engg maths
DESCRIPTION
Engineering MathematicsTRANSCRIPT
-
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