NIT I DIFFERENTIAL QUATION - upsingh.inupsingh.in/Students/bt2/unit1_nas203.pdf · Series solution of second order ordinary differential equations with variable coefficient (Frobenius

SYLLABUS

ENGINEERING MATHEMATICS – II : NAS 203

UNIT - I : DIFFERENTIAL EQUATION Linear differential equations of nth order with constant coefficients, Complementary function and Particular integral, Simultaneous linear differential equations, Solution of second order differential equations by changing dependent & independent variables, Normal form, Method of variation of parameters, Applications to engineering problems (without derivation).

UNIT – II : SERIES SOLUTION AND SPECIAL FUNCTIONS Series solution of second order ordinary differential equations with variable coefficient (Frobenius method), Bessel and Legendre equations and their series solutions, Properties of Bessel function and Legendre polynomials.

UNIT – III : LAPLACE TRANSFORM Laplace transform, Existence theorem, Laplace transforms of derivatives and integrals, Initial and final value theorems, Unit step function, Dirac- delta function, Laplace transform of periodic function, Inverse Laplace transform, Convolution theorem, Application to solve simple linear and simultaneous differential equations.

UNIT – IV : FOURIER SERIES AND PARTIAL DIFFERENTIAL EQUATIONS Periodic functions, Fourier series of period 2 , Euler’s Formulae, Functions having arbitrary periods, Change of interval, Even and odd functions, Half range sine and cosine series, Harmonic analysis. Solution of first order partial differential equations by Lagrange’s method, Solution of second order linear partial differential equations with constant coefficients.

UNIT – V : APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS Classification of second order partial differential equations, Method of separation of variables for solving partial differential equations, Solution of one and two dimensional wave and heat conduction equations, Laplace equation in two dimension, Equation of transmission lines.

TEST BOOKS 1. E. Kreyszig, : Advanced Enginnering Mathematics,Volume-II, John Wiley&Sons 2. B. V. Ramana, Higher Engineering Mathematics, Tata Mc Graw- Hill Publishing Company Ltd. 3. R.K.Jain & S.R.K. Iyenger, Advance Engineering Mathematics, Narosa Publishing House.

REFERENCE BOOKS 1. Chandrika Prasad, Advanced Mathematics for Engineers, Prasad Mudranalaya 2. Peter V. O’ Neil, Advanced Engineering Mathematics, Thomas (Cengage) Learning. 3. A. C. Srivastava & P. K. Srivastava, Engineering Mathematics, Vol. – II, PHI Learning Pvt. Ltd.

DR. U. P. SINGH REC SONBHADRA http://upsingh.in/Students

1

UNIT 1 : DIFFERENTIAL EQUATIONS

CONTENTS 1.01 Differential Equation ....................................................................................................................................2

order and degree of a differential equation.........................................................................................................2

1.02 Linear Differential Equations ........................................................................................................................2

1.03 Linear Differential Equations Of nth Order ....................................................................................................3

linear differential equation...................................................................................................................................3

1.04 Linear Differential Equations of Second Order with Constant Coefficients .................................................4

LINEAR INDEPENDENCE AND DEPENDENCE .........................................................................................................4

METHOD FOR FINDING THE COMPLEMENTARY FUNCTION ................................................................................4

TO FIND THE VALUE OF 1

sinnx ax.f(D)

.......................................................................................................... 11

GENERAL METHOD OF FINDING THE PARTICULAR INTEGRAL OF ANY FUNCTION (x) ................................ 12

1.05 Cauchy Eular Homogeneous Linear Differential Equations ....................................................................... 12

1.06 Legendre’s Homogeneous differential Equations ..................................................................................... 14

1.07 Simultaneous Linear Differential Equations .............................................................................................. 15

simultaneous differential equations ................................................................................................................. 15

METHOD OF REDUCTION .................................................................................................................................. 19

RULE TO FIND OUT PART OF THE COMPLEMENTARY FUNCTION ..................................................................... 23

1.08 Method of reduction of order ................................................................................................................... 24

Method 1. To Find the Complete Solution of RQyPyy '" when part of Complementary Function

is known (Method of reduction of order) ......................................................................................................... 24

Certain rules to find one part of CF(Remember !) ............................................................................................ 26

1.09 Reduced to Normal Form (Removal of first derivative) ............................................................................ 27

Method 2. To Find the Complete Solution of RQyPyy '" when it is Reduced to Normal Form

(Removal of first derivative) .............................................................................................................................. 27

1.10 Changing the Independent Variable .......................................................................................................... 29

Method 3. To Find the Complete Solution of " 'y Py Qy R Changing the Independent Variable ....... 29

1.11 Method of Variation of Parameters .......................................................................................................... 31

Method 4 : To Find the Complete Solution of " 'y Py Qy R by the Method of Variation of Parameters

........................................................................................................................................................................... 31

1.12 Applications of differential equations to engineering problems .............................................................. 35

Application to Electric Circuits ........................................................................................................................... 37

Assignment-I .......................................................................................................................................................... 38


2

UNIT 1 : DIFFERENTIAL EQUATIONS

1.01 DIFFERENTIAL EQUATION

An equation which involves differential co-efficient is called a differential equation.

For Example

1. 2

2

1

1

y

x

dx

dy

2. 082

2

2

ydx

dy

dx

yd 3.

2

22/3

2

1dx

ydk

dx

dy

ORDER AND DEGREE OF A DIFFERENTIAL EQUATION

The order of a differential equation is the order of the highest differential co-efficient present in the equation. Consider

1. .sin2

2

wtEc

q

dt

dqR

dt

qdL 2. .tan8sincos

2

2

2

xydx

dyx

dx

ydx

3.

2

2

23

2

1

dx

yd

dx

dy

The order of the above equation is 2.

The degree of a differential equation is the degree of the highest derivative after removing the radical sign and fraction.

The degree of the equation (1) and (2) is 1. The degree of the equation (3) is 2.

Q.1 The order and degree of the differential equation 06

8

2

4

3

3

dx

dyx

dx

yd are …..and …..

Sol. The highest order of the derivative in the given differential equation is 3. So, the order is 3.

The power of the highest order derivative is 4. So the degree is 4.

1.02 LINEAR DIFFERENTIAL EQUATIONS

A differential equation of the form

Qpydx

dy

(1)

is called a linear differential equation, where P and Q are functions of x (but not of y ) or

constants.

In such case, multiply both sides of (1) by Pdx

e


3

.

Pdx

Pdxdye Py Q e

dx

(2)

The left hand side of (2) is

Pdx

eydx

d. , but

PdxPdx

eQeydx

d

..

Integrating both sides, we get

CdxeQeyPdx

Pdx

..

This is the required solution.

Note : Pdx

e is called the integrating factor.

Solution is CdxFIQFIy .].[.].[

Q.2 Solve : 2( 1) ( 1)xdyx y e x

dx

Sol. We have, )1(1

xex

y

dx

dy x

Integrating factor 1log( 1) log( 1)1

1

1

dx

x xxIF e e ex

The required solution is

1 1. .( 1).

1 1

x xy e x dx c e dx cx x

( 1)( )xy x e c

1.03 LINEAR DIFFERENTIAL EQUATIONS OF NTH

ORDER

LINEAR DIFFERENTIAL EQUATION

If the degree of the dependent variable and all derivatives is one, such differential equations are called linear differential equation e.g.

22

25 6 1

d y dyy x x

dx dx ,

222

22 xd y dy

y edx dx

, )1(

2

2

fxdt

dx

dt

xd

The order of a differential equation is the highest order of the derivative involved. All the above differential equations are of second order.

Fourier and Laplace transforms are mathematical tools to solve the differential equations.


4

1.04 LINEAR DIFFERENTIAL EQUATIONS OF SECOND ORDER WITH CONSTANT COEFFICIENTS

The general form of the linear differential equation of second order is

RQydx

dyP

dx

yd

2

2

Where P and Q are constants are R is a function of x or constant.

Differential Operator . Symbol D stands for the operation of differential i.e.,

22

2,

dy d yDy D y

dx dx

D

1 stands for the operation of integration.

2

1

D stands for the operation of integration twice.

RQydx

dyP

dx

yd

2

2

can be written in the operator form.

2 2( )D y P Dy Qy R D PD Q y R

LINEAR INDEPENDENCE AND DEPENDENCE

Two solutions )(1 xy and )(2 xy are said to be linearly independent if

0)()( 21 xByxAy

A and B are not equal to zero.

COMPLETE SOLUTION = COMPLEMENTARY FUNCTION + PARTICULAR INTEGRAL

METHOD FOR FINDING THE COMPLEMENTARY FUNCTION

(1) In finding the complementary function, R.H.S. of the given equation is replaced by zero.

(2) Let mxeCy 1 be the C.F. of

2

20

d y dyP Qy

dx dx (1)

Putting the values of dx

dyy, and

2

2

dx

yd in (1)

0)( 2

1 QPmmeC mx

.02 QPmm [This equation is called Auxiliary Equation.]

(3) Solve the auxiliary equation :


5

Case I : Roots, Real and Different. If 1m and 2m are the roots, then the C.F. is

xmxmeCeCy 21

21

Case II : Roots, Real and Equal. If both the roots are 1m , 1m then the C.F. is

xmexCCy 1)( 21

Q.3 Solve : 2

26 9 0

d y dyy

dx dx

Sol. Here, we have

0962

2

ydx

dy

dx

yd 0)96( 2 yDD

A.E. : 2 26 9 0 ( 3) 0 3, 3.m m m m

xexCCFC 3

21 )(.. .

Q.4 Find the general solution of the differential equation 5 3

5 30

d y d y

dx dx

Sol. 03

3

5

5

dx

yd

dx

yd

5 3 0D y D y

5 3( ) 0D D y

3 2( 1) 0D D y

A.E.: is 3 2( 1) 0m m

0, 0, 0, 1, 1m

Hence, the solution is

1 2 3 4 5( ) x xy C C x C x C e C e .

Case III : Roots, Imaginary. If the roots are , i then the solution will be

].[ 21

)(

2

)(

1

xixixxixi eCeCeeCeCy

)]sin(cos)sin(cos[ 21 xixCxixCe x

]sin)(cos)[( 2121 xCCixCCe x

]sincos[ xBxAe x

Q.5 Solve : 2

24 5 0

d y dyy

dx dx


6

Sol. Here the auxiliary equation is

2 4 5 0m m 4 16 20

22

m i

The complementary function is

)sincos(2 xBxAey x

Rules to find particular integral

(i) xaxa eaf

eDf )(

1

)(

1

If 0)( af then xaxa eaf

xeDf

)(

1

)(

1

If 0)(' af then xaxa eaf

xeDf

)("

1

)(

1 2

(ii) 11[ ( )]

( )

n nx f D xf D

expand 1)]([ Df and then operate.

(iii) axaf

axDf

sin)(

1sin

)(

122

and axaf

axDf

cos)(

1cos

)(

122

If 0)( 2 af then axaf

xaxDf

sin)(

1sin

)(

122

and axaf

xaxDf

cos)('

1cos

)(

12

(iv) )()(

1)(

)(

1x

aDfexe

Df

axxa

(v)

dxxeexaD

axxa )()(1

Q.6 Solve the following differential equation:

2

26 sin 3 cos 2 .

d y dyy x x

dx dx

Sol. Here, we have

xxydx

dy

dx

yd2cos3sin6

2

2

xxyDD 2cos3sin)65( 2

A.E.: 0652 mm

0662 mmm


7

0)6(1)6( mmm

0)6()1( mm

6,1 m

xx eCeCFC 6

21..

)2cos3(sin65

1..

2xx

DDIP

xDD

xDD

2cos65

13sin

65

122

xD

xD

2cos652

13sin

653

122

xD

xD

2cos105

13sin

155

1

xDD

Dx

DD

D2cos

)2)(2(5

)2(13sin

)3)(3(5

)3(

xD

Dx

D

D2cos

)4(5

23sin

)9(5

322

xD

xD

2cos)42(5

23sin

)93(5

322

xD

xD

2cos)8(5

23sin

)18(5

3

)2)(cos2(40

1)3(sin

90

)3(xDx

D

]2cos22sin2[40

1]3sin33cos3[

90

1xxxx

)2sin2(cos20

1)3sin3(cos

30

1xxxx

Complete solution : .... IPFCy

6

1 2

1 1(cos3 sin 3 ) (cos 2 sin 2 )

30 20

x xy C e C e x x x x

Q.7 Solve : 3 2

3 23 4 2 cosxd y d y dy

y e xdx dx dx

Sol. We have, xeyDDD x cos)243( 23

A.E. is 0243 23 mmm


8

imeimmm 1,1.,.,0)22)(1( 2

xx exCxCeCFC )sincos(.. 321

xDDD

eDDD

IP x cos243

1

)22)(1(

1..

232

xDD

eD

x cos24)1(3)1(

1

)221)(1(

1

xD

eD

x cos13

1

)1(

1

xD

D

Dex cos

19

133.

11

12

19

)cossin3(1.

1

xx

Dex

)cossin3(10

1. xxxex

Hence, complete solution is

y )cossin3(10

1)sincos( 321 xxexexCxCeC xxx

Q.8 Solve : 2

2sin sin 2

d yy x x

dx

Sol. We have,

xxydx

yd2sinsin

2

2

xxyD 2sinsin)1( 2

A.E. is 012 m im

C.F. xCxC sincos 21

P.I. ]3cos[cos2

1

1

12sinsin

1

122

xxD

xxD

x

Dx

D3cos

1

1cos

1

1

2

122

xx

xxx

Dx 3cos

8

1sin

22

13cos

19

1cos

2

1

2

1


9

]3cossin4[16

1xxx

Complete Solution is y C.F.+P.I.

)3cossin4(16

1sincos 21 xxxxCxCy

Q.9 Obtain the general solution of the differential equation

" 2 ' 2 cosxy y y x e x

Sol. We have, xexyyy x cos2'2"

A.E. : 0222 mm im 1

)sincos(.. xBxAeFC x

xeDD

xDD

IP x cos22

1

22

1..

22

where,

xD

D

xD

D

xDD

I

212

1

212

1

22

12221

xD

DxD

D

...2

12

1

21

2

1 21

2

12

1...

22

1 2

xx

DDxx

And

xeDD

I x cos22

122

xD

xexD

exDD

e xxx cos2

1.cos

1

1cos

2)1(2)1(

122

)(

)('

1)(

)(

1,0)(sin

2

12

2thenIf x

Dfxx

Dfafxxex

.... IPFCy

.sin2

1)1(

2

1)sincos( xxexxBxAe xx

Ans.


10

Q.10 Solve : 2 3 2( 4 4) xD D y x e

Sol. We have, xexyDD 232 )44(

A.E. is 0442 mm 0)2( 2 m 2,2m

C.F. xexCC 2

21 )(

P.I. 3

2

223

2 4)2(4)2(

1.

44

1x

DDeex

DD

xx

20

.4

1.

1 52

423

2

2 xe

x

Dex

De xxx

The complete solution is 20

.)(5

22

21

xeexCCy xx Ans

Q.11 Solve 4( 1) cosxD y e x

Sol. Here, we have

xeyD x cos)1( 4

A.E. is 014 m 0)1)(1)(1( 2 mmm

iim ,,1,1

C.F. )sincos( 4321 xCxCeCeC xx

P.I. xeD

x cos1

14

xDDDD

exD

e xx cos646

1cos

1)1(

12344

xDD

ex cos6)1(4)1(6)1(

12

3

coscos

6461

1 xex

DDe

xx

Complete solution is .... IPFCy

3

cos)sincos( 4321

xexCxCeCeCy

xxx Ans.


11

TO FIND THE VALUE OF 1

sinnx ax.f(D)

Now niaxiaxnn xiaDf

eexDf

axiaxxf(D) )(

1

)(

1)sin(cos

1

axxf(D)

n sin1

Imaginary part of iaxenx

iaDf

)(

1

axxf(D)

n cos1

Real part of niax xiaDf

e )(

1

Q.12 Solve the differential equation 2( 2 1) cosD D y x x

Sol. xxyDD cos)12( 2

Auxiliary equation is

0122 mm 0)1( 2 m 1,1 m

C.F. xexCC )( 21

P.I.

xxD

cos)1(

12

Real part of ]sin[cos)1(

12

xixxD

Real part of

ixexD 2)1(

1Real part of x

iDe xi

2)1(

1

Real part of

xiDiD

e xi

22 )1()1(2

1Real part of x

iDiDe xi

2)1(2

12

Real part of

x

i

DD

i

ii

e xi

2

11

1

22

Real part of xi

DD

i

i

i

e xi1

2

2

11

2

Real part of

xD

i

i

i

e xi

...1

12

Real part of

i

ixxix

i

1)sin(cos

2

1

Real part of )1)(sincos(2

1 ixxxi

Real part of xxxixxxi cos2

1)1(sin

2

1)1)(sincos(

2

1

Complete solution is .... IPFCy

xxxexCCy x cos2

1sin)1(

2

1)( 21


12

Q.13 Solve 2

22 sin

d y dyy x x

dx dx

Sol. Auxiliary equation is 0122 mm

xexCCFC )(.. 21

P.I. xxDD

sin12

12

)sincos( xixeix

Imaginary part of

)sin(cos12

12

xixxDD

Imaginary part of xiexDD

12

12

Imaginary part of xiDiD

eix 1)(2)(

12

Imaginary part of xiDiD

eix 2)1(2

12

Imaginary part of xDi

Dii

eix

1

2

2

1)1(1

2

1

Imaginary part of xDii

xx ])1(1[2

)sin(cos

Imaginary part of ]1)[sincos(2

1ixxxi

P.I. xxxx sin2

1cos

2

1cos

2

1

Complete solution is )sincoscos(2

1)( 21 xxxxexCCy x

GENERAL METHOD OF FINDING THE PARTICULAR INTEGRAL OF ANY FUNCTION (x)

1 ax ax(x) e e (x)dx

D a

1.05 CAUCHY EULAR HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS

1

1

1 01... ( )

n nn n

n nn n

d y d ya x a x a y x

d x d x

where ...,,, 210 aaa are constants, is called a homogeneous equation.

Put ,zex ,log xz e Ddz

d


13

Dydx

dyx

dz

dy

dx

dyx

dz

dy

xdx

dz

dz

dy

dx

dy

1.

Again, dx

dz

dz

yd

xdz

dy

xdz

dy

xdx

d

dx

dy

dx

d

dx

yd2

2

22

2 111

yDDxdz

dy

dz

yd

xxdz

yd

xdz

dy

x)(

11111 2

22

2

22

2

2

yDDdx

ydx )( 2

2

22

yDDdx

ydx )1(

2

22

Similarly. yDDDdx

ydx )2)(1(

3

33

Q.14 Solve : 2

2

24 cos(log ) sin(log ).

d y dyx x y x x x

dx dx

Sol. We have,

).sin(log)cos(log42

22 xxxy

dx

dyx

dx

ydx

(1)

Putting dz

dDxzex z ,log and Dy

dx

dyxyDD

dx

ydx ,)1(

2

22 in (1), we get

zezyDDD z sincos]4)1([

i.e. zezyDD z sincos)42( 2

A.E. is 0422 mm2

164)2( m

im 31

C.F. ]3sin3cos[ 21 zCzCe z (2)

P.I. )sin(cos42

12

zezDD

z

zeDD

zDD

z sin42

1cos

42

122


14

zDD

ezD

z sin4)1(2)1(

1cos

421

12

zDDD

ezD

z sin42212

1cos

23

12

zezD

zD

ezD

D zz sin31

1cos

49

23sin

3

1cos

49

2322

3 2 1 1 1 1cos sin sin (3cos 2sin ) sin

13 2 2 13 2

z zDz e z z z z e z

(3)

Complete solution is

.... IPFCy

zezzzCzCey zz sin2

1)sin2cos3(

13

1]3sin3cos[ 21 (4)

Replacing xz log and xez in (4), we get

1 2[ cos 3(log ) sin 3(log )]y x C x C x 3 2 1

cos(log ) sin(log ) sin(log )13 13 2

x x x x

1.06 LEGENDRE’S HOMOGENEOUS DIFFERENTIAL EQUATIONS

A linear differential equation of the form

11

1 1( ) ( ) ...

n nn n

nn n

d y d ya bx a a bx a y X

dx dx

(1)

Where naaaba ...,,,, 21 are constants and X is a function of ,x is called Legendre’s linear equation.

Put

zexba )log( bxaz (2)

Dybdx

dybxa )( (3)

)1()( 2

2

22 DDb

dx

ydbxa (4)

yDDDbdx

ydbxa )2)(1()( 3

3

33 (5)

Q.15 Solve 2

2

2(2 1) 2(2 1) 12 6

d y dyx x y x

dx dx

Sol. Here, we have

22

2(2 1) 2(2 1) 12 6

d y dyx x y x

dx dx (1)


15

Put zex 12 )12(log xz

Dydx

dyx 2)12( and yDD

dx

ydx )1(2)12( 2

2

22 in (1), we get

)1(

2

161222)1(4 zeyDyyDD

)1(312444 2 zeyDyDyyD

A.E. is 01284 2 mm 0322 mm

0)1)(3( mm 1,3 mm

zz eCeCFC 2

3

1..

)33(1284

1..

2

ze

DDIP

zz eDD

eDD

)0(

22 1284

133

1284

1

)1(1200

13

12)1(8)1(4

13

2

ze

4

1

16

3

ze

Complete Solution is .... IPFCy

4

1

16

32

3

1 zzz eeCeCy

4

1)12(

16

3)12()12( 1

2

3

1 xxCxCy .

1.07 SIMULTANEOUS LINEAR DIFFERENTIAL EQUATIONS

SIMULTANEOUS DIFFERENTIAL EQUATIONS

If two or more dependent variables are functions of a single independent variable, the equations involving their derivatives are called simultaneous equations, e.g.,

tydt

dx 4 , tex

dt

dy 2

The method of solving these equations is based on the process of elimination, as we solve algebraic simultaneous equations.


16

Q.16 The equations of motions of a particle are given by

0dx

w ydt

, 0dx

w xdt

Find the path of the particle and show that it is a circle.

Sol. On putting Ddt

d in the equations, we have

0wyDx

0 Dywx

On multiplying (1) by w and (2) by D, we get

02 ywwDx

02 yDwDx

On adding (3) and (4), we obtain

022 yDyw 0)( 22 ywD

Now, we have to solve (5) to get the value of .y

A.E. is 022 wD 22 wD iwD

wtBwtAy sincos

wtBtADy cossin

On putting the value of Dy in (2), we get

0cossin wtBwwtAwwx

wtBwwtAwwx cossin

wtBwtAx cossin

On squaring (6) and (7) and adding, we get

)sin(cos)sin(cos 22222222 wtwtBwtwtAyx

2222 BAyx

This is the equation of circle.


17

Q.17 Solve the following

3 8dx

x ydt

3dx

x ydt

With (0) 6x and (0) 2.y

Sol. Here we have

yxdt

dx83

yxdt

dx3

On putting Ddt

d in (1) and (2), we get

08)3(083 yxDyxDx

and 0)3(03 xyDyxDy

Multiplying (3) by )3( D and (4) by 8 adding them we get

0)1( 2 xD

A.E. is 012 m 12 m 1m

tt eCeCFC 21..

0.. IP

tt eCeCx 21

From (3) we get yeCeCD xx 8])[3( 21

tttt eCeCeCeCy 2121 338

tt eCeCy 21 428

)2(2

121

tt eCeCy

Initially when 0t then .2x

From (5); 0

2

0

12 eCeC 221 CC

Also when 0t , 2y .


18

From (6), )2(4

12 0

2

0

1 eCeC

21 28 CC

Solving (7) and (8), we get

41 C and 62 C

Hence, the required solution is

tt eex 64

and )124(4

1 tt eey

Q.18 Solve : 2

2sin ,

d xy t

dt

2

2cos

d yx t

dt

Sol. Here, we have

2 sinD x y t (1)

tyDx cos2 (2)

Operating equation (1) by ,2D we get

tyDxD sin24

Subtracting (2) from (3), we get

ttxD cossin)1( 4


014 m 0)1)(1( 22 xm

im ,1,1

tctct

ect

ecFC sin4cos321..

)cos(sin14

1.. tt

DIP

)cos(sin34

1. tt

Dt

)sincos(4

ttt

1

D


19

)cos(sin4

1sin

4cos

321tttctctectecx

From first equation (1), 2

2sin

dt

xdty

)cos(sin

4

1)sin

4cos

3()

21(

2

2sin tttctctectec

dt

dt

)cos(sin

4

1)sin(cos

4

1cos

4sin

321sin tttttctctectec

dt

dt

)sin(cos

4

1)sin(cos

4

1)cossin(

4

1sin

4cos

321sin tttttttctctectect

)cos(sin2

1)cos(sin

4

1sincos 4321 tttttctcececy tt

and )cos(sin4

1sincos 4321 tttctcececx tt .

METHOD OF REDUCTION

(WHOSE ONE SOLUTION OF COMPLEMENTARY FUNCTION IS KNOWN)

If uy is given solution be longing to the complementary function of the differential equation.

Let the other solution be .vy

Then vuy . is complete solution of the differential equation. Let

2

2

d y dyp Qy R

dx dx (1)

be the differential equation and u is the complementary function of (1)

02

2

Qudx

dup

dx

ud (2)

vuy . so that dx

dvu

dx

duv

dx

dy

2

2

2

2

2

2

2dx

ydu

dx

du

dx

dv

dx

udv

dx

yd

Substituting the values of 2

2

,,dx

yd

dx

dyy in (1), we get


20

RvQudx

dvu

dx

duvP

dx

vdu

dx

du

dx

dv

dx

udv

.2

2

2

2

2

On arranging, we get

Rdx

dv

dx

du

dx

dvP

dx

vduQu

dx

duP

dx

udv

2

2

2

2

2

The first bracket is zero by virtue of relation (2), and the remaing is divided by .u

u

R

dx

dv

dx

du

udx

dvP

dx

vd

22

2

u

R

dx

dv

dx

du

uP

dx

vd

22

2

Let ,zdx

dv so that

dx

dz

dx

vd

2

2

Equation (3) becomes

u

Rz

dx

du

uP

dx

dz

2

This is the linear differential equation of first order and can be solved ( z can be found), which will contain one constant.

On integration ,dx

dvz we can get v .

Having found v , the solution is .uvy

Q.19 Solve 2" 4 ' (4 2) 0y xy x y given that 2xy e is an integral included in the

complementary function.

Sol. Here, we have 0)24('4" 2 yxxyy

On putting 2

. xevy in (1), the reduced equations as in the article 4.7.

02

2

2

dx

dv

dx

du

uP

dx

vd ]0,24,4[ 2 RxQxP

0)2(2

42

22

2

dx

dvxe

ex

dx

vd x

x

0]44[2

2

dx

dvxx

dx

vd 0

2

2

dx

vd 1c

dx

dv 21 cxcv


21

uvy ][2xeu

)( 21

2

cxcey x Ans.

Q.20 Solve 2 2 3" ( 2 ) ' ( 2) xx y x x y x y x e given that y x is a solution.

Sol. Here, we have xexyxyxxyx 3)2(')22("2

xexyx

xy

x

xxy

2

2'

2

22" (1)

On putting vxy in (1), the reduced equation as in the article 4.7

u

R

dx

dv

dx

du

uP

dx

vd

2

2

2

x

xex

dx

dv

xx

xx

dx

vd

)1(

2

2

22

2

2

xezdx

dzxedx

dv

dx

vd

2

2

dx

dvz

Which is a linear differential equation

xdx

eeFI

..

Its solution is cdxeeez xxx .

xxx cexezcxze .

xx ecxedx

dv .

1. ceceexv xxx

1)1( cecexv xx

xcecxxxvxy x

1

2 )( Ans.


22

Q.21 Solve 2

2(2 1) ( 1) 0

d y dyx x x y

dx dx

Given that xey is an integral included in the complementary function.

Sol. Here, we have 0)1()12(2

2

yxdx

dyx

dx

ydx

0112

2

2

yx

x

dx

dy

x

x

dx

yd (1)

By putting xvey in (1), we get the reduced equation as in the article 4.7.

02

2

2

dx

dv

dx

du

uP

dx

vd (2)

Putting xeu and zdx

dv in (2), we get

0212

ze

ex

x

dx

dz x

x

0212

zx

xx

dx

dz 0

x

z

dx

dz

cxzx

dx

z

dzlogloglog

21111 log cxc

x

dxcvd

x

c

dx

dv

x

cz

)log(. 21 cxcevuy x .


23

RULE TO FIND OUT PART OF THE COMPLEMENTARY FUNCTION

Rule Condition Part of Complementary Function u

1 01 QP xe

2 01 QP xe

3 01

2

a

Q

a

P

xae

4 0QxP x

5 022 2 QxPx 2x

6 0)1( 2 QxPnxnn nx

Q.22. Solve 2

2 3

22 [1 ] 2(1 )

d y dyx x x x y x

dx dx

Sol. 3

2

22 )1(2)1(2 xyx

dx

dyxx

dx

ydx

xx

yx

dx

dy

x

xx

dx

yd

222

2 )1(2)1(2

Here, 0)1(2)1(2

22

x

x

x

x

xxQxP

Hence xy is a solution of the C.F. and the other solution is .v

Putting xvy in (1), we get the reduced equation as in article 4.5

u

x

dx

dv

dx

du

uP

dx

vd

2

2

2

x

x

dx

dv

xx

xx

dx

vd

)1(

2)1(222

2

12122

2

zdx

dz

dx

dv

dx

vd

z

dx

dv


24

Which is a linear differential equation of first order and xdx

eeFI 22

..

Its solution is

1

22 cdxeez xx

12

22 c

xexez

xecz 212

1

dxxecdvxecdx

dv

2

12

1212

1

2

2

2

1

2cxe

cxv

22

2

1

2cxe

cxxuvy Ans.

1.08 METHOD OF REDUCTION OF ORDER

METHOD 1. TO FIND THE COMPLETE SOLUTION OF RQyPyy '" WHEN PART OF

COMPLEMENTARY FUNCTION IS KNOWN (METHOD OF REDUCTION OF ORDER)

Let uy be a part of the complementary function of the given differential equation

2

2

d y dyP Qy R

dx dx (1)

Where u is a function of .x Then, we have

2

20

d u duP Qu

dx dx (2)

Let uvy be the complete solution of equation (1), where v is a function of x .

Differentiating y w.r.t. x ,

vdx

du

dx

dvu

dx

dy.

Again, 2

2

2

2

2

2

.2dx

udv

dx

du

dx

du

dx

vdu

dx

yd

Substituting the values of dx

dyy, and

2

2

dx

yd in equation (1) we get

RuvQdx

duv

dx

dvuP

dx

udv

dx

dv

dx

du

dx

vdu

)(2

2

2

2

2


25

RvQudx

duP

dx

ud

dx

dvPu

dx

du

dx

vdu

2

2

2

2

2

Rdx

dvPu

dx

du

dx

vdu

2

2

2

, using (2)

u

R

dx

dvP

dx

du

udx

vd

22

2

(3)

Put Pdx

dv then,

dx

dp

dx

vd

2

2

Now (3) becomes, u

RPP

dx

du

udx

dp

2 (4)

Equation (4) is a linear differential equation of I order in p and x .

dxPdxPdu

udxp

dx

du

u eueeFI 2

22

..

Solution of (4) is given by

1

22 cdxeuu

Repu

dxPdxP

Where 1c is an arbitrary constant of integration.

12

1cdxeRue

up

dxPdxP

12

1cdxeRue

udx

dv dxPdxP

Integration yields, 212

1cdxcdxeRue

uv

dxPdxP


Hence the complete solution of (1) is given by,

uvy

ucdxcdx

dxpeRu

dxpe

uuy

212

1

To find out the part of C.F. of the linear differential equation of II order given by

.2

2RQy

dx

dyP

dx

yd


26

CERTAIN RULES TO FIND ONE PART OF CF(REMEMBER !)

CONDITION PART OF C.F.

02

1 a

Q

a

P axe

01 QP xe

01 QP xe

02)1( QxmxPmm mx

0QxP x

0222 QxPx 2x

Q.23 Solve : 2

cot (1 cot ) sin .2

d y dy xx x y e xdxdx

Sol. Comparing with the standard form, we get

xxeRxQxP sin),cot1(,cot

0cotcot111 xxQP

A part of xeFC ..

Let xevy be the complete solution of given equation, then

dx

dvxexevdx

dy

2

22

2

2

dx

vdxedx

dvxexevdx

yd

Substituting for dx

dyy, and

2

2

dx

yd in given equation, we get

xdx

dvx

dx

vdsin)cot2(

2

2


27

xpxdx

dpsin)cot2( …(1) where .

dx

dvp

This is a linear differential equation of I order in p and x .

x

eeFI

xdxx

sin..

2)cot2(

Solution of (1) is, 1

2

1

22

2sin.sin

sinc

ecdx

x

ex

x

ep

xxx


xecxp x sinsin2

1 2

1

xecxdx

dv x sinsin2

1 2

1

Integrating, we get 2

2

1 )sin2(cos5

1cos

2

1cxxecxv x

Hence the complete solution is given by,

xxx ecxxecxevy

2

2

1 )sin2(cos5

1cos

2

1.

1.09 REDUCED TO NORMAL FORM (REMOVAL OF FIRST DERIVATIVE)

METHOD 2. TO FIND THE COMPLETE SOLUTION OF RQyPyy '" WHEN IT IS REDUCED TO NORMAL

FORM (REMOVAL OF FIRST DERIVATIVE)

When the part of C.F. can not be determined by the previous method, we reduce the given differential equation in normal form by eliminating the term in which there exists first derivative of the dependent variable.

2

2

d y dyP Qy R

dx dx (1)

Let vuy be the complete solution of eqn. (1), where u and v are the function of x .

dx

dvu

dx

duv

dx

dy

and 2

22

2

2

2

2

dx

vdu

dx

dv

dx

du

dx

udv

dx

yd


28

Substituting the value of 2

2,,

dx

yd

dx

dyy in eqn. (1), we get

2 22 1

2 2

d v du du d u P du RP v Q

u dx dx u u dx udx dx

(2)

Let us choose u such that 02

Pdx

du

u (3)

Which on solving gives,

2

Pdx

u e

(4)

From (3), 2

Pu

dx

du

Differentiating, we get

)(

2

12

2

udx

dP

dx

duP

dx

ud

dx

dPuuP

dx

dPu

PuP

2222

1 2

Coefficient of QPu

u

P

dx

dPuuP

uQ

dx

du

u

P

dx

ud

uv

224

21

2

21

IP

dx

dPQ

4

2

2

1 (say)

Then (2) becomes, SIvdx

vd

2

2 (5)

This is known as the normal form of equation (1).

Solving (5), we get v in terms of x . Ultimately, uy is the complete solution.

Q.24 Solve : 22

24 (4 1) 3 sin 2 .2 xd y dy

x x y e xdxdx

Sol. Here, xeRxQxP x 2sin3,14,422

Let uy be the complete solution.

Now, 2)4(

2

1

xdxx

eeu


29

.1)16(4

1)4(

2

114

42

1 222

xxP

dx

dPQI

Also, xe

xe

u

RS

x

x

2sin32sin3

2

2

Hence normal form is,

xdx

d2sin3

2

2


imm 012

xcxcFC sin2

cos1

..

where 1c and 2c are arbitrary constants of integration.

xxxD

IP 2sin2sin)14(

3)2sin3(

1

1..

2

Solution is, xxcxc 2sinsincos 21

Hence the complete solution of given differential equation is

)2sinsincos( 21

2xxcxceuy x .

1.10 CHANGING THE INDEPENDENT VARIABLE

Method 3. To Find the Complete Solution of " 'y Py Qy R Changing the Independent Variable

2

2

d y dyP Qy R

dx dx (1)

Let us relate x and z by the relation,

( )z f x (2)

dy dy dz

dx dz dx (3)

2

2

d y d dy d dy dz

dx dx dx dx dz dx

22 2 2

2 2 2

dy d z dz d dy dz dy d z dz d y

dz dx dx dz dz dx dz dx dz dx

(4)

Substituting in (1), we get


30

RQydx

dz

dz

dyP

dx

yd

dx

dz

dx

zd

dz

dy

2

22

2

2

1112

2

RyQdz

dyP

dz

yd (5)

where 21212

2

2

1 ,,

dx

dz

RR

dx

dz

QQ

dx

dz

dx

dzP

dx

zd

P

Here 111 ,, RQP are functions of x which can be transformed into functions of z using the

relation )(xfz .

Choose z such that 1Q constant 2a (say)

2

2a

dx

dz

Q

Q

dx

dza

dxa

Qdz

Integration yields, dxa

Qz

If this value of z makes 1P as constant then equation (5) can be solved.

Q.25 Solve : )1(logcos4)1()1(2

22 xy

dx

dyx

dx

ydx .

Sol. )1log(cos)1(

4

)1(1

1222

2

xxx

y

dx

dy

xdx

yd

(1)

Choose z such that,

2

2

)1(

1

xdx

dz

xdx

dz

1

1 (2)

Integration yields, )1(log xz (3)

From(2), 22

2

1

1

xdx

zd


31

0

)1(

1

1

1.

1

1

1

1

2

2

1

x

xxxP

121

dx

dz

QQ

zx

dx

dz

RR cos4)1log(cos4

21

From (3)

Reduced equation is

zydz

ydcos4

2

2

Auxiliary equation is 012 m im

zczcFC sincos.. 21

zzzz

zD

IP sin2sin2

.4)cos4(1

1..

2

Complete solution is

zzzczcy sin2sincos 21

)1log(sin)1(log2)1log(sin)1(logcos 21 xxxcxcy .

1.11 METHOD OF VARIATION OF PARAMETERS

METHOD 4 : TO FIND THE COMPLETE SOLUTION OF " 'y Py Qy R BY THE METHOD OF VARIATION

OF PARAMETERS

RQydx

dyP

dx

yd

2

2

(1)

Let the complementary function of (1) be

BvAuy (2)

u and are part of C.F.

02

2

Qudx

duP

dx

ud (3)


32

and 02

2

Qvdx

dvP

dx

vd (4)

Let the complete solution of (1) be

BvAuy (5)

where A and B are not constants but suitable functions of x to be so chosen that (5) satisfies (1). Now,

vBuABvAuy 11111

)( 11111 vBuABvAuy (6)

Let us choose A and B such that

011 vBuA (7)

Now (6) becomes, 111 BvAuy (8)

2112112 BvvBAuuAy (9)

Substituting the values of 21,, yyy from (5), (8) and (9) in (1) respectively, we get

RBvAuQBvAuPBvvBAuuA )()()( 11211211

RQvPvvBQuPuuAvBuA )()( 12221111

RvBuA 1111 (10)

|Using (3) and (4)

Solving (7) and (10) for 1A and 1B , we get

011 vBuA

01111 RvBuA

vuuvRu

B

Rv

A

11

11 1

)(11

1 xvuuv

RvA

| say (11)

)(11

1 xvuuv

RuB

| say (12)

Integrating (11), we get adxxA )( (13)

Where a is an arbitrary constant of integration.


33

Integrating (12), we get bdxxB )( (14)

where b is also an arbitrary constant of integration.

Putting the above values in (5), we get

vbdxxuadxxy )()(

bvaudxxvdxxuy )()(

This gives the complete solution of (1).

Q.26 Solve by the method of variation of parameters :

2

2

2sec .

d ya y ax

dx

Sol. Here, ,cos axu axv sin are two parts of C.F.

Also, .sec axR

Let the complete solution be

axBaxAy sincos

where A and B are suitable functions of x .

To determine the values of A and B, we have

1

11

cdxvuuv

RvA

1

}sin)sin(cos.{cos

sin.seccdx

axaxaaxaax

axax

1

tancdx

a

ax

12coslog

1cax

aA

where 1c is an arbitrary constant of integration.

2

11

cdxvuuv

RuB

2}sin)sin(cos.{cos

cos.seccdx

axaxaaxaax

axax


34

22

1c

a

xcdx

a

where 2c is an arbitrary constant of integration.

Hence the complete solution is given by

axBaxAy sincos

axca

xaxc

a

axsincos

coslog212

Q.27 Solve by method of variation of parameters:

2

22 logxd y dy

y e xdx dx

.

Sol. Parts of C.F. are xx xeveu , and xeR x log

Let xx BxeAey be the complete solution where A and B are some suitable functions of x . To

determine A and B, we have

121

11 )(

.logcdx

xexeee

xexecdx

vuuv

RvA

xxxx

xx

1

22

14

log2

log cx

xx

cdxxx

222

11

.logcdx

e

execdx

vuuv

RuB

x

xx

22 loglog cxxxcdxx

Hence the complete solution is

xx BxeAey

xx xecxxxecx

xx

)log(

4log

221

22

Q.28 Using variation of parameters method, solve :

2

2 3

22 12 log .

d y dyx x y x x

dx dx

Sol. Consider the equation

01222

22 y

dx

dyx

dx

ydx for finding parts of C.F.


35

Put zex so that xx log and Let dz

dD then the given equation reduces to

0]122)1([ yDDD

0)12( 2 yDD


4,30122 mmm

4

2

3

1

4

2

3

1.. xcxcececFC zz

Hence, parts of C.F. are 3x and 4x

Let BvAuy be the complete solution, where A and B are some suitable functions of x . A and B

are determined as follows :

14253

4

1

11 )(3)4(.

.logcdx

xxxx

xxxcdx

vuuv

RvA

1

2

112

3

)(log14

1log

7

1

7

logcxcdx

x

xcdx

x

xx

and

22

3

2

11 7

.logcdx

x

xxxcdx

vuuv

RuB

2

77

2

6

7.

1

7.log

7

1log

7

1cdx

x

x

xxcdxxx

2

7

2

77

log7

1

4977

1

7

log

7

1cx

xc

xxx

Hence the complete solution is given by

4

2

73

1

243 log7

1

49)(log

14

1

xcx

xxcxBxAxy

1.12 APPLICATIONS OF DIFFERENTIAL EQUATIONS TO ENGINEERING PROBLEMS

Q.29 A particle of unit mass falls under gravity in a resisting medium whose resistance varies with velocity. Find the relation between distance and velocity if initially the particle starts from rest.

Sol. Let v be the velocity when the particle has fallen distance s in time t from rest so that the resistance due to velocity is vk per unit of mass. Hence the force of resistance on the particle of mass

m is mkv against the direction of motion. Also, the force due to gravity, mg is acting vertically

downwards.


36

Equation of motion of the particle is

mkvmgdt

dvm

kvgdt

dv

dtkvg

dv

Integration gives,

1)log(1

ctkvgk

(1)

Since 0v when 0t

gk

c log1

1

tkvg

g

k

log

1

)1( ktek

g

dt

dsv (2)

Integrating (2), we get

22ce

k

gt

k

gs kt

Since 0s when 0t ,

2

2

k

gc

)1(2

ktek

gt

k

gs (3)

Eliminating t between (2) and (3), we get

k

v

kvg

g

k

gs

log

2

which gives a relation between distance and velocity.


37

APPLICATION TO ELECTRIC CIRCUITS

If Q is the electrical charge on a condenser of capacity C and i is the current, then

(a) dt

dQi or dtiQ

(b) the potential drop across the resistance R is Ri.

(c) the potential drop across the inductance L is dt

diL .

(d) the potential drop across the capacitance C is C

Q.

Also, by Kirchhoff’s Law, the total potential drop (voltage drop) in the circuit is equal to the applied voltage (E.M.F.).

Q.30 An inductance L of 2.0 H and a resistance R of 20 are connected in series with an e.m.f. E volt. If the current I is zero when t=0, find the current i at the end of 0.01 second if E=100V, using the following differential equation.

diL Ri E

dt

Sol. The given equation is

L

Ei

L

R

dt

di (1)

LRtdtLR

eeFI //

..

Solution of (1) is

1

/// ceR

Edte

L

Eie LRtLRtLRt

(2)

at ,0t 0i From (2), R

Ec 1

So, from (2), )1( / LRteR

Ei

Putting the values of E, R, L and t , we get

4758.0]1[20

100 1.0 ei ampere.


38

ASSIGNMENT-I

LINEAR DIFFERENTIAL EQUATION WITH CONSTANT COEFFIECIENTS

Solve the following differential equations:

1. D2y – 3 Dy + 2y = Cosh x + Cos3x, 2. (D2 + 4)y = ex + Sin2x, 3. (D2 + 4D + 3)y = e-3x. 4. D3y + 3D2y + 3Dy + y = e-x 5. (D3 – 2D – 4)y = x4 + 3x2 6. (D2-a2)y = eax – e-ax 7. (D+2)(D-1)2y = e-2x+2 sinhx

8. (D-1)2(D2+1)2y = sin2xe

2

x

9. (D2 – 4 D + 2)y = Sin2x, given that y = 1/8 and Dy = 4 when x = 0. also the find y when x = 4

10. (D2+4D+8)y=sin(2x+3) 11.(D2+5D-6)y=sin4x.sinx 12.(D3+8)y= x4+2x+1 11. (D-2)2y=8(e2x+sin2x+x2), 14.D2- 4y =xsinhx 15.(D2 – 4D + 1)y = e2x Sinx

HOMOGENOUS LINEAR DIFFERENTIAL EQUATION:


1. 2 3 xx y xy y x e

2. 2 5 4 log .x y xy y x x

3. 3

4

3

d yx

dx

23 2

22 1

d y dyx x xy

dx dx

4. 3 2

3 2

3 2

12 2 10 .

d y d yx x y x

dx dx x

5. 2

2

2

(log )sin(log ) 13

d y dy x xx x y

dx dx x

6. 2

2

2(2 3) 2(2 3) 12 6

d y dyx x y x

dx dx


39

SIMULTANEOUS LINEAR DIFFERENTIAL EQUATION:


1. 2

( ) 1dx

x ydt t

, 1

( 5 )dy

x y tdt t

2. 2 , 2x t x dt tdy tx ty x t dt

3. 0t x y and 0,ty x given x(1) = 1, y(-1) = 0.

4. sin , y x cost.x y t

5. 2

2

24 5

d xx y t

dt ,

2

25 4 1

d yy y t

dt

6. , 55 , 5t tdx dyx y e x y e

dt dt

METHOD OF VARIAION OF PARAMETERS:


1. .nxsecyny 2

2. .x2yy2y

3. .ey6yy x

4. .x2tany4y

5. .xe64y3y2y x

6. .x2secey5y2y x

7. .xlogeydx

dy2

dx

yd x

2

2

8. .e)x1(y2dx

dy)5x2(

dx

yd)2x( x

2

2

NORMAL FORM OR BY REMOVING THE FIRST DERIVATIVE:

1. .0y)2x2x(y)xx(2yx 222

2. .esecx 5yy xtan2y x

3. .ey)2x(yx2y)x2x(

2

1

22

4. xsecxtanydx

dy2xcoty

dx

yd2

2


40

5. 0)xx8(x4

y

dx

dy

x

1

dx

yd22

2

BY CHANGING THE INDEPENDENT VARIABLE:

1. ).x1log(cos4yy)x1(y)x1( 2

2. .0xcosy y xtany 2

3. .0y4y)x1(x2y)x1( 222

4. 332

2

2

x2yx4dx

dy)1x4(

dx

yd

5. .xyx4yx

1y 42

SECOND ORDER DIFFERENTIAL EQUATION:

1. y2yxSin 2 , given that y = cot x is a solution of it.

2. 0xcosyyxcosxy)xcosxsinx( of which y = x is a solution.

3. x322 exy)2x(y)x2x(yx , given that y = x is a solution

4. 0y)2x4(yx4y 2 given that 2xey is a solution

Mathematics (NAS-203) PracticeTest Unit-I(Diff. equations.)

Time- 60 min M.M. 30

Note: Attempt any Five Questions. All questions carry equal marks

1. Solve the differential equation D2y – 3 Dy + 2y = Cosh x + Cos3x,

2. Solve the differential equation3 2

4 3 2

3 22 1

d y d y dyx x x xy

dx dx dx

3. Normal form or by Removing the first Derivative solve 2 2 22( ) ( 2 2) 0.x y x x y x x y

4. By Changing the Independent Variable solve 2(1 ) (1 ) 4coslog(1 ).x y x y y x

5. By method of Variaion of parameters solve 2 sec .y n y nx

6. Solve the differential equation 2

2

24 5

d xx y t

dt ,

2

25 4 1

d yy y t

dt


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NIT I DIFFERENTIAL QUATION - upsingh.inupsingh.in/Students/bt2/unit1_nas203.pdf · Series solution of second order ordinary differential equations with variable coefficient (Frobenius