 0

The solution will be:

ˆf(x)dx +

ˆf(y)dy = c

3In these notes, solutions to ODE′s are explained with the help of examples.

5

Example: Solve

(x + 1)dy

dx= x

(y2 + 1

)

Separating terms, we get;

⇒ dyy2+1 = xdx

(x+1)

Now, Intergrating both sides;

⇒´

dyy2+1 =

´xdx

(x+1)

´dy

y2+1 =´ (

x+1−1x+1

)dx =

´ (1− 1

x+1

)dx

tan−1y = x− ln(x + 1) + c

2 Homogeneous DEQ's

Such DEQ's in which each term has the same degree are calledHomogeneous DEQ's.

Consider

dy

dx=

f (x,y)

g (x,y)

Put y = Vx where V is dierentiable.Then,

dy

dx= V+x

dV

dx

and simplify!

6

Example: Solve

(x2 + y2

)dx + 2xydy = 0

⇒dydx = − (x2+y2)

2xy

Put y = Vx, we get;

⇒V + xdVdx = − (x2+V2x2)

2Vx2 =(1+V2)

2V

⇒xdVdx = − (1+3V2)

2V

⇒ 2VdV1+3V2 = −dx

x

Integrating both sides, we get;

⇒´

2VdV1+3V2 = −

´dxx

⇒13 ln

(1 + 3V2

)= −lnx + lnc

⇒(1 + 3V2

) 13 x = c

Putting V = yx , we get;

⇒(

1 +3y2

x2

) 13

x = c

2.1 Equations Reducible to homogeneous

As,

dy

dx=

f (x,y)

g (x,y)

Now, it becomes of the type followed as:

(a1x + b1y + c1) + (a2x + b2y + c2) = 0

7

Consider

l1 : (a1x + b1y + c1)

l2 :(a2x + b2y + c2)

There are two cases:

if (l1 ⊥ l2) if (l1 || l2)a1

a26= b1

b2

a1

a2= b1

b2

Put x = X + h and y = Y + k Put z = a1x + b1ywhere h and k are constants. Just simplify!

2.1.1 if (l1⊥l2)

Example:

(−x + y − 3)dy + (x + y + 1)dx = 0

⇒dydx = − (x+y+1)

(−x+y−3)

As, a1

a26= b1

b2. Therefore,

Put x = X + h and y = Y + k, we get;

dYdX = − (X+h+Y+k+1)

(−X−h+Y+k−3) =(X+h+Y+k+1)(X−Y+h−k+3)

For homogeneous, we have;

h + k + 1 = 0;

h− k + 3 = 0;

⇒h = −2 and k = 1

⇒dYdX=X+Y

X−Y

Put Y = Vx, we get;

⇒V + XdVdX = 1+V

1−V

⇒XdVdX = 1+V2

1−V

⇒dV(1−V)(1+V2) = dX

X

8

Integrating both sides;

⇒´ dV(1−V)

(1+V2) =´

dXX

⇒tan−1V − 12 ln

(1 + V2

)= lnX + lnc

⇒tan−1 YX − ln

√1 + Y2

X2 = lnX + lnc

Putting this in its original form, we get;

⇒ tan−1(

y − 1

x + 2

)− ln

√1+

(y − 1)2

(x + 2)2 = ln (x + 2) + c

2.1.2 if (l1 || l2)

Example

(2x + y + 1)dx + (4x + 2y − 1)dy =0

Put z = 2x + y

dz = 2dx + dy

⇒ dy = dz− 2dx

Thus,

(z + 1)dx + (2z− 1)dy = 0⇒ (z + 1)dx + (2z− 1) (dz− 2dx) =0

⇒(2z− 1)dz = − (3z− 3)dx

⇒ dxdz = − (2z−1)

3(z−1)

⇒3´

dx = −´ (2z−1)

(z−1) dz

⇒ 3x = zln (1− z)− ln (1− z) + lnc

⇒ 3x = (z− 1) ln (1− z) + lnc

Put z = 2x + y, we get;

3x = (2x + y − 1) ln (1− 2x− y) + lnc

9

3 Linear DEQ's

A DEQ of the form:

(dy

dx

)+ p(x)y1 = Q(x)

or

(dx

dy

)+ p(y)x1 = Q(y)

How to Solve it! First of all, we 'll nd the (I.F.)Integrating factor.Then, multiplyyour equation by (I.F.). The L.H.S. of the equation becomes the derivative of theproduct.And the last one is to take the integral on bothe sides and simplify yourexpression.

Example:

dy

dx+ y =ex

Here we have;

p(x) = 1 and I.F. =e´p(x)dx=e

´1.dx=ex

Now,multiplying on both sides;

ex dydx + exy = ex.ex

The L.H.S. becomes the derivative of (ex.y), we get;

ddx (ex.y) =e2x

Taking Integral on both sides:

´d (ex.y) =

´e2xdx⇒ex.y =e2x

2 + c

Hence,

y =ex

2+ ce−x

10

3.1 Reduction to Linear DEQ's

A DEQ of the following form:

dy

dx+ p(x)y = Q(x)yn

where nεR

It becomes linear if n = 0 or n = 1. Suppose n 6= 0 or n 6= 1.Then divide theabove equation by ynon both sides:

y−n dydx + p(x).y(1−n) = Q(x)

Put

V = y(1−n)

We get;

⇒dVdx = (1− n)y−n dy

dx

⇒ 1(1−n) .

dVdx + p(x).V = Q(x)

⇒ dVdx + (1− n)p(x)V = Q(x)(1− n)

Further it can be solved by the method descirbed above.

Example:

dy

dx+ y = xy3

It becomes linear if n = 0 or n = 1. Suppose n 6= 0 or n 6= 1.Then divide theabove equation by yn on both sides:

⇒y−3 dydx+y−2= x

Put

V = y−2

⇒ dVdx = −2y−3 dy

dx

Then,

11

−12 .

dVdx + V = x⇒ dV

dx − 2V = −2x

Here we have;

p(x) = −2 and I.F. =e−´2dx =e−2x

⇒ e−2x.dVdx − 2V.e−2x = −2x.e−2x

⇒ ddx (e

−2x.V) = −2x.e−2x


⇒´

ddx (e

−2x.V) = −´

2x.e−2x

⇒e−2x.V = −2´

xe−2xdx = −2(−1

2xe−2x − 14e−2x

)+ c

⇒ e−2x.V = xe−2x + e−2x

2 + c

V =(x + 1

2

)+ ce2x

Put V = y−2. Then,

1

y2=

(x +

1

2

)+ ce2x

4 Exact DEQ's

A DEQ is of the form:

M(x,y)dx + N(x,y)dy = 0

is said to be exact if and only if:

σMσy = σN

σx .

How to solve!

M(x,y)dx +´(TermsInN(x,y)FreeOfx) = 0

12

Example:

(x + y)dx + (x− y)dy = 0

Here we have;

⇒M(x,y) = x+y and N(x,y) = x− y

As,

My = σMσy = 1 and Nx = σN

σx = 1

My =Nx

⇒´(x + y)dx +

´(−y)dy = 0

Hence,

x2

2+ xy − y2

2= c

Example: If

My 6= Nx

Lets take an example, we have:

xdy − ydx = −x3dx

My = −1 and Nx = 1

⇒My 6= Nx

Then you must follow the following procedure:

My−Nx

N = p(x)⇒ I.F.=e´p(x)dx

Multiplying on both sides so that it becomes an exact. Moving on the example:

p(x) =My−Nx

N = −1−1x = −2

x

Here we have;

13

I.F. = e´p(x)dx= e−2

´1xdx=e−2lnx =elnx−2

= 1x2

Then, multiplying both sides, we get;

(x3−y)x2 dx+dy

x2 = 0⇒´ (

x− yx2

)dx +

´dyx = c

⇒´ (

x− yx2

)dx +

´dy = c..............4

Hence,

x2

2+

y

x+ y = c

Initial and Final value Problems

→While solving 1stOrderDEQ′s, we get one arbitrary constant. The valueof the constant can be determined by the initial condition provided to you inthe problem.

→While solving 2ndOrderDEQ′s and HigherOrderDEQ′s requires num-ber of conditions equal to the 5Order of the DEQ.

Part II

2nd Order DEQ's

2nd Order DEQ's are the most important one because many phenomenon in thenature and many physical systems are characterized by such equations. We 'll grad-ually dive in and will solve such equations with the help of marvellous techniques.

Homogeneous DEQ's-Constant Coecients:

A (DEQ) of the form:

aodny

dxn+a1

dn−1y

dxn−1 + .....+an−1dy

dx+any = 0

4We 'll consider only´dy instead of

´ (dyx

)because we want´

(TermsInN(x,y)FreeOfx)dy.5

An nthOrderDEQ must have n− InitialConditions

14

For n = 2, we get;

aod2ydx2 + a1

dydx + a2y = 0

⇒d2ydx2 + a1

ao

dydx + a2

aoy = 0

Let

ddx = D ⇒D2y + a1

aoDy + a2

aoy = 0

⇒ y(D2 + a1

aoD + a2

ao

)= 0⇒ f(D)y = 0

Either f(D) = 0 or y = 0. So, we take f(D) = 0

⇒(

D2 +a1

aoD +

a2

ao

)= 0

.

Then we 'll nd the roots of the above equation. These roots correspond to r1and r2.

It is known as Auxiliary or Characteristic equation. There are dierenttypes of roots, depending upon their nature. Let r1and r2 be the roots of the aboveequation. yc denotes the solution to homogeneous equation which is normally calledComplementarySolution.

Cases Types Of Roots Solution

1. Real and Distinct(r1,r2) yc = Aer1x + Ber2x

2. Real and Equal(r1 = r2 = r) yc = (A + Bx) erx

3. Complex(α± iβ) yc = eαx [Acosβx + Bsinβx]4. Imaginary(±iβ) yc = Acosβx + Bsinβx

Non-Homogeneous DEQ's-Constant Coecients:

A DEQ of the form:

aod2ydx2 + a1

dydx + a2y = f(x)

⇒d2ydx2+

a1

ao.dydx+

a2

aoy = f(x)

ao

In case of such DEQ's, we have the solution of the following type:

15

y=yc+yp

yc ⇒6Complementary Solution/ Transient Response/ Zero-Input Responseor Natural Response of the system which you have modelled.

yp ⇒7Particular Solution/ Steady State Response/ Zero-Output Response orForced Response.

Depending upon the excitation(Input), we have 5 dierent cases as followed:

Case Excitation f(x)

1. eαx

2. xn

3. sinαx or cosαx4. eαxxn

5. eαxsinβx or eαxcosβx

If f (x) = eαxwhere α is a constant

1. Homogeneous Solution/ Complementary Solution yc can be found easily asdescribed above.

2. Particular Solution: yp = eαx

f(D) ; Replace D by α and solve. If f(D) becomes

0. Then multiply the numerator with x and dierentiate the denominatorw.r.t D. Apply the rule again and so on.

Example:

d2y

dx2− 5

dy

dx+ 6y = ex

There are two parts of solution:

Complementary Solution (yc)

For that we 'll put:

6For stability of the system, we check the roots of the yc.7It is a function of both the system and the excitation(Input) but independent of the initial

conditions.

16

f(x) = 0

⇒d2ydx2 − 5dy

dx + 6y = 0

Put

ddx = D⇒ D2y − 5Dy + 6y = 0⇒ (D2 − 5D + 6)y = 0

⇒ f(D)y = 0

Either

f(D) = 0 or y = 0

So, we take;f(D) = 0.

⇒D2 − 5D + 6 = 0

We get;

D = 3,2

i.e. r1 = 3 and r2 = 2.

∴ yc =Ae3x + Be2x

Particular Solution (yp)

⇒yp = f(x)f(D) =

ex

f(D) =ex

D2−5D+6

Replace D by (α = 1), we get;

⇒ yp = ex

12−5(1)+6 = ex

2

Then,

y = yc + yp

Since,

∴ y =Ae3x + Be2x +ex

2

17

If f (x) = xn where n = 0 or nεZ+


2. Particular Solution: yp = xn

f(D) = xn[f(D)]−1, Then apply the Binomial Se-

ries Expansion, expand it upto n− terms and simplify!

Example:

d2y

dx2− 2

dy

dx+ y = x2 + 2x− 4



For that we 'll put:f(x) = 0

⇒ d2ydx2 − 2dy

dx + y = 0

Put

ddx = D⇒ D2y − 2Dy + y = 0⇒ (D2 − 2D + 1)y = 0

⇒ f(D)y = 0

Either

f(D) = 0 or y = 0

So, we take

f(D) = 0

⇒ D2 − 2D + 1 = 0

⇒ (D− 1)2= 0

We get;

⇒ D = 1,1

i.e. r = r1 = r2 = 1

∴ yc = (A + Bx) ex

18


⇒ yp = f(x)f(D)

⇒ x2+2x−4D2−2D+1=

(x2 + 2x− 4

).(1−D)−2

Applying Binomial Expansion, we get;

⇒(x2 + 2x− 4

).(1−D)−2 =

(x2 + 2x− 4

).(1 + (−2)(−D)

1! + (−2)(−2−1)(−D)2

2!

).................8

⇒(x2 + 2x− 4

).(1 + 2D

1! + 6D2

2!

)=[

x2 + 2x− 4 + 2D(x2 + 2x− 4

)+ 3D2

(x2 + 2x− 4

)]=x2 + 2x− 4 + 2 (2x + 2) + 3(2) = x2 + 6x + 6

Thus,∴ yp = x2 + 6x + 6

Then,

y = yc + yp = (A + Bx) ex + x2 + 6x + 6

If f(x) = sinαx or cosαx where α is a constant


2. Particular Solution: yp = sinαxf(D) or cosαx

f(D) ; Replace D2by(−α2

). If f(D)

becomes 0, case fails. Then multiply the numerator with x and dierentiatethe denominator w.r.t. D. Apply the rule again and so on.

8As the degree of polynomial (x2 + 2x− 4) is 2 so we 'll expand (1−D)−2upto D2.

19

Example:

d2y

dx2+ y = cosx




f(x) = 0

⇒ d2ydx2 + y = 0

Put

ddx = D⇒ D2y + y = 0⇒ y(D2 + 1) = 0

⇒ f(D)y = 0

Either

f(D) = 0 or y = 0

So, we take

f(D) = 0

D2 + 1 = 0⇒ D2 = −1⇒ D = ±i

i.e., r1 = i and r2 = −i.Here

β = 1

So,

∴ yc = Acosx + Bsinx


⇒yp = f(x)f(D)=

cosxD2+1

20

Replace D2 by (−α2)= (−12)= −1.

We get;

⇒ cosxD2+1 = cosx

−1+1 =∞

Here case fails.

Now, multiply numerator with x and dierentiate the denominator w.r.t. D, weget:

⇒ xcosx2D = xcosx

2D2 .D

Applying rule again, we get:

⇒ xcosx2D2 .D = xcosx.D

2(−12) = −x2D(cosx)

⇒ −x2D(cosx) = −x

2 .(−sinx) = x2sinx.................9

∴ yp =x

2sinx

Then,

y = yc + yp = yc = Acosx + Bsinx +x

2sinx

If f(x) = eαxxn where α = contant and n = 0 or nεZ+


2. Particular Solution: yp = eαxxn

f(D) ; Replace D by (D + α), we get;

⇒ eαxxn

f(D) = eαxxn

f(D+α) = eαxxn[f(D + α)]−1. Apply Binomial Series Expan-

sion, expand it upto n− terms and operate on xn!

Example:

9

We 'll dierentiate cosx only. If it would be sinx the same thing will be done.

21

d2y

dx2− 5

dy

dx= xe5x

There are two parts of solutions:



f(x) = 0⇒d2ydx2 − 5dy

dx = 0

Put

ddx = D⇒ D2y − 5Dy = 0⇒ y(D2 − 5D) = 0

⇒ f(D)y = 0

Either

f(D) = 0 or y = 0

So, we take;

f(D) = 0

⇒ D2 − 5D = 0⇒ D(D− 5) = 0

We get;

D = 0,5

i.e., r1 = 0 and r2 = 5.

∴ yc= A + Be5x


⇒ yp = f(x)f(D)=e5x x

D2−5D

Replace D by (D + α) = (D + 5).Then,

22

⇒ e5x xD2−5D = e5x x

(D+5)2−5(D+5)=e5x x

D2+5D =e5x x

5D(1+D5 )

=e5x

5D .x

(1+D5 )

⇒ e5x

5D .x

(1+D5 )

= e5x

5D x(1 + D

5

)−1As the degree of polynomial x is 1. Then we 'll expand

(1 + D

5

)−1upto one

term only!

⇒e5x

5D .x(1 + D

5

)−1=e5x

5D .x(1 + (−1)D5

)= e5x

5D .x(1− D

5

)Now, we 'll operate on x, we get;

∵ 1D =

´dx

⇒ e5x

5D

(x−Dx

5

)= e5x

5D .(x− 1

5

)= e5x

5 .(

x2

2 −x5

)= e5xx2

10 − xe5x

25

∴ yp =e5xx2

10− xe5x

25

Then,

y = yc + yp = A + Be5x +e5xx2

10− xe5x

25

If f(x) = eαxsinβx or eαxcosβx where α and β are constants.


2. Particular Solution: yp = eαxsinβxf(D) or eαxcosβx

f(D) ; Replace D by (D + α).

Then, replacing D2 by(−β2

). Just Simplify!

23

Example:

d2y

dx2+ 6

dy

dx+ 73y = 80excos4x




f(x) = 0⇒ d2ydx2 + 6dy

dx + 73y = 0

Put

ddx = D⇒ D2y + 6Dy + 73y = 0⇒ y(D2 + 6D + 73) = 0

⇒ f(D)y = 0

Either

f(D) = 0 or y = 0.

So, we take;

f(D) = 0

⇒ D2 + 6D + 73 = 0

⇒ D = −6±√255i

2 = −3±√2552 i = α± iβ

∴ yc = e−3x

(Acos(

√255

2)x + Bsin(

√255

2)x

)


⇒ yp = f(x)f(D)=

80excos4xD2+6D+73

Replace D by (D + α) = (D + 1), we get;

⇒ 80excos4xD2+6D+73 = 80excos4x

(D+1)2+6(D+1)+73 = 80excos4xD2+8D+80

Now, replacing D2 by(−β2

)= (−42) = −16. We get;

24

⇒ 80excos4xD2+8D+80 = 80excos4x

−16+8D+80 =80excos4x8(D+8)

Rationalizing by (D− 8), we get;

⇒80excos4x8(D+8) =80excos4x

8(D+8) . (D−8)(D−8) =10excos4x(D−8)

D2−64

Replacing D2 by(−β2

)= (−42) = −16. We get;

⇒10excos4x(D−8)D2−8 = 10excos4x(D−8)

−16−64 = 10excos4x(D−8)−80 = −1

8excos4x(D− 8)

Now, operating on cos4x, we get;

−18excos4x(D− 8) = −1

8ex[−4sin4x− 8cos4x] = ex

2 sin4x + excos4x = ex(12sin4x + cos4x

)

∴ yp = ex

(1

2sin4x + cos4x

)Then,

y = yc + yp =e−3x

(Acos(

√255

2)x + Bsin(

√255

2)x

)+ ex

(1

2sin4x + cos4x

)

10Variation of Parameters

There is another method to solve 2ndOrder non-homogeneous DEQ′s.The pro-cedure to solve such type of DEQ's of the form:

aod2ydx2 + a1

dydx + a2y = f(x)

⇒d2ydx2 + a1

ao.dydx+

a2

aoy= f(x)

ao

Solution will be of the form:

y = u1v1 + u2v2

Solve the associated homogeneous equation:

10This method becomes handy in some probelms.

25

⇒ D2 + a1D + a2 = 0

where r1and r2 are the roots of the above equation.

Let

u1 = er1x and u2 = er2x

Apply: D =

∣∣∣∣ u1 u2up1 up2

∣∣∣∣ = u1up2 − u2up

1

For v1: v1 = −´ u2(x)f(x)

D dx

For v2: v2 =´ u1(x)f(x)

D dxThus, y = u1v1 + u2v2

Example:

d2y

dx2+ 2

dy

dx− 3y = 6

Solution will be of the form:

y = u1v1 + u2v2

Solve the associated homogeneous equation and for that we 'll put:

f(x) = 0

⇒ d2y

dx2+ 2

dy

dx− 3y = 0

Put

ddx = D⇒ D2y + 2Dy − 3y = 0⇒ (D2 + 2D− 3)y = 0

⇒ f(D)y = 0

Either

f(D) = 0 or y = 0

26

So, we take

⇒f(D) = 0

D2 + 2D− 3 = 0⇒ D = −3,1

i.e., r1 = −3 and r2 = 1Let

u1 = er1x = e−3x and u2 = er2x = ex

Apply:

D =

∣∣∣∣ e−3x ex

−3e−3x ex

∣∣∣∣ = e−2x + 3e−2x = 4e−2x

For v1:

v1 = −´

ex64e−2x dx = −3

2

´e3xdx = −3

2 .e3x

3 + c1 = −e3x

2 + c1

For v2:

v2 =´

e−3x64e−2x dx = 3

2

´e−xdx = 3

2 .e−x

−1 + c2 = −32e−x + c2

Then,

y = u1v1 + u2v2 = e−3x(−e3x

2 + c1

)+ ex

(−3

2e−x + c2)= −2 + c1e−3x + c2ex

Thus,

∴ y = −2 + c1e−3x + c2ex

Part III

Higher Order DEQ's

The methods of solving DEQ's can also be extended to equations of 11higher order

DEQ's12. The characteristic or homogeneous algebraic equation associated withthe DEQ:

11Higher Order non-homogeneous DEQ's can be solved in the same way as we did for the2ndOrderDEQ′s but it would be tedious.

12Higher Order DEQ's are mostly solved with the help of LaplaceTransform.

27

aodnydxn + a1

dn−1ydxn−1 + .......+ an−1

dydx + any = 0

⇒ dnydxn + a1

ao

dn−1ydxn−1 + .......+ an−1

ao

dydx + an

aoy = 0

is

ddx = D⇒

(Dn + a1

a0Dn−1 + .......+ an−1

aoD + an

ao

)y = 0(

Dn + a1

a0Dn−1 + .......+ an−1

aoD + an

ao

)= 0

Let r1, r2, .......rnbe the roots of the above equation. The same technique, we'll apply here as in the above cases which have discussed.

Example:

d4y

dx4− 3

d3y

dx3+ 3

d2y

dx2− dy

dx= 0

Put,

ddx = D⇒ (D4 − 3D3 + 3D2 −D)y = 0

⇒ D4 − 3D3 + 3D2 −D = 0⇒ D(D− 1)3

The roots of the characteristic equation are:

r1 = 0 and r2 = r3 = r4 = 1

Thus the solution is:

⇒ya = C1er1x = C1e0x = C1

⇒ yb =(C2 + C3x + C4x2

)ex

Hence,

y = ya + yb = C1 +(C2 + C3x + C4x2

)ex

Part IV

Applications

Orthogonal Trajactories

In Physics or Geometry, we have to nd a family of curves that intersect a givenfamily of curves at right angles. Thus, it becomes crucial to nd these trajactories.

28

Example:

y = cx2

First we 'll nd the tangent to that curve;

ddx

(yx2

)= d

dx (c)

⇒yp.x2−2xyx4 = 0

⇒ yp = 2yx

The trajactor will be ⊥ to yp(tangent).Thus,

yp1 = − x1

2y1⇒ yp

12y1 = −x1

Now, Integrate to get whole family of curves:

y21 +

x21

2= c1

Hence, above ellipses are the orthogonal trajactories of parabolas y = cx2.

29

Growth and Decay problems

Let N(t) be the amount of substance or population either growing / or decaying.Suppose N(t) is dierentiable. Then, rate of change of N(t)is directly

proportional to the amount present.

d

dtN(t) ∝ N(t)

⇒ N(t) = cektwherec = N(InitialCondition)

Example:In a culture, bacteria increases at the rate proportional to the numberof bacteria present. If bacteria are 100 initially and are doubled in 2 hours, nd thenumber of bacteria present 4 hours later. (CollegeMathematics)

As,

⇒ N(t) = cektwherec = N(InitialCondition)

Initially means:

⇒ N(0) = cek0⇒ c = N(0) = 100

⇒ N(t) = N(0)ekt = 100ekt.................(a)

After 2 hours, there amount is doubled; we get:

⇒ 2(100) = 100e2k ⇒ 2 = e2k

Taking Natural logarithm on both sides, we get;

⇒ ln2 = 2k⇒ k = 12 ln2

Put it in (a), we get;

⇒ N(t) = 100et2 ln2

After 4 hours, we get;

⇒N(t) = 100e42 ln2 =100eln(2)2 = 100(4) = 400.

30

13Law of Cooling and Temperature Problems

It states that the time rate of change of temperature is directly proportional todierence in ”Body” and ”Surrounding” or ”Ambient” temperature.

⇒ d

dtT ∝ (T−TS)

where T > TS and TS is constant

.

⇒ ddtT = k (T−TS)

⇒ dT(T−TS)

= kdt


⇒´

dT(T−TS)

=´

kdt

⇒ ln |T−TS| = kt + c⇒ T−Ts = ekt+c = ec.ekt = cekt

⇒ T = TS + cekt

For t = 0, we get;

⇒ To = TS + c⇒ c = To −TS

The solution will be of the type:

T(t) = TS + (To −Ts)ekt

where To = InitialTemperature

Example:A body at a temperature of 50oF is placed in an oven whosetemperature is kept at 150oF. If after 10minutes the temperature of the body is75oF, then nd the time required for the body to reach a temperature of 100oF.

Initial temperature of body =To = 50oFSurrounding temperature =TS = 150oFAs,

13This method is used by detectives who investigate at what time person has been murderedor died.

31

⇒ d

dtT = k (T− 150)

⇒ dT(T−150) = kdt


⇒´

dT(T−150) =

´kdt

⇒ ln |T− 150| = kt + c⇒ T− 150 = e−kt+c = ec.ekt = cekt

⇒ T = 150 + cekt

For t = 0, we get;

⇒ To = TS + c

⇒50 = 150 + c⇒ c = 50− 150 = −100

The solution will be of the type:

T(t) = 150− 100ekt................. (a)

After 10min. and T(t) = 75oF. we get;

⇒ 75 = 150− 100e10k⇒ −75 = −100e10k

Taking Natural Logarithm on both sides, we get;

⇒ ln(0.75) =10k⇒ k = −0.0287

Putting this value in (a), we get;

⇒ T(t) = 150− 100e−0.0287t

To reach 100oF, the time required is given by:

⇒ 100 = 150− 100e−0.0287t⇒ 12 = e−0.0287t

Taking Natural Logarithm on both sides, we get;

⇒ ln (0.5) = −0.0287t

Hence,

⇒ t = 24.75mins

32

Circuit Order Network Equation

RC 1 dVC

dt + VC

RC = E(t)RC

RL 1 dIdt +

RL I = E(t)

L

RLC 2 d2Idt2 + R

L .dIdt +

ILC = E′(t)

L

14Electric Circuit Problems15

ODEQ's are widely used in the eld of engineering and mathematics. These equa-tions do marvellous jobs and keep account of the changes occuring due to theversatile nature of derivatives and dierentials. These are applied equally to bothparallel and series circuits. DEQ's for series circuits are given above.

Example: Consider an RC-circuit containing resistance 10Ω and capacitance of10−1F. The input voltage is 5V . The initial voltage of capacitor is zero. Findthe voltage of the capacitor at any time t.

As,

⇒ dVC

dt+

VC

RC=

E(t)

RC

Putting R = 10Ω and C = 10−1F.

We get;

⇒ dVC

dt + VC

10.10−1 = E(t)10.10−1

⇒ dVC

dt + VC = 5

Here we have;

⇒ p(t) = 1 and I.F. = e´p(t)dt = e

´1.dt = et

Now, multiplying with I.F. on both sides, we have;

⇒ et dVC

dt + VC.et = 5et

⇒ ddt

(VC.e

t)= 5et

14RC and RL circuits,1stOrderSystems, are similar in the way they are solved.15RLC circuits are 2nd Order Systems i.e. these circuits are modelled as 2nd Order

DEQ′s which can be solved as described in PART II Non-Homogeneous DEQ's.

33

Now, integrating both sides, we get;

⇒´

ddt

(VC.e

t)=´

5et

⇒ VC.et = 5et

1 + c

⇒ VC = 5 + ce−t

Since,

InitialVoltage = c =Vc(0−) = 0

Putting values, we get;

⇒ 0 = 5 + ce−0.t

⇒ c = −5

Thus,

⇒ VC = 5− 5e−t = 5(1− e−t

)

Approximation Techniques

In mathematics, there are some situations where we cannot realize physical systems.On the other hand, we have some techniques to approximation such systems.

16Power Series

Numerical Methods

Series Solution of DEQ's

Consider a 2ndOrder Non-Homogeneous ODEQ:

16Only Power Series is discussed.

34

b2 (x)y′′ + b1 (x)y′ + bo (x)y = g (x)

where b2 (x), b1 (x) and bo (x); all are not constants.

Dividing by b2 (x), we get;

⇒ y′′ + b1(x)b2(x)

y′ + bo(x)b2(x)

y = g(x)b2(x)

.................17

⇒ y′′ + p (x)y′ + q (x)y = Q (x)

where p (x) and q (x) must be 18analytic.

Example:

y′ − y = 0................(1) On the nextpage, Recom-mended booksare awaiting foryou or you canfollow the link.J. M. Cargel

Here we have;

p (x) = 1 and q (x) = −1

Let

y =

∞∑m=0

amxm

is a solution of equation (1).

Putting values of y′and y in (1), we get;

⇒ y′ − y =

∞∑m=1

mamxm−1 −∞∑

m=0

amxm = 0

⇒(a1 + 2a2x + 3a3x2 + 4a4x3 + ...+ ......

)−(ao + a1x + a2x2 + a3x3 + .......

)= 0

Equating coecients, we get;

17The coecent of y′′ must be 118A function is said to be analytic if it is dierentiable at each point of domain Or A function is

said to be analytic if it can be expressed in a series (nite or innite) with non-negative increasingpower of x.

35

http://www.cargalmathbooks.com/

xo ⇒ a1 − ao = 0

x1 ⇒ 2a2 − a1 = 0

x3 ⇒ 3a3 − a2 = 0

x4 ⇒4a4 − a3 = 0

x5⇒5a1−a4 = 0

By solving and putting values in the above equation, we get;

= ao + aox + aox2

2!+ ao

x3

3!+ ......+ .........

= ao

(1 + x +

x2

2!+

x3

3!+ ......+ .........

)= aoex

36

19Like in some other areas, many books on dierential equations are clones. Thestandard text is often little more than a cookbook containing a large variety of toolsfor solving DEQ's.

Part V

References

Most people use only a few of these tools. Moreover, after the course, mathmajors usually forget all the techniques. Engineering students on the otherhand can remember a great deal more since they often use these techniques.A good example of the standard text is: Ross, Shepley L. Introduction toOrdinary Dierential Equations, 4th ed. Wiley.1989. 04710-9881-7

Given the nature of the material one could much worse for a text than to usethe Schaum Outline Series book for a text, and like all of the Schaum OutlineSeries it has many worked examples. Bronson, Richard. Theory and Problemsof Dierential Equations, 2nd ed. Schaum (McGraw-Hill). 1994. 070080194

Still looking at the standard model, a particularly complete and enthusiasticvolume is: Braun, Martin. Dierential Equations and Their Applications, 3rded. S-V . 1983. 0387908471

An extremely well written volume is: Simmons, George F. Dierential Equa-tions with Applications and Historical Notes, 2nd ed. McGraw-Hill. 1991.070575401

The following book is the briefest around. It covers the main topics verysuccinctly and is well written. Given its very modest price and clarity I recom-mend it as a study aid to all students in the basic d.e. course. Many otherswould appreciate it as well. Bear, H. S. Dierential Equations: A ConciseCourse. Dover. 1999. 0486406784

Of the volumes just listed if I were choosing a text to teach out of, I wouldconsider the rst two rst. For a personal library or reference I would preferthe Braun and Simmons.

An introductory volume that emphasizes ideas (and the graphical underpin-nings) of d.e. and that does a particularly good job of handling linear systemsas well as applications is: Kostelich, Eric J., Dieter Armbruster. IntroductoryDierential Equations From Linearity to Chaos. A-W . 1997. 0201765497Note that this volume sacrices the usual compendium of techniques foundin most rst texts.

Another book that may be the best textbook here which is strong on modelingis Borrelli and Coleman. Dierential Equations: A Modeling Perspective.

19These references has been taken from: J. M. Cargel

37

http://www.cargalmathbooks.com

Wiley. 1996. 0471433322 Of these last two books I prefer to use Borelli andColeman in the classroom, but I think Kostelich and Armbruster is a betterread. Both are quite good.

The following book can be considered a supplementary text for either thestudent or the teacher in d.e. Braun, Martin, Courtney S. Coleman, DonaldA. Drew. ed's. Dierential Equation Models. S-V . 1978. 0387906959

The following two volumes are exceptionally clear and well written. Similar tothe Kostelich and Armruster volume above these emphasize geometry. Thesevolumes rely on the geometrical view all the way through. Note that thesecond volume can be read independently of the rst. Hubbard, J. H., B. H.West. Dierential Equations: A Dynamical Systems Approach. S-V. Part 1.1990. 0-387-97286-2 (Part II) Higher-Dimensional Systems. 1995. 0-387-94377-3

The following text in my opinion is a fairly good d.e. text along traditionallines. What it does exceptionally well is to use complex arithmetic to sim-plify complex problems. Redheer, Raymond M. Introduction to DierentialEquations. Jones and Bartlett. 1992. 08672-0289-0

The following rather small book is something of a reader. Nonetheless, itis aimed at roughly the junior level. O'Malley, Robert E. Thinking AboutOrdinary Dierential Equations. Cambridge. 1997. 0521557429

An undergraduate text that emphasizes theory and moves along at a fairclip is: Birkho, Garrett. Gian-Carlo Rota. Ordinary Dierential Equations.Wiley. 1978. 0471860034

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