1

UNIT-I

ORDINARY DIFFERENTIAL EQUATIONS

Higher order differential equations with constant coefficients Method of variation of

parameters Cauchys and Legendres linear equations Simultaneous first order

linear equations with constant coefficients.

The study of a differential equation in applied mathematics consists of three phases.

(i) Formation of differential equation from the given physical situation, called

modeling.

(ii) Solutions of this differential equation, evaluating the arbitrary constants

from the given conditions, and

(iii) Physical interpretation of the solution.

HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH

CONSTANT COEFFICIENTS.

General form of a linear differential equation of the nth order with constant

coefficients is

XyKdx

ydK

dx

ydK

dx

ydnn

n

n

n

n

n

=++++

..............2

2

21

1

1 .. (1)

Where KnK

K ...............,.........21 , are constants.

The symbol D stands for the operation of differential

(i.e.,) Dy = ,dx

dy similarly D ...,,

3

33

2

22 etc

dx

ydyD

dx

dy

y

==

The equation (1) above can be written in the symbolic form

(D XyKDK nnn

=+++ )..........11 i.e., f(D)y = X

Where f (D) = D nnn KDK +++ ...........11

Note

1. = XdxXD1

2. dxXeeXaD

axax

=

1

3. dxXeeXaD

axax

=

+

1

(i) The general form of the differential equation of second order is

2

(1)

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

(ii)Differential operators:

The symbol D stands for the operation of differential

(i.e.,) Dy = ,dx

dy D

2

22

dx

ydy =

D

1Stands for the operation of integration

2

1

DStands for the operation of integration twice.

(1) can be written in the operator form

RQyPDyyD =++2 (Or) ( RyQPDD =++ )2

(iv) Complete solution = Complementary function + Particular Integral

PROBLEMS

1. Solve (D 0)652 =+ yD

Solution: Given (D 0)652 =+ yD

The auxiliary equation is m `0652 =+ m i.e., m = 2,3

xx BeAeFC 32. += The general solution is given by

xx BeAey 32 +=

2. Solve 0362

2

=+ ydx

dy

dx

yd

Solution: Given ( ) 0362 =+ yDD The auxiliary equation is 01362 =+ mm

i.e., 2

52366 =m

= i23 Hence the solution is )2sin2cos(3 xBxAey x +=

3. Solve (D 0)12 =+ given y(0) =0, y(0) = 1

RQydx

dyP

dx

yd=++

2

2

3

Solution: Given (D 0)12 =+

A.E is 012 =+m M = i Y = A cosx + B sinx

Y(x) = A cosx + B sinx

Y(0) = A =0

Y(0) =B =1

A = 0, B = 1 i.e., y = (0) cosx + sinx

y = sinx

3. Solve xeyDD 22 )134( =+

Solution: Given xeyDD 22 )134( =+

The auxiliary equation is 01342 =+ mm

im 322

364

2

52164 ===

( )xBxAeFC x 3sin3cos. 2 += P.I. = xe

DD

2

2 134

1

+

= xe2

1384

1

+

= xe2

9

1

y = C.F +P.I.

y = ( )xBxAe x 3sin3cos2 + + xe29

1

5. Find the Particular integral of y- 3y + 2y = exx e2

Solution: Given y- 3y + 2y = e xx e2

xx eeyDD 22 )23( =+

P.I xeDD 23

121 +

=

= xe431

1

+

= xe0

1

=x xeD 32

1

= x xe32

1

= xxe

4

P.I ( )xeDD

2

22 23

1

+=

= - xe2

264

1

+

= - x xeD

2

32

1

= - x xe2

34

1

= - xe x2

P.I. = P.I 21 .IP+

= xxe + (- xe x2 )

= -x( xe + e x2 )

6. Solve xydx

dy

dx

ydcosh254

2

2

=+

Solution: Given xydx

dy

dx

ydcosh254

2

2

=+

The A.E is m 0542 =+ m

m = i= 22

20164

C.F = e ( )xBxAx sincos2 + P.I = ( )

+

++=

++

254

12cosh2

54

122

xx ee

DDx

DD

= xx eDD

eDD

++

+++

54

1

54

122

=541541 +

++

xx ee

=210

xx ee

IPFCy .. +=

= e ( )xBxAx sincos2 + - 210

xx ee

Problems based on P.I = ( ) axDf

oraxDf

cos)(

1sin)(

1

Replace D 22 aby

7. Solve xydx

dy

dx

yd3sin23

2

2

=++

5

Solution: Given xydx

dy

dx

yd3sin23

2

2

=++

The A.E is m 0232 =++ m (m+1)(m+2) = 0

M = -1, m = -2

C.F = Ae xx Be 2 +

P.I = xDD

3sin23

12 ++

= xD

3sin233

12 ++

(Replace D 22 aby )

= xD

D

Dx

D3sin

)73(

)73(

73

13sin

73

1

+

+

=

= xD

D3sin

)7()3(

7322

+

= xD

D3sin

499

732

+

= xD

3sin49)3(9

732

+

= xD

3sin130

73

+

= ( )xxD 3sin7)3(sin3130

1+

= ( )xx 3sin73cos9130

1+

IPFCy .. +=

Y = Ae xx Be 2 + ( )xx 3sin73cos9130

1+

8. Find the P.I of (D xsin)12 =+

Solution: Given (D xsin)12 =+

P.I. = xD

sin1

12 +

= xsin11

1

+

= x xDsin

2

1

= xD

xsin

1

2

= xdxx

sin2

P.I =-2

cos xx

6

9. Find the particular integral of (D xxy sin2sin)12 =+

Solution: Given (D xxy sin2sin)12 =+

=- ( )xx cos3cos2

1

= - xx cos2

13cos

2

1+

P.I 1 =

+x

D3cos

2

1

1

12

= x3cos19

1

2

1

+

= x3cos16

1

P.I 2 =

+x

Dcos

2

1

1

12

= xcos11

1

2

1

+

= xD

x cos2

1

2

1

= xdxx

cos4

= xxsin

4

P.I = x3cos16

1 + x

xsin

4

Problems based on R.H.S = e axeorax axax cos)(cos ++

10. Solve (D xeyD x 2cos)44 22 +=+

Solution: Given (D xeyD x 2cos)44 22 +=+

The Auxiliary equation is m 0442 =+ m (m 2 ) 2 = 0

m = 2 ,2

C.F = (Ax +B)e x2

P.I 1 = xe

DD

2

2 44

1

+

= xe2

484

1

+

= xe2

0

1

= x xeD

2

42

1

= x xe2

0

1

7

= x xe22

2

1

P.I 2 = x

DD2cos

44

12 +

= xD

2cos442

12 +

= xD

2cos4

1

=

x

D2cos

1

4

1

= 8

2sin

2

2sin

4

1 xx=

IPFCy .. +=

y = (Ax +B)e x2

8

2sin x

Problems based on R.H.S = x

Note: The following are important

.......1)1( 321 ++=+ xxxx

.................1)1( 321 ++++= xxxx

..............4321)1( 322 ++=+ xxxx

..............4321)1( 322 ++++= xxxx

11. Find the Particular Integral of (D xy =+ )12

Solution: Given (D xy =+ )12

A.E is (m 0)12 =

m = 1 C.F = Ae xx Be+

P.I = xD 1

12

= xD 21

1

= [ ] xD 121 = ( )[ ]xDD .........1 22 +++ = [ ]...........000 +++ x = - x

12. Solve: ( ) 3234 2 xyDDD =+

8

Solution: Given ( ) 3234 2 xyDDD =+ The A.E is m 02 234 =+ mm m 0)12( 22 =+ mm

0)1( 22 =mm

m = 0,0 , m = 1,1

C.F = (A + Bx)e xx eDxC )(0 ++

P.I = 3234 2

1x

DDD +

= ( )[ ] 322 211

xDDD +

= ( )[ ] 3122

211

xDDD

+

= ( ) ( ) ( )[ ]...............22211 322222

++ DDDDDDD

= [ ] 34322

43211

xDDDDD

++++

= [ ]241861 232

+++ xxxD

=

+++ x

xxx

D24

2

18

3

6

4

1 334

= 2

24

6

18

12

6

20

2345 xxxx+++

= 2345

123220

xxxx

+++

IPFCy .. +=

y = (A + Bx)e xx eDxC )(0 ++ + 2345

123220

xxxx

+++

Problems based on R.H.S = xeax

P.I = xaDf

exeDf

axax

)(

1

)(

1

+=

13. Obtain the particular integral of ( xeyDD x 2cos)522 =+

Solution: Given ( xeyDD x 2cos)522 =+

P.I = xeDD

x 2cos52

12 +

= ( ) ( ) ( )1Re2cos51211

2=

+++placeDbyDx

DDe x

9

= xDDD

e x 2cos52212

12

+++

= xD

e x 2cos4

12

+

= xe x 2cos44

1

+

= xD

xe x 2cos2

1

= xdxxe x

2cos2

P.I = 4

2sin xxe x

14. Solve ( xeyD x sin)2 22 =+

Solution: Given ( xeyD x sin)2 22 =+

A.E is (m 0)12 =+

m = -2, -2

C.F. = (Ax + B)ex2

P.I = ( ) xeDx sin

2

1 22

+

= e ( ) xDx sin

22

12

2

+

= e xD

x sin12

2

= e xx sin1

12

= -e xx sin2

IPFCy .. +=

y = (Ax + B)e x2 - e xx sin2

Problems based on f(x) = x axxorax nn cos)(sin

To find P.I when f(x) = x axxorax nn cos)(sin

P.I = axxoraxxDf

nn cos)(sin)(

1

VDfdD

dV

DfxxV

Df

+=

)(

1

)(

1)(

)(

1

i.e., ( )

VDfDf

DfV

DfxxV

Df

=

)(

1

)()(

1)(

)(

1 '

[ ] VDfDf

VDf

xxVDf 2

'

)(

)(

)(

1

)(

1=

10

15. Solve (xx xexeyDD 32 sin)34 +=++

Solution: The auxiliary equation is m 0342 =++ m m= -1,-3

C.F =A xx Bee 3 +

P.I ( )xeDD

x sin)34(

121

++=

= ( ) ( )[ ] xDD sin3141 12 ++ = e ( ) xDDx sin2

12 +

= e( )

xD

Dx sin41

212+

+

= [ ]xxe x sincos25

+

P.I ( ) xDDe x 3)3(4)31

2

3

2++++

=

= e ( ) xDDx 24101

2

3

++

= xDDe x

123

24

101

24

++

= xDe x

12

51

24

3

=

12

5

24

3

xe x

General solution is IPFCy .. +=

y = Axx Bee 3 + [ ]xxe x sincos2

5+

+

12

5

24

3

xe x

16 Solve xxydx

dy

dx

ydcos2

2

2

=++

Solution: A.E : m 0122 =++ m m = -1,-1

C.F = xeBxA + )(

P.I = )cos()1(

12

xxD +

= ( ) ( ) ( )xDDD

x cos1

1

1

)1(222 +

+

+

11

= ( ) ( )( )xDDDx cos121

1

22 ++

+

= ( ) ( )xDDx cos1211

1

2

++

+

= 2

sin

1

2 x

Dx

+

= 2

sin xx

= 1

sin

2

sin

+

D

xxx

= ( )

1

sin1

2

sin2

D

xDxx

= 2

sincos

2

sin xxxx +

The solution is y = xeBxA + )( +

2

sincos

2

sin xxxx +

17. Solve ( ) xyD 22 sin1 =+ Solution: A.E : m 012 =+ m = i C.F =A cosx +B sinx

P.I = xD

2

2sin1

1

+

=

+ 2

2cos1

1

12

x

D

=

+

+x

De

D

x 2cos1

1

1

1

2

12

0

2

=

+ x2cos3

11

2

1

= x2cos6

1

2

1+

= y A cosx +B sinx + x2cos6

1

2

1+

18. Solve ( ) xexxxydx

yd++= 1sin

2

2

Solution: A.E : m 012 =

m = 1 C.F = A xx Bee +

P.I ( ) )()(

1

)(

)('

)(

11 V

DfDf

DfxxV

Df

==

12

= ( )( )xDDD

x sin1

1

1

222

= 2

sin

1

22

x

D

Dx

= ( )

+12

cos2

2

sin2D

xxx

= 2

cos

2

sin xxx

P.I ( ) xexD

2

221

1

1+

=

= ( )[ ] )1(111 22 xDe x ++ = e ( ) )1(2

1 22

xDD

x ++

= ( )

12

932 23 xxxe x +

y = A xx Bee + +( )

12

932 23 xxxe x +

19. Solve xxeydx

yd x sin2

2

=

Solution: A.E : m 012 =

m = 1 C.F = A xx Bee +

P.I = ( ) xxeD x sin112

= e ( )[ ]( )xxDx sin111 2 + = e ( )( )xxDDx sin2

12 +

= e ( ) ( )

+

+x

D

Dx

DDxx sin

12

22sin

2

122

sin ,sin xx = 1,1 22 === putD

= e( )( )

+

xD

Dx

Dxx sin

12

22sin12

12

= e( )( )

( )( )

+

+

+

144

sin22sin

41

2122 DD

xDx

D

Dxx

Put D 12 =

= e( ) ( )( )

++

+ x

D

DDx

Dxx sin

169

4322sin

5

212

13

= e [ ]

+++

xD

DDxx

xx sin169

628cos2sin

5 2

2

= e ( ) ( )

++

xD

xxxx sin

25

214cos2sin

5

P.I = e ( ) ( )

++

25

cos2sin14cos2sin

5

xxxx

xx

Complete Solution is

y = A xx Bee + +

e ( ) ( )

++

25

cos2sin14cos2sin

5

xxxx

xx

METHOD OF VARIATION OF PARAMETERS

This method is very useful in finding the general solution of the second order

equation.

Xyadx

dya

dx

yd=++ 212

2

[Where X is a function of x] .(1)

The complementary function of (1)

C.F = c 2211 fcf +

Where c 21 ,c are constants and f 21andf are functions of x

Then P.I = Pf 21 Qf+

P =

dxffff

Xf

2

'

1

1

21

2

Q =

dxffff

Xf

2

'

1

'

21

1

IPfcfcy .2211 ++=

1. Solve ( ) xyD 2sec42 =+ Solution: The A.E is m 042 =+ m = i2 C.F = C xCx 2sin2cos 21 +

f x2cos1= f x2sin2 =

f x2sin2'1 = f x2cos2'

2 =

f xxfff 2sin22cos2 222'

1

'

21 +=

= 2 [ ]xx 2sin2cos 22 + = 2 [ ]1 = 2

P =

dxffff

Xf

2

'

1

1

21

2

14

= dxxx

2

2sec2sin

= dxx

x2cos

12sin

2

1

=

dxx

x

2cos

2sin2

4

1

= )2log(cos4

1x

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxx

2

2sec2cos

= dxx

x2cos

12cos

2

1

= dx21

= x2

1

P.I = Pf 21 Qf+

= )2log(cos4

1x (cos2x) + x

2

1sin2x

2. Solve by the method of variation of parameters xxydx

ydsin

2

2

=+

Solution: The A.E is m 012 =+ m = i C.F = C xCx sincos 21 +

Here xf cos1 = xf sin2 =

xf sin'1 = xf cos'

2 =

1sincos 222'

1

'

21 =+= xxffff

P =

dxffff

Xf

2

'

1

1

21

2

= dxxxx

1)sin(sin

= xdxx2sin

= ( )

dxx

x2

2cos1

= ( ) dxxxx 2cos21

= + xdxxxdx 2cos21

2

1

15

=

+

4

2cos)1(

2

sin

2

1

22

1 2 xxx

x

= xxxx

2cos8

12sin

44

2

++

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxxx

1

)(sin)(cos

= xdxxx cossin

= dxx

x2

2sin

= xdxx 2sin21

=

4

2sin)1(

2

2cos

2

1 xxx

= xxx

2sin8

12cos

4+

P.I = Pf 21 Qf+

= xxxx

xxxxx

sin2sin8

12cos

4cos2cos

8

12sin

44

2

+

+

++

4. Solve (D xeyD 22 )44 =+ by the method of variation of parameters.

Solution: The A.E is 0442 =+ mm

( ) 02 2 =m m = 2,2

C.F = ( ) xeBAx 2+ = xx BeAxe 22 +

xxef 21 = xef 22 =

xx exef 22'1 2 += , xef 2'2 2=

( ) ( )xxxx exeeexffff 22222'12'21 22 += = 2x ( ) ( ) ( )222222 2 xxx eexe = ( ) [ ]12222 xxe x = ( )22xe = xe4

P=

dxffff

Xf

2

'

1

1

21

2

= dxeeex

xx

4

22

16

= xdx =

Q =

dxffff

Xf

2

'

1

'

21

1

=

dxe

exex

xx

4

22

= xdx

= 2

2x

P.I = xxx ex

ex

ex 22

22

22

22=

y = C.F + P.I

= (Ax +B) e x2 + xex 22

2

5. Use the method of variation of parameters to solve ( ) xyD sec12 =+ Solution: Given ( ) xyD sec12 =+ The A.E is 012 =+m m = i C.F = xcxc sincos 21 +

= 2211 fcfc +

xfxf sin,cos 21 ==

xfxf cos,sin '2'

1 =

1sincos 222'

1

'

21 =+= xxffff

P=

dxffff

Xf

2

'

1

1

21

2

= dxxx

1secsin

= dxxx

cos

sin

= xdxtan = log (cos x)

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxx

1seccos

= dx = x

P.I = 21 QfPf +

= xxxx sincos)log(cos +

y = xcxc sincos 21 + + xxxx sincos)log(cos +

17

6. Solve ( ) axyaD tan22 =+ by the method of variation of parameters. Solution: Given ( ) axyaD tan22 =+ The A.E is 022 =+ am aim = C.F = axcaxc sincos 21 +

axfaxf sin,cos 21 ==

axafxaf cos,sin '2'

1 ==

)sin(sincoscos'12'

21 axaaxaxaxaffff =

= axaaxa 22 sincos +

= )sin(cos 22 axaxa +

= a

P.I = 21 QfPf +

P=

dxffff

Xf

2

'

1

1

21

2

= dxa

axax

tansin

= dxaxax

a cos

sin1 2

=

dxax

ax

a cos

cos11 2

= ( ) dxaxaxa cossec1

= ( )

+

a

axaxax

aa

sintanseclog

11

= ( )[ ]axaxaxa

sintanseclog12

+

= ( )[ ]axaxaxa

tanseclogsin12

+

Q =

dxffff

Xf

2

'

1

'

21

1

= dxaaxax tancos

= axdxasin

1

= axa

cos12

21. QfPfIP +=

= ( )[ ] [ ]axaxa

axaxaxaxa

cossin1

tanseclogsincos1

22+

18

= ( )[ ]axaxaxa

tanseclogcos12

+

IPFCy .. +=

= axcaxc sincos 21 + ( )[ ]axaxaxa

tanseclogcos12

+

7. Solve xydx

ydtan

2

2

=+ by the method of variation of parameters.

Solution: The A.E is 012 =+m im = C.F = xcxc sincos 21 +

Here xfxf sin,cos 21 ==

xfxf cos,sin '2'

1 ==

1sincos 22'12'

21 =+= xxffff

P=

dxffff

Xf

2

'

1

1

21

2

= dxxx

1

tansin

= dxxx

cos

sin 2

= dxx

x

cos

cos1 2

= ( )dxxx cossec = xxx sin)tanlog(sec ++

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxx

1tancos

= dxsin = xcos

21. QfPfIP +=

= )tanlog(seccos xxx +

IPFCy .. +=

= xcxc sincos 21 + )tanlog(seccos xxx +

8. Solve by method of variation of parameters 144 22

''' +=+ xyx

yx

y

Solution: Given 144 22

''' +=+ xyx

yx

y

i.e., 24'''2 44 xxyxyyx +=+

i.e., [ ] 2422 44 xxyxDDx +=+ .(1) Put x = ze

19

Logx = log ze

= z

So that XD = D '

( )1''2 = DDDx (1) ( )[ ] ( ) ( )24''' 441 zz eeyDDD +=+

[ ] zeeyDD z 22 4'' 45 +=+ A.E is 0452 =+ mm ( )( ) 014 = mm 4,1=m

zz ececFC 24

1. +=

Here zz efef == 24

1 ,

zz efef == '24'

1 ,

zzz eeeffff 555'12'

21 34 ==

21. QfPfIP +=

P=

dxffff

Xf

2

'

1

1

21

2

= [ ]

+ dx

e

eeez

zzz

5

24

3

= +

dxe

eez

zz

5

35

3

= [ ]dze z + 2131

=

+

23

1 2 zez

= zez 2

6

1

3

1

Q =

dzffff

Zf

2

'

1

'

21

1

= ( )

+dz

e

eeez

zzz

5

344

3

= +

dze

eez

zz

5

68

3

1

= ( ) + dzee zz331

=

+ z

z

ee

33

1 3

= zz ee3

1

9

1 3

zzzzz eeeeezIP

+

=

3

1

9

1

6

1

3

1. 342

20

= zzzz eeeze 2424

3

1

9

1

6

1

3

1

IPFCy .. +=

= zz ecec 24

1 + +zzzz eeeze 2424

3

1

9

1

6

1

3

1

= ( ) ( ) ( )3

1

9

1

6

1

3

1)(

424

2

4

1 ++zzzzz eeezecec ( )2ze

= ( ) ( )29

log3

244

2

4

1

xxx

xecec zz ++

DIFFERENTIAL EQUATIONS FOR THE VARIABLE COEFFICIENTS

(CAUCHYS HOMOGENEOUS LINEAR EQUATION)

Consider homogeneous linear differential equation as:

)1(............................1

11

1 Xyadx

ydxa

dx

yda nx

nn

n

nn

=+++

(Here as are constants and X be a function of X) is called Cauchys

homogeneous linear equation.

)(.............. 011

1

1 xfyadx

dyxa

dx

yda

dx

ydxa

n

n

nn

nn

n =++++

is the homogeneous linear equation with variable coefficients. It is also known

as Eulers equation.

Equation (1) can be transformed into a linear equation of constant Coefficients

by the transformation.

xzorex z log)(, ==

Then

dz

dy

xdx

dz

dz

dy

dx

dyDy

1===

==

dz

d

dx

dD ,

===dz

dxD

dx

dx

Similarly

=

=

dz

dy

xdx

d

dx

dy

dx

d

dx

yd 12

2

=dx

dz

dz

dy

dz

d

xdz

dy

xdz

dy

dx

d

xdz

dy

x

+=

+

111122

2

2

222

2 11

dz

yd

xdz

dy

xdx

yd+=

dz

dy

dz

yd

dx

ydx =

2

2

2

22

21

( ) ( )1222 == Dx Similarly,

( )( )2133 = Dx

( )( )( )32144 = Dx and soon.

= xD ( )122 = Dx ( )( )2133 = Dx

and so on.

1. Solve )sin(log42

22 xy

dx

dyx

dx

ydx =++

Solution: Consider the transformation

xzorex z log)(, ==

= xD ( )122 = Dx ( ) )sin(log4122 xyxDDx =++ ( ) ( )zy sin412 =+ R.H.S = 0 : ( ) 012 =+ y A.E : imm ==+ ,012 C.F = A cosz + B sinz

C.F = A cos (log x) +B sin (log x)

P.I = ( )zsin41

12 +

= zzzz

cos22

cos4 =

P.I = )cos(loglog2 xx

Complete solution is: y = A cos (log x) +B sin (log x) )cos(loglog2 xx

y = ( ) )cos(loglog2)cos(loglog2 xxxxA

2. Solve xxydx

dyx

dx

ydx log24

2

22

=++

Solution: Given ( ) xxyxDDx log2422 =++ (1) Consider: xzorex z log)(, ==

= xD ( )122 = Dx

(1) : ( )( ) zey z=++ 241

22

( ) zzey =++ 232 A.E : 0232 =++ mm M = - 2, -1

C.F = 12 loglog2 +=+ xxzz BeAeBeAe

C.F = x

B

x

A+

2

P.I = ( )( )ze z231

2 ++

=e ( ) ( ) zz

2131

12 ++++

= ( ) ze z 651

2 ++

= ze z

12

6

51

6

++

=

=

6

5

66

51

6z

ez

e zz

=

6

5log

6

log

xe x

=

6

5log

6x

x

Complete solution is y = C.F +P.I

= x

B

x

A+

2+

6

5log

6x

x

3. Solve ( giventhatxyxDDx ,)43( 222 =+ y(1) = 1 and y(1) = 0

Solution: Given ( ) 222 43 xyxDDx =+ (1) Take xzorex z log)(, ==

= xD ( )122 = Dx

(1) : ( ) zey 22 44 =+ A.E : 2,2,0442 ==+ mmm

C.F = ( ) ( ) 22 log xxBAeBzA z +=+ P.I = ( ) ( ) ( ) )1(22

1

2

12

22

2+

=

zz ee

= ( )112

2

ze

P.I = 2

22 ze z

23

= ( )2

log22 xx (2)

Complete solution is : y = ( ) 2log xxBA + + ( )2

log22 xx

Apply conditions: y(1) =1,y(1) = 0 in (2)

A = 1, B = -2

Complete solution is y = ( ) + 2log21 xx ( )2

log22 xx

EQUATION REDUCIBLE TO THE HOMOGENEOUS LINEAR

FORM (LEGENDRE LINEAR EQUATION)

It is of the form:

( ) ( ) )(...........1

11

1 xfyAdx

ydbxaA

dx

ydbxa nn

nn

n

nn

=+++++

..(

1)

nAAA ....,..........,.........21 , are some constants

It can be reduced to linear differential equation with constant

Coefficients,

by taking: )log()( bxazorebxa z +==+

Consider givesdz

dD

dx

d,, ==

( ) ( ) ( )ybDybxadz

dyb

dx

dybxa =+=+

Similarly ( ) ( )ybyDbxa 1222 =+ (2) ( ) ( )( )21333 =+ byDbxa and so on Substitute (2) in (1) gives: the linear differential equation of

constant Coefficients.

Solve ( ) ( ) xyyxyx 612'32''32 2 =++ Solution: This is Legendres linear equation:

( ) ( ) xyyxyx 612'32''32 2 =++ .(1) Put z = log (2x + 3) , 32 += xe z ( ) 232 =+ Dx ( ) ( )

dz

dDx ==+ ,432 222

Put in (1) : ( ) 931264 2 = zey R.H.S =0

A.E : 01264 2 = mm

4

573,

4

57321

=

+= mm

C.F =zmzm

BeAe 21 + C.F = 21 )32()34(

mmxBxA +++

24

P.I 1=1264

32

ze= ( )32

14

3+ x

P.I 2 = 1264

92

ze

= 4

3

12

9=

Solution is

y = ( )( ) ( ) + +++ 45734/573 3232 xBxA ( )3214

3+ x

4

3

SIMULTANEOUS FIRST ORDER LINEAR EQUATIONS WITH

CONSTANT COEFFICIENTS

1. Solve the simultaneous equations, 023,232 2 =++=++ yxdt

dyeyx

dt

dx t

Solution: Given 023,232 2 =++=++ yxdt

dyeyx

dt

dx t

Using the operator D = dt

d

( ) teyxD 2232 =++ (1) ( ) 023 =++ yDx .(2) Solving (1) and (2) eliminate (x) :

( ) )3....(..........654)2()2()1(3 22 teyDDD =++ A.E : 0542 =+ mm m = 1,-5

C.F = tt BeAe 5+

P.I = t

t

eDD

e 22

2

7

6

54

6=

+

y = tt BeAe 5+ te2

7

6

put in (1) : x = ( )[ ]yD 23

1+

x = ttt eBeAe 25

7

8++

solution is :

x = ttt eBeAe 25

7

8++

y = tt BeAe 5+ te2

7

6

2. Solve giventhattxdt

dyty

dt

dx,cos,sin =+=+ t=0, x = 1, y =0

Solution: Dx + y = sin t ..(1)

x + Dy = cost (2)

Eliminate x : (1) (2)D ttyDy sinsin2 +=

25

( ) )3.....(..........sin212 tyD = 1,012 == mm C.F = Ae

tt Be+

P.I = tt

D

tsin

11

sin)2(

1

sin2

2=

=

y = Aett Be+ + sint

(2) : x = cost D(y)

x = cost - ( )tBeAedt

d tt sin++

x = tBeAet tt coscos + x =

tt BeAe + Now using the conditions given, we can find A and B

BAxt +=== 11,0 BAyt +=== 00,0

B = 2

1, A =

2

1

Solution is

x = tee tt cosh2

1

2

1=+

y = tttee tt sinhsinsin2

1

2

1=++

3. Solve txdt

dyty

dt

dxcos2,sin2 ==+

Solution: Dx + 2y = -sin t ..(1)

- 2x +Dy = cos t ...(2)

(1) )(cossin24)2(2 2 tDtyDyD +=++

( ) tD sin342 =+ 2,042 imm ==+ C.F = tBtA 2sin2cos +

P.I = tt

D

tsin

41

sin3

4

sin32

=

+

=

+

y = tBtA 2sin2cos + - sin t

(2) : x = [ ]tDy cos2

1

x = ( )

+ tttBtA

dt

dcossin2sin2cos

2

1

x = A cos2t +Bsin2t cost

Solution is :

x = A cos2t +Bsin2t cost

y = tBtA 2sin2cos + - sin t

26

4. Solve txdt

dy

dt

dxty

dt

dy

dt

dx2sin2,2cos2 =+=+

Solution: Dx + (-D +2)y = cos 2t ..(1)

(D 2)x +Dy = sin 2t (2)

Eliminating y from (1) and (2)

( ) ttxDD 2cos2sin2222 +=+ R.H.S = 0 0222 =+ mm m = i1 C.F = ( )tBtAe t sincos + P.I 1=

( )D

t

DD

t

+=

+

1

2sin

22

2sin22

= ( )

41

)2(sin2sin2sin

1

12 +

=

tDtt

D

D

= 5

2cos22sin tt

P.I 2 = ( )tDD

2cos22

12 +

= ( )

10

2sin22cos tt +

x = ++ )sincos( tBtAe t 5

2cos22sin tt ( )10

2sin22cos tt +

(1) +(2) ttxydt

dx2sin2cos222 +=+

2y = cos2t +sin 2t + 2x -dt

dx2

y =

++

dt

dxxtt 222sin2cos

2

1(3)

Substitute x in (3)

y = ( )2

2sinsincos

ttBtAe t

Solution is :

x = ++ )sincos( tBtAe t 5

2cos22sin tt ( )10

2sin22cos tt +

y = ( )2

2sinsincos

ttBtAe t

UNIT-II

VECTOR CALCULUS

Directional derivative

The derivative of a point function (scalar or vector) in a particular

direction is called its directional derivative along the direction.

The directional derivative of a scalar point function in a given direction is the rate of change of in the direction. It is given by the component of grad in that direction. The directional derivative of a scalar point function

(x,y,z) in the direction of a is given by

a

a..

Directional derivative of is maximum in the direction of . Hence the maximum directional derivative is grador Unit normal vector to the surface

If (x, y, z) be a scalar function, then (x, y, z) = c represents A surface and the unit normal vector to the surface is given by

Equation of the tangent plane and normal to the surface

Suppose

a is the position vector of the point ),,( 000 zyx

On the surface (x, y, z) = c. If ++= kzjyixr is the position vector of any point (x,y,z) on the tangent plane to the surface at

a , then the

equation of the tangent plane to the surface at a given point a on it is given by 0. =

gradar

If

r is the position vector of any point on the normal to the surface

at the point

a on it. The vector equation of the normal at a given point

a on the surface is 0=

gradar The Cartesian form of the normal at ),,( 000 zyx on the surface

(x,y,z) = c is

z

zz

y

yy

x

xx o

=

=

00

Divergence of a vector

If ),,( zyxF

is a continuously differentiable vector point function in

a given region of space, then the divergences of

F is defined by

z

Fk

y

Fj

x

FiFdivF

+

+

==

.

=x

Fi

If

++= kFjFiFF 321 ,then ).( 321

++= kFjFiFFdiv

i.e., z

F

y

F

x

FFdiv

+

+

=

321

Solenoidal Vector

A vector

F is said to be solenoidal if 0=

Fdiv (ie) 0. =

F

Curl of vector function

If ),,( zyxF

is a differentiable vector point function defined at each

point (x, y, z), then the curl of

F is defined by

= FFcurl

= z

Fk

y

Fj

x

Fi

+

+

= x

Fi

If

++= kFjFiFF 321 ,then )( 321

++= kFjFiFFcurl

321 FFF

zyx

kji

Fcurl

=

=

+

y

F

x

Fk

z

F

x

Fj

z

F

y

Fi 121323

Curl

F is also said to be rotation

F

Irrotational Vector

A vector

F is called irrotational if Curl 0=

F

(ie) if 0=

F

Scalar Potential

If

F is an irrotational vector, then there exists a scalar function Such that =F . Such a scalar function is called scalar potential of F

Properties of Gradient

1. If f and g are two scalar point function that ( ) gfgf = (or) ( ) gradggradfgfgrad =

Solution: ( ) ( )gfz

ky

jx

igf

+

+

=

= ( ) ( ) ( )

+

+

gfz

kgfy

jgfx

i

= z

gk

z

fk

y

gj

y

fj

x

gi

x

fi

+

+

=

+

+

+

+

z

gk

y

gj

x

gi

z

fk

y

fj

x

fi

= gf

2. If f and g are two scalar point functions then ( ) fggffg += (or) ggradffgradgfggrad +=)(

Solution: ( ) = fg ( )fgz

ky

jx

i

+

+

= ( ) ( ) ( )

+

+

fgz

kfgy

jfgx

i

=

+

+

+

+

+

z

fg

z

gfk

y

fg

y

gfj

x

fg

x

gfi

=

+

+

+

+

+

z

fk

y

fj

x

fig

z

gk

y

gj

x

gif

= fggf +

3. If f and g are two scalar point function then 2g

gffg

g

f =

where

0g

Solution: =

g

f

+

+

g

f

zk

yj

xi

=

g

f

xi

=

2g

x

gf

x

fg

i

=

x

gif

x

fig

g 21

= [ ]gffgg

2

1

4. If

++= kzjyixr such that rr =

,prove that

= rnrr nn 2

Solution: nn r

zk

yj

xir

+

+

=

=

+

+

z

rk

y

rj

x

ri

nnn

= z

rnrk

y

rnrj

x

rnri nnn

+

+

111

=

++

r

zk

r

yj

r

xinr n 1

=

++

kzjyixr

nr n 1

=

rr

nr n 1

5. Find a unit normal to the surface 422 =+ xzyx at (2,-2, 3)

Solution: Given that xzyx 22 +=

)2( 2 xzyxz

ky

jx

i +

+

+

=

= ( ) ( ) ( )xkxjzxyi 222 2 +++ At (2,-2, 3)

( ) )4()4(68 +++= kji =

++ kji 442

63616164 ==++= Unit normal to the given surface at (2,-2,3)

=6

442

++ kji

=

++

kji 223

1

6. Find the directional derivative of xyzxzyzx ++= 22 4 at (1,2,3) in the direction of

+ kji2

Solution: Given xyzxzyzx ++= 22 4

)4( 22 xyzxzyzxz

ky

jx

i ++

+

+

=

= ( ) ( ) ( ) +++++++ kxyxzyxjxzzxiyzzxyz 842 222 At (1, 2, 3)

++= kji 28654 Given:

+= kjia 2

6114 =++=

a

=a

aDD ..

= 6

22.28654

+

++kji

kji

= [ ] [ ]866

1286108

6

1=+

7. Find the angle between the surface 5222 =++ zyx and

52222 =++ xzyx at (0,1,2)

Solution: Let 222

1 zyx ++= and xzyx 22222 ++= z

zy

yx

x2,2,2 111 =

=

=

zz

yy

xx

2,2,22 222 =

=

=

++= kzjyix 2221

++= kzjyix 22)22(2 At (o,1,2)

+= kj 421

++= kji 4222

Cos644416

422.42.

21

21

+++

++

+=

=

kjikj

2420

20

2420

164cos =

+=

=

2420

20cos 1

=

24

20cos 1

8. Find the angle between the surfaces 1log 2 = yzx and zyx = 22 at the

point (1,1,1)

Solution: let zxy log21 = and zyx += 22

z

x

zy

yz

x=

=

=

111 ,2,log

1,,2 2222 =

=

=

zx

yxy

x

+= kz

kjyiz 2)log(1

= kj22

Cos65

1

11414

12.

21

21=

+++

=

=

=

65

1cos 1

9. Find ( )nr2 Solution: ( )nr2 = ( )nr. = ( ) ( ) ( )nnn r

zkr

yjr

xi

+

+

= z

rnrk

y

rnrj

x

rnri nnn

+

+

111

++= kzjyixr

222 zyxrr ++==

2222 zyxr ++=

r

x

x

rx

x

rr =

=

22

r

y

y

ry

y

rr =

=

22

r

z

z

rz

z

rr =

=

22

( )= nr2

++

r

zk

r

yj

r

xinr n 1

=

++

kzjyixnr n 2

=

rnr

n 2

Since

+=

udivuu .

( )

=

rnrr nn 22

= ( ) +

rnrrnr nn .. 22

++

+

+

=

kzjyixz

ky

jx

ir.

=1+1+1 = 3

( ) ( ) += rrnnrr nnn .3 222 = ( )( ) 242 .23 rrnnnr nn + = ( )( )22 23 + nn rnnnr

( ) [ ] ( ) 2222 1 +=+= nnn rnnnnrr

10. If

++= kzjyixr and rr =

.Prove that

rr n is solenoidal if 3=n and

rr n is irrotational for all vectors of n.

Solution:

rr n

++= krjyrixr nnn

div ( ) ( ) ( )zrz

yry

xrx

rr nnnn

+

+

=

(1)

Now 2222 zyxr ++=

Differentiating partially w.r.to x,

r

x

x

rx

x

rr =

=

22

Similarly, r

y

y

ry

y

rr =

=

22

r

z

z

rz

z

rr =

=

22

Now ( ) ( ) nnn rx

rr

rxxr

x+

=

.

= x.nnn r

r

xr +1

( ) nnn rynryry

+=

22

( ) nnn rznrzrz

+=

22

From (1) we have

( ) nnn rzyxnrrrdiv 32222 +++=

= nn rnr 3+

= ( ) nrn 3+ The vector

rr n is solenoidal if

rrdiv n = 0

( ) 03 =+ nrn 03 =+ n 3= n

rr n is solenoidal only if n = -3

Now

zryrxr

zyx

kji

rrcurl

nnn

n

=

= ( ) ( )

yrz

zry

i nn

=

z

rynrz

y

rnri nn 11

=

r

zynrz

r

ynri nn 11

= ( ) yznryznri nn 22 = 0

Curl (

rr n ) =

++ kji 000 =0

Curl (

rr n ) = 0 for all values of n

Hence

rr n is irrotational for all values of n.

11. Prove that ( ) ( ) +++= kxzjxyizxyF 232 34sin2cos is irrotational and find its scalar potential

Solution:

232 34sin2cos xzxyzxy

zyx

kji

Fcurl

+

=

= [ ] [ ] [ ] 0cos2cos23300 22 =+ xyxykzzji

F is irrotational.

To Find such that gradF = ( ) ( )

zk

yj

xikxzjxyizxy

+

+

=+++ 222 34sin2cos

Integrating the equation partially w.r.to x,y,z respectively

),(sin 132 zyfxzxy ++=

),(4sin 22 zxfyxy +=

),(33 yxfxz +=

,4sin 32 Cyxzxy ++= is scalar potential

12. Prove that ).().(

=

BcurlAAcurlBBAdiv

Proof : ).(

=

BABAdiv

=

BAx

i

=

+

Bx

Ai

x

BAi

=

+

Bx

AiA

x

Bi

=

+

Bx

AiA

x

Bi ..

=

+

BAcurlABcurl ..

13.Prove that

=

FFFcurlcurl 2

Solution:

=

FFcurlcurl

By using

=

cbabcacba ..

= ( )

FF ..

=

FF 2.

VECTOR INTEGRATION

Line, surface and Volume Integrals

Problems based on line Integral

Example 1:

If ( ) ++= kxzjyziyxF 22 201463 Evaluate C

drF . from (0,0,0) to

(1,1,1) along the curve 32 ,, tztytx ===

Solution: The end points are (0, 0, 0) and (1, 1, 1)

These points correspond to t = 0 and t = 1

23,2, tdztdydtdx ===

C

drF . = ( ) ++C

dzxzyzdydxyx 22 201463

= ( ) ( ) ( ) ++1

0

27522 32021463 dttttdttdttt

= ( ) +1

0

962 60289 dtttt

= ( )101073 643 ttt + = ( )[ ] 50643 =+

Example 2:

Show that

++= kzjyixF 222 is a conservative vector field.

Solution: If

F is conservative then 0=

F

Now 0000

222

=++=

=

kji

zyx

zyx

kji

F

F is a conservative vector field.

Surface Integrals

Definition: Consider a surface S. Let n denote the unit outward normal to the

surface S. Let R be the projection of the surface x on the XY plane. Let

f be

a vector valued defined in some region containing the surface S. Then the

surface integral of

f is defined to be dydx

kn

nfdsnf

RS

.

.

..

=

Example 1;

Evaluate dsnfS

. where

+= kzyjxizF 2 and S is the surface of

the cylinder 122 =+ yx included in the first octant between the planes z = 0

and z = 2.

Solution: Given

+= kzyjxizF 2

122 += yx

+= jyix 22 22 44 yx += =2

22 yx +

=2

The unit normal

n to the surface =

= yjxiyjxi

+=+

2

22

xyxzjyixkzyjxiznF +=

+

+=

.. 2

INTEGRAL THEOREMS

(i) Gausss divergence theorem

(ii) Stokes theorem

(iii) Greens theorem in the plane

Greens Theorem

Statement:

If M(x,y) and N(x,y) are continuous functions with continuous

partial derivatives in a region R of the xy plane bounded by a simple closed

curve C, then

dxdyy

M

x

NndyMdx

Rc

=+ , where C is the curve described in the

positive direction.

Verify Greens theorem in a plane for the integral ( ) xdydxyxc

+ 2

taken around the circle 422 =+ yx

Solution: Greens theorem gives

dxdyy

M

x

NNdyMdx

Rc

=+

Consider ( ) xdydxyxc

+ 2

M = x 2y N = x

1,2 =

=

x

N

y

M

dxdyy

M

x

N

R

( ) =+RR

dxdydxdy 321

= 3[Area of the circle]

= 32rpi

=3. 4.pi = pi12 (1)

Now + NdyMdx We know that the parametric equation of the circle 422 =+ yx

x = 2 cos y = 2 sin ddx sin2= , ddy cos2=

( ) xdydxxxNdyMdx +=+ 2 = ( )( ) ( ) dd cos2cos2sin2sin4cos2 + = d22 cos4sin8sincos2 ++ Where various from 0 to pi2

( ) ++=+pi

2

0

2 4sin4sincos2 dNdyMdxC

=

+

+

pi

2

0

42

2cos142sin d

= ( ) +pi

2

0

2cos262sin d

=

pi2

02

2sin26

2

2cos

+

= pipi 122

112

2

1=+ .(2)

From (1) and (2)

dxdyy

M

x

NNdyMdx

Rc

=+

Hence Greens Theorem is verified.

Example 2

Using Greens theorems find the area of a circle of radius r.

Solution: By Greens theorem we know that

Area enclosed by C = C

ydxxdy2

1

The parametric equation of a circle of radius r is x = sin,cos ryr = Where pi 20

Area of the circle = pi

2

0

)sin(sin)cos(cos2

1drrrr

= ( ) pi drr +2

0

2222 sincos2

1

= pi

2

0

2

2

1dr

= [ ] 22022

1rr pi pi =

Example 3:

Evaluate ( )[ ] c

xdydxyx cossin where c is the triangle with

vertices (0,0) ,( )0,2

pi and )1,

2(pi

Solution: Equation of OB is

02

0

01

0

=

pixy

pi

xy

2=

By Greens theorem dxdyy

M

x

NNdyMdx

Rc

=+

Here 1,sin =

=

y

MyxM

N xx

Nx sin,cos =

=

( )[ ] ( )dxdyxxdydxyxRC

+= 1sincossin

In the region R, x varies from x = 22

pipitoy

and y varies from y = 0 to y = 1

( ) = xdydxyxC

cossin ( ) +1

0

2

2

1sin dxdyxy

pi

pi

= [ ] +1

0

2

2

cos

pi

piyxx dy

= dyyy

+

1

0222

cospipipi

=

1

0

2

422sin2

+y

yy pipipi

pi

= 2

2

42

2 pi

pi

pipi

pi+=+

Example 4

Verify Greens theorem in the plane for

( ) ( ) +C

dyxyydxyx 6483 22 where C is the boundary of the region defined

by

X = 0 , y= 0, x + y =1

Solution: We have to prove that

dxdyy

M

x

NNdyMdx

Rc

=+

M = xyyNyx 64,83 22 =

yx

Ny

y

M6,16 =

=

By Greens theorem in the plane

dxdyy

M

x

NNdyMdx

Rc

=+

= ( ) 1

0

1

0

10 dydxy

x

=

1

0

1

0

2

210

x

y

= ( ) 1

0

215 dxx

= ( )

3

5

3

15

1

0

3

=

x

Consider ++=+BOABOAc

NdyMdx

Along OA, y=0 , x varies from 0 to 1

[ ] 13 1031

0

2===+ xdxxNdyMdx

OA

Along AB, y = 1 - x dxdy = and x varies from 1 to 0 .

( ) ( ) ( )[ ]dxxxxxxNdyMdxAB

+=+0

1

22 1614183

= ( ) ( ) 0

1

32

232

232

14

3

18

3

3

+

xxxxx

= 3

82312

3

8=++

STOKES THEOREM

If S is an open surface bounded by a simple closed curve C and if a vector

function

F is continuous and has continuous partial derivatives in S and on

C, then

=

c

rdFdsnFcurl .. where

n is the unit vector normal to the

surface (ie) The surface integral of the normal component of

Fcurl is equal

to the integral of the tangential component of

F taken around C.

Example 1

Verify Stokes theorem for ( ) = kzyjyziyxF 222 where S is the upper half of the sphere 1222 =++ zyx and C is the circular boundary on z = 0

plane.

Solution: By Stokes theorem

=

sc

dsnFcurlrdF ..

( ) = kzyjyziyxF 222

zyyzyx

zyx

kji

Fcurl

222

=

= [ ] ( ) ( ) =+++ kkjyzyzi 100022 Here

= kn since C is the circular boundary on z = 0 plane

= S

area of the circle =

S

dxdydsnFcurl .

= pipi =2)1( .(1)

ON z = 0,

=

sc

dsnFcurlrdF ..

On C, x = cos sin, =y ddyddx cos,sin ==

varies from 0 to pi2

( )( ) pi

drdFc

sinsincos2.

2

0

=

= ( ) +pipi

2

0

2

2

0

sinsincos2 dd

= ( )

+

pipi

2

0

2

02

2cos12sin dd

=

pipi 2

0

2

0 2

2sin

2

1

2

2cos

+

= pipi =++2

1

2

1 (2)

From (1) and (2)

=

sc

dsnFcurlrdF ..

Hence stokes theorem is verified

Example 2

Verify stokes theorem for ( ) ( ) +++= kxzjyzizyF 42 where s is the surface of the cube x = 0, x = 2, y = 0, y = 2, z = 0 and z = 2 above the xy

plane.

Solution:

By Stokes theorem

=

sc

dsnFcurlrdF ..

Given ( ) ( ) +++= kxzjyzizyF 42 xzyzzy

zyx

kji

Fcurl

++

=

42

= [ ] [ ] [ ]1010 ++ kzjyi = [ ] + kzjiy 1

Hence Stokes theorem is verified.

Example 3:

Verify Stokes theorem for

++= kxjziyF where S is the upper half

surface of the sphere 1222 =++ zyx and C is its boundary.

Solution: By stokes theorem

=

sc

dsnFcurlrdF ..

Gauss Divergence theorem

Statement:

The surface integral of the normal component of a vector

function F over a closed surface S enclosing volume V is equal to the volume

integral of the divergence of F taken throughout the volume V ,

dvFdsnFVS

= ..

Evaluate zdxdyxydzdxxdydzx 223 ++ over the surface bounded by z = 0 ,z = h,

222 ayx =+

Solution:

16

3

22

1

4

3cos

2

0

4 pipi

pi

== d

2

3.

4adsnF

S

=

UNIT IV

COMPLEX INTEGRATION

CAUCHYS INTEGRAL THEOREM

Statement:

If f(z) is analytic and f(z) is continuous at all points inside and

on a simple closed curve c, then 0)( =c

dzzf .

Proof: Let R be the region by c,

ivuzfidydxdziyxz +=+=+= )(,,

Now ++=cc

idydxivudzzf ))(()(

= )1.(..........)()( ++ udyvdxivdyudx

Since f(z) is continuous, the four partial derivatives y

v

x

v

y

u

x

u

,,,

are also exists and continuous in R and on c

By Greens theorem in the plane

=+c R

dxdyy

M

x

NNdyMdx

(1) Becomes,

+

=

c R R

dxdyy

v

x

uidxdy

y

u

x

vdzzf )( (2)

Since f(z) = u+iv is analytic by the CR Equation,

From (2)

+

=

c R R

dxdyx

u

x

uidxdy

y

u

y

udzzf )(

= 0 + i0 =0

CAUCHYS INTEGRAL FORMULA (OR) CAUCHYS

FUNDAMENTAL FORMULA

Statement:

If f(z) is analytic inside and on a simple closed curve c and if a is

any point within c, then,

f(a) =

c

dzaz

zf

i

)(

2

1

pi, the integration around c being taken in the

positive direction.

Proof:

Given f(z) is analytic inside and on c. Now az

zf

)( is analytic inside

and on c except at z =a

Draw a circle razc =:1 with center at z = a and radius r units such

that 1c lies entirely inside c.

Now az

zfz

=

)()( is analytic in the region enclosed between c and 1c

=1

)()(cc

dzzdzz

(ie)

=

1

)()(

cc

dzaz

zfdzaz

zf

On 1c , any point z is given by ireaz += , pi diredz i= 20

+

=

pi

2

0

)()(dire

re

reafdzaz

zf ii

i

c

= )(2 aifpi

=c

dzaz

zf

iaf

)(

2

1)(

pi

CAUCHYS INTEGRAL FORMULA FOR DERIVATIVES OF AN

ANALYTIC FUNCTION:

Statement:

If f(z) is analytic inside and on a simple closed curve c and z = a is any

interior point of the region R enclosed by c, then

=

c

dzaz

zf

iaf

2)(

)(

2

1)('

pi

=

c

dzaz

zf

iaf

3)(

)(

2

!2)(''

pi

In general,

+

=

c

n

n dzaz

zf

i

naf

1)(

)(

2

!)(

pi

Problems:

1. Evaluate

+

c

dzz

z

1

12

2

where c is circle

(i) 11 =z ( ) 11 =+zii ( ) 1= iziii

Solution:

Given

+

c

dzz

z

1

12

2

= +

+

c

dzzz

z

)1)(1(

12

Consider 11)1)(1(

12

++

=

+

+

z

B

z

A

zz

z

)1()1(12 ++=+ zBzAz

Put z = 1,B = 1, put z =-1 , A = -1

++

=

+

+ dz

zdz

zdz

zz

z

c1

1

1

1

)1)(1(

12

= )]()([2 afafi +pi

Here the point are z = -1 and z = 1

(i) c is the circle 11 =z . The point z = -1 lies outside and z = 1 lies inside the

circle 11 =z

)](0[2)1)(1(

12afidz

zz

z

c

+=+

+ pi

= ipi2 [since f(z) = 1, f(-1) = 1]

(ii) c is the circle 11 =+z . The point z = -1 lies inside and z = 1 lies outside the

circle 11 =+z

]0)([2)1)(1(

12+=

+

+ afidzzz

z

c

pi

= ipi2 [ since f(z) = 1]

(iii) c is the circle 1= iz

When Z = 1, 121 >= i lies outside c

When z = -1, 121 >= i lies outside c

=c

dzzf 0)(

2. Evaluate +++

c

dzzz

z

52

42

where c is the circle

(i) 21 =++ iz (ii) 21 =+ iz (iii) 1=z

Solution:

Consider +++

c

dzzz

z

52

42

The singular points of f(z) are given by

Z = i212

2042 =

Z = izi 21,21 =+

( )( )( ) ++

=

++

+

c c

dziiz

zdz

zz

z

2121

4

52

42

(i) 21 =++ iz is the circle

When z = -1+ 2i, 23121 >=+++ iii lies outside c

,21 iz = 21121

= [ ]ii

ii 23

24

232 +=

+ pipi

Where )21(

4)(

iz

zzf

+=

i

i

ii

iif

4

23

2121

421)21(

+=

+++

++=

(iii) 1=z is the circle

When 121,21 >++= iiz lies outside c

When 121,21

This is known as Taylors series of f(z) about z = a .This expression is valid at

all points interior to any circle. Having its centre at z = a and within which the

function is analytic.

The circle of convergence of the Taylor series is the largest circle that can be

drawn around z = a such that within which f(z) is analytic. The radius of this circle is

called the redius of convergence.

Maclaurins Series :

Taking a = 0 in the Taylor series for f(z), then we have

..............!3

)0('''

!2

)0(''

!1

)0(')0()( ++++=

ffffzf

The series is called Maclaurins series of f(z).

Laurents Series:

If 1C and 2C are two concentric circles with centre at z = a and radii

1r and 2r ( )21 rr < and if f(z) is analytic inside and on the circles and within the annulus between 1C and 2C , then for any in the annulus , we have ,

( ) ( )

=

+= nnn

n

n azbazazf0

)(

Where ( ) += C nn azdzzf

ia

1

)(

2

1

pi and ( ) += C nn az

dzzf

ib

1

)(

2

1

pi

Where C is any circle lying between 1C and 2C with centre at z = a for all n

and the integration being taken in positive direction.

Some Important Results: If ,1

2)1('',)2(

2)(''

3=

= fz

zf

6)1(''',)2(

6)('''

4=

= fz

zf

The Taylor series at z = 1 is given by

( ) ( ) ...............!2

)('')(

!1

')()( 2 +++=

afaz

afazafzf

= ........)6(!3

)1()2(

!2

)1()1)(1(1

32

+

+

++zz

z

= [ ]..................)1()1()1(1 32 ++++ zzz

2. Expand ( )( )321

)(2

++

=

zz

zzf as a Laurents series if (i) 2z

(iii) 32

=

+

++ .......

331

3

8......

221

2

31

22zzzz

(ii) 3>z 12

,32,13

3

12)1)(2(

27

++

+=+

z

C

z

B

z

A

zzz

z

)2()1()1)(2(27 ++++= zCzzBzzzAz

Put 1,0 == Az

2,2 == Bz

3,1 == Cz

F (z) =)1)(2(

27

+

zzz

z

1

3

2

21)(

+

+=zzz

zf

Given 311

If f (a) = 0 and f (a) 0 then z = a is called a simple zero of f (z) (or) a zero of the first order.

If 0)(&0)(.....)(')( 1 ==== afafafaf nn then z = a is a zero of order n.

Example; Let f (z) = 2z

Then 2)('',2)(' == zfzzf

02)0('',0)0('),0( == fff 0= z is a zero of order 2.

Singular point: A point z = a is said to be a singular point (or) singularity of f (z) is not

analytic at z =a.

Types of Singular Point:

Isolated Singular Point: A point z = a is said to be an isolated singular point of f (z) if

(i) f (z) is not analytic at z = a

(ii) f (z) is analytic at all points for some neighbourhood of z = a

Example: ( ) )2(1)( = zzz

zf

Then z = 1, 2 are isolated points.

Pole: A point z = a is said to be a pole of f(z) of order n if we can find a positive

integer such that 0)()(lim

zfaz naz

Essential singular point: A singular point z = a is said to be an essential point f (z) if the Laurents

series of f (z) about z = a possesses the infinite number of terms in the principal

part (terms containing negative powers).

Example

Let 21)( ezf =

Clearly z = 0 is a singular point

Also ( )

.......!2

1

!1

11)(

2

1

+++== zz

ezfz

= .......1

!2

111

2+

++

zz

z = 0 is an essential singular point

Removable singular point: A singular point z = a is said to be a removable singular point of f(z) if the

Laurents series of f(z) about z = a does not contain the principal part.

Example

Let z

zzf

sin)( =

Clearly z = 0 is a singular point

z

zzf

sin)( = =

+ ........

!5!3

1 53 zzz

z

= .........!5!3

142

+zz

Z = 0 is a removable singular point.

1. Classify the nature of the singular point of f ( )z =z

ztan

Solution : f (z ) = z

ztan

=

++ ......

3

1 3zz

z

= 1+ .....3

2

+z

This is the Laurents series of f ( )z about z=0 and there is no principal part.

Z=0 is a removable singular point.

Also f ( )z = z

ztan=

zz

z

cos

sin

Poles of f(z) are z cos z = 0

,....1,0,,0 === nnzz pi

pinz ,0= are simple poles (pole of order 1)

2. Consider the function f ( )4

sin

z

zz = . Find the pole and its order.

Solution: 4

sin)(

z

zzf =

=

+ ......

!5!3

1 53

4

zzz

z

= .......1206

113

+z

zz

0= z is a pole of order 3.

Definition

A function )(zf is said to be an entire function or integral function if it is

analytic everywhere in the finite plane except at infinity.

Example zzezf z cos,sin,)( =

Definition

A function )(zf is said to be a meromorphic function if it is analytic

everywhere in the finite plane except at finite number of poles.

Example

Consider ( )( )21cos

)(2

+=

zz

zzf

pi

Then )(zf is not analytic at z=1,-2

0(zf is a meromorphic function.

Definition

The residue of a function )(zf at a singular point az = is the coefficient

1b of az

1 in the Laurents series of )(zf about the point az =

EVALUATION OF RESIDUES

1. Suppose z = a is a pole of order 1

Then { } )()()(Re zfazzsf Limaz

az=

=

2. Suppose z = a is a pole of order n

Then { } ( ) ( ) ( )zfazdzd

nLimZsf

n

n

n

azaz

=

== 1

1

!1

1)(Re

3. Suppose z =a is a pole of order 1 and ( ) ( )( )zQzP

zf =

Then { Res ( )zf } ( )( )aQaP

az '=

=

CAUCHYS RESIDUE THEOREM

If )(zf is analytic at all point inside and on a simple closed curve C

Except at a finite number of point nzzzz .....3,2,1 inside C

Then

=c

idzzf pi2)( [Sum of residues of f(z) at nzzzz ....,3,2,1 ]

Proof

Given that )(zf is not analytic

Only at nzzzz ...,, ,321

Draw the non intersecting small

Circles ncccc ...,, 321 with centre at

nzzzz ,...,, 321 and radii n ...,, 321 Then )(zf is analytic in the region

Between c and ncccc ...,, 321

+++=nccc c

dzzfdzzfdzzfdzzf )(...)()()(

21

..(1)

Now nzzzz ...,, 321 are the singular points of )(zf .

Res == izz

zf )( the coefficient of izz

1in the Laurents series of

)(zf about izz = (by definition of residues)

( )

( ) dzzzzf

ib

c

n

==

1

1

1

12

1

pi

Since ( )

= 1 1

)(

2

1

c

nzndz

zz

zf

ib

pi

( )

( ) =1

0

12

1

c

dzzz

zf

ipi

( )dzzfic

=1

2

1

pi

( ) ( )izz

c

zsfidzzf=

= Re21

pi ..(2)

From (1) and (2)

( ) ( ) +== 1

1

Re2zz

c

zsfdzzf pi ( ) += 2

Re2zz

zsfpi ( )nzz

zsf=

Re2pi

( ) +== 1

Re2zz

zsfpi ( ) += 2

Rezz

zsf ( )nzz

zsf=

+ Re...

ipi2= {Sum of residues of ( )zf at }...,, 321 nzzzzz =

Example 1 Find the residue of f(z) ( )( )2122

+

+=

zz

z about each singularity.

Solution: The poles of f (z) are given by

( ) 01,02 =+= zz 1,2 == zz

The poles of f (z) are z = 2 is a simple poles and z = -1 is a pole of order 2. [ ] ( ) )(2lim)(Re

22 zfzzsf

zz =

=

= ( )2lim2

z

z ( )( ) 94

12

22

=

+

+

zz

z

[ ] ( ) )](2[lim)(Re 21

1 zfzdz

dzsf

zz =

=

= ( ) ])1)(2(

22[lim

2

2

1 +

+

zz

zz

dz

d

z

=

+ 2

2lim

1 z

z

dz

d

z

= ( )( ) ( )( )

( ) 94

2

1212lim

21=

+ z

zz

z

Example 2. Evaluate ( ) +c zdz

22 4 where c is the circle 2= iz

Solution: Given ( )22 41

)(+

=

zzf

iz 2= is a pole of order 2 Here iz 2= lies inside the circle 2= iz

And iz 2= lies outside the circle and is of order 2.

( )[ ] ( ) ( )zfizdz

dzsf

iz

ziz

2

2

2limRe =

=

( ) ( ) ( )222

222

12lim

iziziz

dz

d

iz+

=

( )22

2lim izdz

d

iz

=

( ) ( ) ( )

( )42

2 2

2202lim

iz

iziz

iz +

++=

256

8i=

By Residue theorem

( ) 62568

24

42

pipi =

=

+

ii

z

dz

c

CONTOUR INTEGRATION

TYPE: I

Integrals of the type ( )pi

2

0

sin,cos df where ( ) sin,cosf is A rational function of cos&sin . In this case we take unit circle

1=z as the contour. On 1=z .

( )

+=+== z

zeeez iii1

2

1

2

1cos,

( )

===

zz

iee

idiedz iii

1

2

1

2

1sin,

iz

dzd =

Also various from 0 to 2pi

( )

+

=pi

2

0

22

2

1,

2

1sin,cos

iz

dz

iz

z

z

zfdf

c

Now applying Cauchys residue theorem, we can evaluate the

Integral on the right side.

Example 1. Evaluate +pi

2

0sin513

dby using contour Integration.

Solution: +=

pi

2

0sin513

dI limit : 0 to 2pi

Contour: 1=z

Put iez =

( )

===

zz

iee

idiedz iii

1

2

1

2

1sin,

iz

dzd =

I =

+c

iz

ziz

dz

2

1513

2

= +

cizz

dz

52652

2

Where 5265

1)(

2+

=

izzzf

=c

dzzfI )(2 .(1)

To find Residue:

The poles of f (z) are 5265 2 + izz

Z = ( )10

10026262 + i

= ii

5,510

2426

=

The pole z = 5

ilies inside the circle 1=z and z = -5i lies outside the circle 1=z

Now [ ] )(5

lim)(Re5

5zf

izzsf

iziz

+=

=

=

( )izizi

ziz

55

5

1

5lim

5 +

+

+

= i

ii 24

1

55

5

1=

+

By Cauchys Residue theorem

= Ridzzfc

pi2)(

= 1224

12

pipi =

ii

(1) becomes

612

2 pipi==I

Example 2: Evaluate >>+pi

2

0

2

0,cos

sinbad

bausing contour integration.

Solution: Let +=pi

2

0

2

,cos

sind

baI

= +

pi

2

0cos22

2cos1d

ba

We can write 2cos = Real part of 2sin2cos, 22 iee ii +=Q

+

=pi

2

0

2

cos22

1. d

ba

ePRI

i

Put iez =

z

z

iz

dzd

1cos,

2 +==

++

=pi2

0

2

2

12

1.

iz

dz

z

zba

zPRI

++

=

pi2

0

2

2

2

11. dz

bazbz

z

iPR

= pi2

0

)(1

. dzzfi

PR (1)

Where bazbz

zzf

++

=

2

1)(

2

2

To find Residues:

Poles of f(z) are given by 022 =++ bazbz

Z = b

baa

2

442 22

=b

baa 22

Let b

baa 22 += and

b

baa 22 = , since a>b,

1&1 >< The simple pole z = lies inside C,

[ ] ( ) )(lim)(Re zfzzsfz

z

=

=

= ( ) ( )( )

zzb

zz

z

21lim

= ( ) 22221

b

baa

b

=

2

22

b

baaR

=

By Cauchys Residue theorem

= Ridzzfc

pi2)(

=

2

22

2b

baaipi

(1) becomes

I= pi2

0

)(1

. dzzfi

PR

=

2

22

2.b

baaPR pi

= ( )222

2baa

b

pi

Type II Integration around semi-circular contour

Consider the integral

Improper integrals of the form

dxxQ

xP

)(

)(,where P(x) and Q(x)

Are polynomials in x such that the degree of Q exceeds that of P atleast by two

and Q(x) does not vanish for any x.

Example 1: Prove that

=

++

+

12

5

910

224

2 pidx

xx

xxusing contour integration.

Solution: Let 910

2)(

24

2

++

+=

zz

zzzf

Consider C

dzzf )( where C is the closed contour consisting of , semi- large

circle of radius R and the real axis from R to R.

Then

+=R

Rc

dxxfdzzfdzzf )()()( (1)

Now 0910

2)(

24

2

++

+=

zz

zzzf as z

0)(lim =

zzfz

Hence from (1) =

C

dzzfdxxf )()(

By using residue theorem, = RidzzfC

pi2)(

)(Re2)( zsfidxxf pi=

The poles of f(z) are given by

( )( )iziz

zz

zz

3,

019

0910

22

24

===++

=++

The poles z = 3i, z = i lies in the upper half of the z plane

[ ] ( ) )(lim)(Re zfizzsfiz

iz=

=

= ( ) ( )( )( )92

lim2

2

++

+

ziziz

zziz

iz

= ( ) ii

i

i

16

1

28

1 =

[ ] ( ) )(3lim)(Re3

3zfizzsf

iziz

=

=

= ( ) ( )( )( )9332

3lim2

2

3 ++

+

ziziz

zziz

iz

= ( )( )932

lim2

2

3 ++

+ ziz

zz

iz

= ( )( ) ii

i

i

48

37

86

37 +=

i

i

i

izsf

48

37

16

1)(Re

++

=

= i48

10

= i24

5

12

5

24

52)(

pipi =

=

i

idxxf

Evaluate ( )( )

++

dxbxax

x2222

2

by using contour

Solution: Consider the integral C

dzzf )( where

( )( )22222

)(bzaz

zzf

++=

And C is the closed contour consisting of , the upper semi large Circle Rz = and the real axis from R to R

+=R

Rc

dxxfdzzfdzzf )()()(

When R

dzzf )(,

Hence =

C

dzzfdxxf )()(

By using residue theorem, = RidzzfC

pi2)(

Poles of f(z) are given by 0,0 2222 =+=+ bzaz

bizaiz == , The pole z = ai and z = bi lies in the upper half plane

[ ] ( ) )(lim)(Re zfaizzsfaiz

aiz =

=

= ( ) ( )( )( )222

limbzaizaiz

zaiz

aiz ++

= ( )( )222

limbzaiz

z

aiz ++

= ( )

( )222

2 baai

ai

+

= ( )222 baia