MATHEMATICS II
UNIT-1 ORDINARY DIFFERENTIAL EQUATIONS
1. What is Inverse operator?
Inverse operator
ANSWER:
(i)
1
X   Xdx
D
(iii)
1
e aX  xeax
Da
(ii)
1
X  e aX
Da
(iv)
1
x n ax
aX
e

e
r!
( D  a) r
e
 ax
Xdx
2. What is meant by Singular solution?
The Solution of a differential equation which is not obtained from the general
solution is known as Singular solution.
ANSWER:
3. What is Exponential shift rule?
Let X= e m , v (x ) , where v is of the form sin x, cos x or x m
ANSWER:
1
1
e aX v( x)  e aX
v( x)
f ( D)
f (D   )
This rule we call it as Exponential shift rule.
Particular integral =


4. Solve D 2  5D  6 y  0
ANSWER:


Given D 2  5D  6 y  0


The Auxillary equation is m 2  5m  6  0
i.e., m  3m  2  0
i.e., m  3 & m  2
 Complementary function = Ae2 x  Be3 x
 The General solution is given by y  Ae 2 x  Be 3 x


5. Solve D 3  3D 2  3D  1 y  0
ANSWER:


Given D 3  3D 2  3D  1 y  0


The Auxiliary equation is m 3  m 2  3m  1  0
i.e., m  1  0 ⇒ m =1(thrice)
3

 Complementary function = e x A  Bx  Cx 2



Hence the solution is y  e x A  Bx  Cx 2


6. Solve D 2  1 y  0
ANSWER:


Given D 2  1 y  0


The Auxiliary equation is m 2  1  0
i.e., m 2  1
i.e., m    1
i.e. m  i
 Complementary function = A cos x  B sin x
Hence the solution is y  A cos x  B sin x


7. Find the Complementry Function of D 3  2 D 2  D y  0
ANSWER:


Given D 3  2 D 2  D y  0


The Auxiliary equation is m 3  2m 2  m  0


i.e., m m 2  2m  1  0
i.e., m  0, m  1  0
2
i.e., m  0, m  1,1
 Complementary function = Ae 0  e x B  Cx
C.F = A  e x B  Cx
8. Solve y ' '4 y '20 y  0
ANSWER:
Given y ' '4 y '20 y  0
The Auxiliary equation is m 2  4m  20  0
i.e., m 
 4  16  80  4   64  4  8i


2
2
2
i.e., m  2  4i
 Complementary function = e 2 x A cos 4 x  B sin 4 x
Hence the solution is y  e 2 x A cos 4 x  B sin 4 x
9. Solve
ANSWER:
d 2 y dy

0
dx 2 dx
d 2 y dy

0
dx 2 dx
Given
i.e.,
D2  D  0
The Auxiliary equation is m 2  m  0
i.e., mm  1  0
i.e., m  0, m  1

 Complementary function = Ae 0 x  Be  x

Hence the solution is y  A  Be  x


10. Solve D 4  8D 2  16 y  0
ANSWER:


Given D 4  8D 2  16 y  0


The Auxiliary equation is m 4  8m 2  16  0




2
i.e., m 2  4  0
i.e., m 2  4  0
i.e., m  2i (twice)
 Complementary function = C1  C 2 x  cos 2 x  C3  C 4 x sin 2 x
 The Complete solution is y  C1  C2 x cos 2 x  C3  C4 x sin 2 x
11. Solve D  2  y  e 2 x
2
ANSWER:
Given D  2  y  e 2 x
2
The Auxiliary equation is m  2  0
2
i.e., m =2(twice)
 Complementary function = e 2 x Ax  B
1
1
e2X  e2X
2
0
( D  2)
1
=x
e2X
D  2.2
Particular Integral =
[Replace D by 2]
[Ordinary rule fails]
x  1  2X
= 
e
2  2  2 
[Ordinary rule fails]
x 1
=  e 2 X
2 1
 x2 
=  e 2 X
2
 x 2e2 X 

 2 
 The Complete solutuion y  e 2 x Ax  B + 
12. Find the Particular Integral of
ANSWER:
Given
d2y
dy
 2  y  cosh 2 x
2
dx
dx
d2y
dy
 2  y  cosh 2 x
2
dx
dx
i.e., ( D 2  2 D  1) y  cosh 2 x
Particular Integral =
1
cosh 2 x
( D  2 D  1)
2
 e 2 x  e 2 x
1

= 2
2
( D  2 D  1) 




1
1
1
e2x  2
e 2 x 
=  2
2  ( D  2 D  1)
( D  2 D  1)


1
1
1
e2x 
e 2 x 
= 
2  (4  4  1)
(4  4  1)

1  1 2 x 1 2 x 
e  e 
= 
2  (13)
(1)

[Replace D by 2]

1  e2x
= 
 e 2 x 
2  13

 The Particular solution is =

1  e2x
 e 2 x 

2  13

13. Find the particular integral of
ANSWER:
Given
d2y
dy
 3  2 y  sin 3x
2
dx
dx
d2y
dy
 3  2 y  sin 3x
2
dx
dx
i.e., ( D 2  3D  2) y  sin 3x
1
sin 3 x
( D  3 D  2)
1
=
sin 3 x
2
(3  3D  2)
Particular Integral = yp=
2
=
1
sin 3x
(3D  7)
=
(3D  7)
sin 3x
(3D  7)(3D  7)
=
3D  7
sin 3x
9 D 2  49
=
3D  7
sin 3 x
9 3 2   49
=
3Dsin 3 x   7 sin 3 x
 130
=
9 Dcos 3x   7 sin 3x
 130
 The Particular solution is =
9 Dcos 3x   7 sin 3x
 130
14. Find the particular integral of
ANSWER:
d2y
 4 y  sin 2 x
Given
dx 2
d2y
 4 y  sin 2 x
dx 2
[Replace D2 by -32]
[Replace D2 by -32]
i.e., ( D 2  4) y  sin 2 x
Particular Integral = yp=
1
sin 2 x
( D  4)
=
1
sin 2 x
(2  4)
=
x 1
sin 2 x
2D
=
x
cos 2 x
4
[Replace D2 by -22]
2
2
[Ordinary rule fails]
x
cos 2 x
4
 The Particular solution is =


15. Find the particular integral of D 4  8D 2  16 y  16 x  10
ANSWER:


Given D 4  8D 2  16 y  16 x  10
Particular Integral = yp=
=
1
16 x  10
D  8 D 2  16


4

1


D 4  8D 2 
161 

16






16 x  10
1
1 
D 4  8D 2 
= 1 
 16 x  10
16 
16


1 
D 4  8D 2
 ...16 x  10
= 1 
16 
16

1
16 x  10
16
1
 The Particular solution is =
16 x  10
16
2
16. Find the particular integral of  D  1 y  e x sin x
=
ANSWER:
Given
 D  1
2
y  e x sin x
1
Particular Integral = yp=  D  1
=
ex
e x sin x
2
1
 D  1  1
2
sin x
1
=𝑒 𝑥 𝐷2 𝑠𝑖𝑛𝑥
1
= 𝑒 𝑥 −1 𝑠𝑖𝑛𝑥
−𝑒 𝑥 𝑠𝑖𝑛𝑥
=
 The Particular solution is = −𝑒 𝑥 𝑠𝑖𝑛𝑥
17. Find the particular integral of (𝑫𝟐 + 𝟒𝑫 + 𝟒)𝒚 =
ANSWER:
Given (𝑫𝟐 + 𝟒𝑫 + 𝟒)𝒚 =
Particular Integral = yp=
=
𝒆−𝟐𝒙
𝒙𝟐
𝒆−𝟐𝒙
𝒙𝟐
𝑒 −2𝑥
1
(𝐷2 +4𝐷+4) 𝑥 2
1
𝑒 −2𝑥
(𝐷+2)2 𝑥 2
= 𝑒 −2𝑥
= 𝑒 −2𝑥
1
1
𝐷
𝑥
2 ( 2)
1
𝐷2
(𝑥 −2 )
= 𝑒 −2𝑥 (−𝑙𝑜𝑔𝑥)
[𝑆𝑖𝑛𝑐𝑒
= −𝑒 −2𝑥 𝑙𝑜𝑔𝑥
 The Particular solution is = −𝒆−𝟐𝒙 𝒍𝒐𝒈𝒙
18. Find the particular integral of(𝑫𝟑 − 𝟑𝑫𝟐 + 𝟑𝑫 − 𝟏)𝒚 = 𝒙𝟐 𝒆𝒙
ANSWER:
Given (𝑫𝟑 − 𝟑𝑫𝟐 + 𝟑𝑫 − 𝟏)𝒚 = 𝒙𝟐 𝒆𝒙
𝑑 2 (−𝑙𝑜𝑔𝑥)
𝑑𝑥 2
=
1
𝑥2
]
1
Particular Integral = yp=
(𝐷3 −3𝐷2 +3𝐷−1)
𝑥 2𝑒 𝑥
1
= (𝐷−1)3 𝑥 2 𝑒 𝑥
= 𝑒𝑥
= 𝑒𝑥
= 𝑒𝑥
1
(𝐷+1−1)3
1
(𝐷)3
ANSWER:
𝒅𝟐 𝒚
𝒅𝒙𝟐
−𝟑
Given
𝒅𝒚
𝒅𝒙
𝒅𝟐 𝒚
𝒅𝒙𝟐
[Replace D by D+1]
𝑥2
𝑥5
[Integrating 𝒙𝟐 three times]
60
 The Particular solution is = 𝒆𝒙
19. Solve
𝑥2
𝒙𝟓
𝟔𝟎
+ 𝟐𝒚 = 𝟏 + 𝒙
−𝟑
𝒅𝒚
𝒅𝒙
+ 𝟐𝒚 = 𝟏 + 𝒙
i.e (𝑫𝟐 − 𝟑𝑫 + 𝟐)𝒚 = 𝟏 + 𝒙
The Auxiliary equation is 𝑚2 − 3𝑚 + 2 = 0
⇒ (𝑚 − 1)(𝑚 − 2) = 0
⇒ 𝑚 = 1,2
 Complementary function =𝑨𝒆𝒙 + 𝑩𝒆𝟐𝒙
Particular Integral = yp=
1
(𝑫𝟐 −𝟑𝑫+𝟐)
=
𝟏
(1 + 𝑥)
1
𝟐 (𝑫𝟐 −𝟑𝑫+𝟏)
𝟐
(1 + 𝑥)
𝟏
𝑫𝟐 −𝟑𝑫
𝟐
𝟐
𝟏
𝑫𝟐 −𝟑𝑫
𝟐
𝟐
= [1 + (
= {1 − (
𝟏
3𝐷
𝟐
2
= (1 +
𝟏
−1
)]
) + ⋯ } (1 + 𝑥)
)(1 + 𝑥)
3
(1 + 𝑥)
= {𝟏 + 𝒙 + }
𝟐
2
𝟏
5
𝟐
2
= {𝒙 + }
𝒙
5
={ + }
𝟐
4
 The Particular solution is = {
𝒙
𝟓
+ }
𝟐
𝟒
 The Complete solution y  𝑨𝒆𝒙 + 𝑩𝒆𝟐𝒙 +
𝒅𝟐 𝒚
𝒙
𝟓
+
𝟐
𝟒
𝒅𝒚
20. Solve 𝒙 𝒅𝒙𝟐 + 𝒅𝒙 = 𝟎
ANSWER:
𝑑2 𝑦
𝑑𝑦
Given 𝑥 𝑑𝑥 2 + 𝑑𝑥 = 0
i.e. [𝑥𝐷2 + 𝐷]𝑦 = 0
⇒ [𝑥 2 𝐷2 + 𝑥𝐷]𝑦 = 0
[Multiplying by x]
𝑥 = 𝑒𝑧
l𝑜𝑔𝑥 = 𝑧
𝑥𝐷 = 𝐷′
2 2
𝑥 𝐷 = 𝐷′ (𝐷′ − 1)
[𝐷′ (𝐷′ − 1) + 𝐷′ ]𝑦 = 0
2
[𝐷′ − 𝐷′ + 𝐷′]𝑦 = 0
𝐷′2 𝑦 = 0
Auxillary Equation is 𝑚2 = 0
𝑚 = 0,0
Complementry Function is y = [𝐴𝑥 + 𝐵]𝑒 0𝑥
 The Complete Solution is y= 𝑨𝒙 + 𝑩
PART - B
Put
1. Solve the equation (i)(D2 – 4D + 3)y = sin3x + x2,(ii) (D2 + 4)y = x4 + cos2x.
2. Solve the equation (i) (D3 + 8)y = x4 + 2x + 1 + cosh2x, (ii) (D4 – 2D3 + D2)y = x2 + ex.
3. Solve the equation (i)(D2 + 1)2y = x4 + 2sinx cos3x, (ii) (D3 – 3D2 + 3D – 1)y = e-xx3.
4. Solve the equation (i) (D4 – 4)y = x2cosh2x,(ii) (D4 – 2D2 + 1)y = (x + 1)e2x.
5. Solve the equation (i) (D2 + 5D + 4)y = e-xsin2x,(ii) (D4 – 1)y = cos2x coshx.
6. Solve the equation (i)(D2 – 4D + 13)y = e2x cos3x,(ii) (D2 + D + 1)y = e-x sin2(x/2).
7. Solve the equation (i) (D2 + 2D + 5)y = ex cos3x, (ii) (D2 + 4D + 8)y = 12e-2x sinx sin2x.
8. Solve the equation (i) (D3 – 1)y = x sinx,(ii) (D2 + 4)y = x2 cos2x.
9. Solve the equation (i)(D2 – 4D + 4)y = 8x2 e2x sin2x,(ii)(D2 + a2)y = secax.
10. Solve the equation (i) (D2 – 4D + 3)y = sin3x cos2x,(ii)(D4 – 1)y = cosx coshx.
11. Solve (i)(x2D2 + xD + 1)y = logx sin(logx),(ii)(x2D2 +4xD + 2)y = sinx.
12. Solve (i)(x2D2 + 2xD – 20)y = (x2 + 1)2,(ii) (x4D3 +- x3D2 + x2D)y = 1.
13. Solve (i) [(3x + 2)2D2 + 3(3x + 2)D – 36]y = 3x2 + 4x + 1,
(ii) [(x + 1)2D2 + (x + 1)D + 1]y =4 cos log(x + 1).
14. Solve the simultaneous equations (dx/dt) + 2x – 3y = 5t,(dy/dt) – 3x + 2y = 2e2t.
15. Solve the simultaneous equations Dx – (D – 2)y = cos2t,(D – 2)x + Dy = sin2t.
Unit – II
Vector Calculus
1. If  (x,y,z) is a constant , then P.T   0 .
Ans :
If  (x,y,z) is a constant , its partial derivatives with respect to x,y,z will be zero .ie.,
  


0
x y z
     
   i
 j
k
0
x
y
z
2. Find r  .
Soln:




We’ve r  xi  yj  zk .

r  r   x 2  y 2  z 2
r2  x2  y2  z2.
r 2 x x r y r z
 
 ,
 ,  .
x 2r r y r z r
 r  r  r
r  i
 j
k
x
y
z
x y z
=i  j  k
r
r
r



i 
r
= xi  yj  zk =
r
r.



3. P.T r n   xrn2 r
Soln :
  n   n   n
r   j y r   k z r 
r n   i
x

r  n 1 r  n 1 r
=i i nr n 1
 j nr
 k nr
x
y
z




= nr n 2 xi  yj  zk

= nr n 2 r
4. Find log r 
Soln:

 x 
 y 
z
= i nr n 1    j nr n 1    k nr n 1  
r
r
r


 

log r   j  log r   k  log r 
log r   i
x
y
z
 1 r  1 r  1 r
 j
k
=i
r x
r y
r z
1 x 1 y 1 z
 j
k
.
=i
r r
r r
rr



1 
r
= 2 xi  yj  zk  2
r
r .
2
2
r2 
 y  z  2e r .



5. P.T e x 2
Soln:




r r  xi  yj  zk
r r 2  x 2  y 2 z 2

  
 
 e x  y 2  z 2   er
  r2
  r2
  r2
=i
e  j
e k
e
x
y
z
 2 r  r 2 r  r 2 r
 i e r 2r
 j e 2r
 k e 2r
x
y
x



2 x
y
z 
 2e r r  i  j  k 
r
r 
r



2
 2e r xi  yj  zk
2 
 2e r r .
6. Find the directional derivative of   x 2 y z  4 xz2  xyz at (1,2,3) in the direction of
2
2

 
 

  
2i  j  k .
Soln:
  x 2 yz  4 xz2  xyz.
  
D.D at (1,2,3) in the direction of 2i  j  k .
  
 (1,2,3)( 2i  j  k )

  
( 2i  j  k )
     
  i
 j
k
x
y
z



 i ( 2 xyz  4 z 2  yz )  j ( x 2 z  xz)  k ( x 2 z  8 xz  xy)



 (1,2,3)  54i  6 j  28k .
  
D.D of  at (1,2,3)is the direction of 2i  j  k .
   


(54i  6 j  28k )( 2i  j  k )

4 11
108  6  28 86

6
6.
7. Find the angle between the surfaces xlogz=y2-1 and x2y=2-z at the point (1,1,1)
Soln.
1  x log z  y 2  1
2  x 2 y  2  z
Angle between 1  c and 2  c is
at (1,1,1) at(1,1,1)
1 (1,1,1)  2 (1,1,1)
cos  
1 (1,1,1)  2 (1,1,1)

 
 
 
1  i 1  j 1  k 1
x
y
z
 x


 i (log z )  j ( 2 y )  k ( )
2



1 (1,1,1)  i log 1  j ( 2)  k (1)
 
 
 
2  i 2  j 2  k 2
x
y
z


 2
 i ( 2 xy)  j ( x )  k (1)



2 (1,1,1)  i ( 2 xy)  j ( x 2 )  k (1)
 
  
( 2 j  k )  (2i  j  k )
cos  
 
  
 2 j  k  2i  j  k
 2 1
1
1


4 1 4 11
5 6
30
 1 
  cos 1 

 30 

8. Find the unit normal to the surface s2+y2-z=10 at (1,1,1).
 (1,1,1)

Soln.
Unit Normal n at (1,1,1) =
 (1,1,1)
  x 2  y 2  z  10
     
  i
j
k
x
y
z



 i 2 x  j 2 y  k ( 1)

 
 (1,1,1)  2i  2 j  k

 

 

 
2i  2 j  k 2i  2 j  k 2i  2 j  k
nˆ  

  
3
4  4 1
2i  2 j  k
2 2  1 
 nˆ  i  j  k
3
3
3





9. If F  x 3i  y 3 j  z 3k . Find div F


Soln.
div curl F  curlF ------------(1)



i
j
k

 


F  F 
x y z
x3 y3 z3



 i (0  0)  j (0  0)  k (0  0)

0


div curlF  (0)  0



10. If v  ( x  3 y )i  ( y  2 z ) j  ( x  z )k is solenodial. Find the value of .
Soln.


Given v is solenodial  V  0



      
j
 k   ( x  3 y )i  ( y  2 z ) j  ( x  z )k
 i
y
z 
 x
=0



 ( x  3 y )  ( y  2 z )  ( x  z )  0
x
y
z
11   0

  2

11. If F  3xyi  y 2 j , Evaluate

 
 Fdr where c is the curve in the xy-plane y=2x2 from
c
(0,0) to (1,2)
 
2
Soln.
 Fdr   3xydx  y dy
c
c
(0,0) to (1,2)
y  2x
dy  4 xdx
y:0 2
x : 0 1
1
 
2
2
F
 dr   3x2 x dx  (2 x2) 4 xdx
2
c
0
1
 6 x 4 16 x 6 
  (6 x  16 x )dx  

4
6  0

0
6 16  7
 

4 6
6




12. If F  ( 4 xy  3x 2 Z 2 )i  2 x 2 j  2 x 3 zk . Check whether the integral
1
3
5
 
F
 dr is
c
independent of the path c
Soln.
T.P.T ,

 
F
d
r
is
independent
of
the
path
c,
we’ve
to
prove
that


F
0

c



i
j
k




 F 
x
y
z
2 2
2
4 xy  3x z 2 x  2 x 3 z



 i (0)  j ( 6 x 2 z  6 x 2 z )  k (4 x  4 x )

0
 
  Fdr is independent of the path.
c
13. P.T. div   
Soln.
div   2
2
               
 i
j
 k   i
j
k 
y
z   x
y
z 
 x
 2  2  2



x
y
z
  2



14. Find the Constants a,b,c so that F  ( x  2 y  az )i  (bx  3 y  z ) j  ( 4 x  cy  2 z )k
is irrational

Soln.
Given,   F  0



i
j
k



0
x
y
z
4 x 2  2 y  0 z bx  3 y  z 4 x  cy  2 z






i ( c  1)  j ( 4  a )  k (b  2)  0i  0 j  0k
c+1=0
4-a=0
b-2=0
c=-1
a=4
b=2

15. If s is any closed surface enclosing a volume v and r is position vector of a point. P.T.

 r nds  3v
S
Soln.
Gauss Div Theorem


f
n
ds


f

 dv
S
V


 r nds   r dv
S
V


         
r   i
j
 k  xi  yj  zk
y
z 
 x
=1+1+1=3

 r nds   2dv  3v
S

V

16. P.T div ( curlA)  0
Soln.





curlA    A    ( A1i  A2 j  A3k )



i
j
k




x y z
A1 A2 A3
  A A    A A    A A 
 i  3  2   j 3  1  k  2  1
z 
z 
 x
 y
 x y 

  A A    A A    A A 
div (curlF )   3  2    3  1    2  1 
x  y
z  y  x
z  z  x
y 
 2 A3  2 A2  2 A3  2 A1  2 A2  2 A1






xy xz xy yz xz zy

div ( curlF )  0
17. Find the unit normal vector to the surface xy3z2=4 at(-1,-1,2)
     
Soln.
  xy3 z 2  4
  i
j
k
x
y
z



 ( 1,1,2)
 4i  12 j  4k
ˆ 

 ( 1,1,2)
176
18. Apply Greens’ theorem at plane P.T. the area enclosed by a simple closed curve c is
1
xdy  ydx
2
Soln.
Green’s Theorem
 Q P 
c Pdx  Qdy  R  x  y dxdy
  ydx  xdy   1  (1)dxdy
R
 2  dxdy
= 2(Area of R)
R
Area of R 
1
 ydx  xdy
2 c
 


19. If A and B are irrotational, P.T A  B is solenoidal
 
Soln.
T.P.  A  B  0
 
 


W.K.T,  A  B  B    A  A    B

 

 
A and B are irrotational   A  0 &   B  0
 
 
 A  B  B 0  A0  0
 
 A  B is solenodial


20. If F  grad ( x 3  y 3  z 3  3xyz) Find the curlF
Soln.

F  grad
















  x 3  y 3  z 3  3xyz

2
 3 x  3 yz
x

 3 y 2  3 xz
y

 3z 2  3xy
z




F  grad    i 3x 2  3 yz  j 3 y 2  3xz  k 3z 2  3 yx

curlF    grad

 
 




i
j
k




x
y
z
2
2
2
3x  3 yz 3 y  3xz 3z  3xy




 i ( 3 x  3 x )  j (  3 y  3 y )  k ( 3 z  3 z )  0
 



21. If F  x 2 i  y 2 j evaluate  Fdr along the line y=x from (0,0) to (1,10
Soln.


c
 




2
2
F
d
r

x
i

y
j

dx
i

dy
j


c

curlF  0

c
  x 2 dx  y 2 dy
c
(0,0)  (1,1)
x=0  1
y=0  1
c: y  x
dy=dx
1
 2x3 
2
  x dx  x dx   2 x dx  


 3 0 3
0
x 0
 
22. Find the directional derivative of xyz at (2,1,10 in the direction j  k
Soln.
Directional derivative in the direction of
  
A j k
 at (2,1,1),   xyz
     
  i
j
k
x
y
z



 i ( yz )  ( xz) j  ( xy)k

D.D in the direction of A
  


( i  2 j  2k )  ( j  k )

  
0i  j  k
1
1
2
2
2
22
4

2 2
2
2
23. State green’s Theorem:
(or)
State relationship between line integral and a double integral
Soln.
If c is a simple close curve enclosing a region R in the xy-plane and p(x,y),
Q(x,y) and its first order partial derivates are continuous in R, then
 Q P 
 ( Pdx  Qdy )   

xy Where c is described in the anticlockwise
x y 
R 
direction.

24. State Stoke’s theorem
(or)
Relationship between line and surface integral
Soln.
If s is an open two sided surface bounded by simple closed curve cand if is a
vector point function with continuous first order partial derivatives on s, then
 
 
 Fdr   curlFds
c
S
Where c is described in the anti-clockwise direction as seen from the positive tip of the
outward drawn normal at any point of the surface S.
25. State Gauss Divergence theorem

Soln.
If s is a close surface enclosing a region of space with volume v and if F is a
vector point function with constants first order partial derivates in v, then
 

F
d
s

(
div
F
)dv


S
V
PART - B
1. Evaluate using Green’s theorem in the plane for  (xy + y2) dx + x 2 dy where C is
the closed curve of the region bounded by y = x and y = x2.
2. Evaluate  [ (sinx –y) dx – cos xdy] where c is the triangle with vertices (0,0), (/2 ,0)
and (/2,1).
3. Verify Green’s theorem in the plane for (3x2 -8y2)dx + (4y-6xy)dy where c is the
boundary of the region defined by x = 0, y = 0, x+y =1.
4. Verify Green’ s theorem in plane for  x2(1+y)dx + (y3 +x3) dy where C is the square
bounded by x = a, y = a.
5. Verify Divergence theorem for F = (x2-yz) i + (y2 –zx) j + (z2 – xy) k taken over the
rectangular parallelepiped 0 x  a, 0 y  b , 0 z  c.
6. Verify Divergence theorem for F = (x2-yz) i + (y2 –zx) j + (z2 – xy) k taken over the
rectangular parallelepiped 0 x  3, 0 y  3 , 0 z  3.
7. Verify Divergence theorem for F = (2x –z) i + x2 y j – xz2 k over cube bounded by x
= 0, x =1, y =0 , y=1, z = 0 and z =1.
8. Verify Divergence theorem for F = (2x –z) i + x2 y j – xz2 k over cube bounded by x
= 0, x =2, y =0 , y=2, z = 0 and z =2.
9. Verify Divergence theorem for the function F = 4xi – 2 y2 j + z2 k taken over the
surface region bounded by the cylinder x2+y2 = 4 and z =0, z =3.
10. Verify Stoke ‘s theorem for a vector field F = (x2 –y2) i + 2xy j in the rectangular
region of the xoy plane bounded by the lines x = -a , x = a, y = 0 and y =b.
11. Verify stoke’s theorem for F = (y-z+2) i +(yz + 4) j – xz k 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.
12. Verify stoke’s theorem for F = y2 i + y j –xz k over the upper half of the sphere
x2+y2+z2 = a2 , z  0.
Unit – III Analytic Functions
1. If u and v are harmonic then can we say u + iv as analytic?
Ans. Yes. But it is not always. Eg., u = x, v = -y.
2. State the sufficient conditions for the function F(z) to be analytic
Ans.
The single valued continuous function w=f(z)=u+iv is analytic in a region R if the
u u v v
four partial derivatives
exist, are continuous and satisfy CR equation at
, , ,
x y x y
u v u
v
each point in R,
 ,
 .
x y y
x
3. Define an enter function?
Ans. A function F(z) which is analytic at every point of the complex plane is called as
entire function. Eg. e z
4. state the Cauchy –Riemann – in polar co-ordinates .
u 1 v
1 u v
Ans.

;


r r 
r  r
5. Give an Example that the C-R equation are necessary but not sufficient for a function
to be analytic?
x 3 (1  i)  y 3 (1  i)
Ans. f ( z ) 
, When z  0
x2  y2
=0, when z = 0
6. If u+iv is analytic, show that v-iu is also analytic
Ans. Given: f(z) = u+iv is analytic
Hence it satisfies C-R Equation (i.e) Ux=Vy

Uy=-Vx

g(z) = v-iu, replace u by v & v by –u in ,  Ux=Vy, Uy=-Vx
7. When f(z)=u+iv is analytic only if Ux=Vy and Uy
8. If f(x)=u+iv is an analytic function. If u  v  e x [cos y  sin y] find f(z)
Ans. Let u  v  e x [cos y  sin y]

u v

 e x [cos y  sin y ]
Diff  partically w.r to x

x x
u v
Diff  partically w.r to y

 e x [ sin y  cos y ]
y y
u v
By C-R Equations 


 e x sin y  e x cos y ]
y y
u v
( 2) 

 e x cos y  e x sin y
x x
u v
(3) 

 e x cos y  e x sin y
x x
------------------------------------------u
u
(2)  (3)  2
 2e x cos y 
 e x cos y
x
x
-------------------------------------------u v

 e x cos y  e x sin y
x x
u v
(3) 

 e x cos y  e x sin y
x x
--------------------------------------------( 2) 
v
v
 2e x sin y 
 e x sin y
x
x
--------------------------------------------u
v
We know f ( z ) 
i
x
x
By Milne Thomson Method, f ( z )   U x ( z ,0)  iV x ( z ,0)dz
(2)  (3)  2
  e z (cos 0)  ie z sin 0dz
  e z dz
 sin 0  0  e z  c
f ( z )  u  iv  e xiy  c
 e x e iy  c
u+iv = excosy+iexsiny+c
9. For what values of a,b,c the function f(z)=x+x+2ay+i(bx-cy) is analytic?
Ans. If f(z) = x – 2ay + i(bx-cy) is analytic
Then it satisfies C-R equations ux=1, Uy=-2a
V2=b, Vy=-c
U x  V y  1  c
U y  V y  2a  b
 c  1, b  2a
Hence for c=-1 and any values of a and b satisfying b=2a then function f(z) is
analytic.
10. If u+iv analytic, then the curves u=constant and v=constant are orthogonal.
11. If Orthogonal f is a fun of z alone then f is analytic whether it is true or false?
Ans. True
12. Choose the correct answer.
W=f(z) is analytic function of z then
w w
2w
w
w
w
(a )
i
(d )
0
(b)

(c )
0
z
x
z
z y
zz
2w
0
zz
13. If u and v are harmonic then u+iv is analytic function?
Ans. True
14. Note: If f(z) is analytic function, then kf(x) is also analytic function
15. If f(z) and g(z) are analytic function f(z)+g(z) is also analytic.
16. verify whether w  z is analytic or not
w  z  x  iy
u  iv  x  iy
Ans. u  x, v   y
U x  1,V x  0,U y  0,V y  1
Ans.
U x  V y ,U y  V x
 w  z is not alnalytic
1
17. f ( z ) 2
is anal;ytic everywhere except
z 1
a)z=1
b)z=±1
c)z=±i
Ans. z=±i because f(z) becomes infinity at this point.
d)z=i
1
is not analytic at z=2i, -2i
z 4
 2
2 
2
19. If f(z) is an analytic function of z. Prove that  2  2  f ( z )  4 f ( z )
y 
 x
18. The Function f ( z ) 
Ans.
2
 2
2 
2
We know that  2  2   4
zz
y 
 x
 2
2 
2
2
2
  2  2  f ( z )  4
f ( z)
zz
y 
 x
2
 f ( z ) f ( z )
zz

 4  f ( z ) f ( z )
z
 4 f ( z ) f ( z )
4
 4 f ( z )
20. State he two properties of analytic functions
1. Both Real Part and imaginary part of an analytic function satisfy Laplace
equation
2. if w=u+iv is an analytic function then the curves of the family
u(x,y)=Constant, cut orthogonally the curves of the family v(x,y)=constant=c2
21. State the necessary Ans sufficient condition for f(z) to be analytic.
u v
Ans. Necessary condition: Let w=f(z)=u+iv be an analytic function then
and

x y
u
v
Sufficient condition: The single valued continuous function u+iv is an analytic

y
x
u u v v
in a region R if the four partial derivatives
, , , exists and continuous and satisfy
x y x y
the C-r Equation at each point in R.
22. Can U=3x2y-y3 be the real part of an analytic function
Ans. u=3x2y-y3
Ux=6xy. Uxx = 6y
Uy=3x2-3y2 Uyy=-6y
U xx  Uyy  6 y  6 y  0   2u  0 u is harmonic and can be the real past of an analytic function.
23. State whether sin(x+iy) is analytic or not?
 sin x cos iy  cos x sin iy
u  iv  sin x cosh y  i cos x sinh y
u  sin x cosh y
v  cos x sinh y
Vx   sin x sinh y
U x  cos x cosh y
2
Uy  sin x sinh y
Uy  cos x cosh yy
Ux=ry and vy=-vx
 sin( x  iy )isanalytic .
24. Show that the function x x 2  6 x 2 y 2  y 4 is harmonic.
u
 12 yx 2  4 y 3
y
Ans: let u  x 4  6 x 2 y 2  y 4
 2u
 12 yx 2  12 y 2
y 2
u
 4 x 3  12 xy 2
x
 2u
 12 x 2  12 y 2
2
x
2
 u  2u

 0  u is harmonic.
x 2 y 2
 y
25. Show that tan 1   is harmonic
x
Ans:
 y
u  tan 1  
x
u
1
y
 x.0  y.1 

.
 2
2 
2
2
x
 x y
 y  x
1  
x
2
 u
x 2  y 2 .0  y.2 x
2 xy


2
2
2
x
x2  y2
x2  y2

u

y





2 xy
 x.1  y,0 

 2
2
2
 x  y2
 y  x
1  
x
2
 u
x 2  y 2 .0  x 2 y
 2 xy

 2
2
2
2 2
y
(x  y )
(x  y 2 )2
1

2



 2u  2u

0
y 2 y 2
U is harmonic .
1
log( x 2  y 2 ) is harmonic.
26. Verify whether
2
1
Ans:
Let u  log( x 2  y 2 )
2
1 1(2 y )
1
2x
uy  . 2
ux  , 2
2
2 x  y2
2 x y
x
y
 2
 2
2
x y
x  y2
uxx  y 2  x 2
uyy  x 2  y 2
x2  y2
( y2  x2 )

0
( x 2  y 2 )2 ( x 2  y 2) 2
27. Verify whether w=sin x cos h y +I cos x sin h y is analytic or not?
Ans:
U  sin x cosh y
V  cos x sinh y
Vx   sin x sinh y
U x  cos x cosh y
uxx  uyy 
Uy  sin x sinh y
Vy  cos x cosh y
u v
u v
&
and all the partial


x y
y x
 W is analytic
28. If f ( z)  r 2cos 2  i sin p is analytic, if the value of p is
1
a)
b)0 c)2 d)1 .
2
Ans:
P=2
U  r 2 cos 2
Proof:
v  r 2 sin p
u
v
 r cos 2
 r sin p
r
r
u
v
 r 2 sin 2
 pr 2 sin p


u 1 v v 1 u

By using
& 
we get
r r  r r 
1
2r cos 2  pr 2 cos p
r
2r cos   pr cos p
2 cos   p cos p
 p2
29. Define conformal mapping?
Ans:
A mapping that preserves angles between any oriented curves both in
magnitude and in sense is called as a conformal mapping.
30. The mapping w=z 2 is not conformal at______________
Ans:
W  z2
w
 2z
z 
2z=0 when z=0 At z=0, It is conformal.
31. Find the points at which the transformation w=sin z is not conformal.
Ans:
w
 3
 cos z; cos z  0; atz  , 
z
2 2
32. Define Bilinear transformation ?(or) Mobius transformation
Ans:
az  b
W
,where a,b,c,d are complex constants ad-bc#0 is known as bilinear
az  d
transformation.
33. Define isogonal transformation
Ans. If a transformation is such that the angle between two curves c1 and c2 intersection
at (x0,y0) in z-plane is equal to the angle between their images intersecting at (U0,V0) in
the w plane then the transformation is called isogonal transformation.
34. Write the cross ratio of the points Z1, Z2, Z3, Z4
Ans.
Cross Ration 
( Z1  Z 2 )( Z 3  Z 4 )
( Z1  Z 4 )( Z 3  Z 2 )
35. Write the invariant point of transformation w 
Ans.
w
1
;
z  2i
z 2  2iz  1  0
36. Find the image of |z-2i|=2 where w  
Ans.
w
1
z  2i
1
z
1
z
1
x  iy
1
x  iy 
u  iv
1
u  iv
u  iv
x  iy 

 2
;
u  iv u  iv u  v 2
u
x 2
u  v2
v
y 2
u  v2
Now |z-2i|=2 menas |s+iy-2i| = 2
 x  i ( y  2)  2
u  iv 
x 2  ( y  2) 2  4
x2  y2  4y  0
 u2   v2   v 
   2
  4 2
  2
0
2 
2 
2 
u v  u v  u v 
 1  4v  0
37. Find the Critical point of w2  ( z   )( z   )
w
z
0
0
Ans. The critical point are obtained by
z
w
w 2  z 2  z (   )  
w
 2 z  (   )
z
w
1
w
 z  (   )
z
2
w
1
1
 0  z  (   )  0  z  (   )
z
2
2
z
w

0
1
w
z     
2
 w  0, (i, e) z   z     z   , 
 2w
z
2i   4  4
i
2
Hence the critical points are z   ,  ,
 
2
38. Find the bilinear transformation that has i and –i as fixed points.
az  b
Ans.
The Bilinear transformation is given by w 
cz  d
 i (ci  d )  ia  b
a (i )  b
ai  b
and  i 
ci  d
c (i )  d
(i )( ci  d )  ai  b
 c  id  ia  b
 c  id  ai  b
Since i and –i are the fixed points we have i 
az  b
where a and b are arbitary.
bz  a
39. Show that the function f(z)=|z|2 is not analytic at any point
Ans.
Let z=u(x,y)+iv(x,y)
2
f ( z)  z  x 2  y 2
 The bilinear transformation is w 
Hence u(x,y)=x2+y2
v(x,y)=0
The Cauchy Riemann equation are ux=vy
uy= -vx
2x=0
2y=0
Which is possible only when z=y =o. Thus the function is differentiated only at ongin.
40. Find the analytic function of f(z)=(x-y)2+2i(x+y).
Ans:
f ( z )  ( x  y ) 2  2i ( x  y )
utir  ( x  y )2  2i ( x  y )
u  ( x  y )2v  2( x  y )
ux  2( x  y )vx  2
uy  2( x  y )( 1)vy  2
 2( x  y )
Ux  Vu
Uu  Vx
2(x-y)=2
-2(x-y)=-2
x-y=1
x-y=1
C.R equation are satisfied only if x-y=1.
41. Define a critical point of the bilinear transformation?
Ans:
The point at which the mapping w=f (z) is not conformal (i,e) f (z) =0 is called a
critical point of the mapping .
The transformation w = f(z) is conformal ,z =f 1 (w) is also confomal its
dz
0
critical point occurs at
dw
42. If f(z) = z 3 analytic ? Justify.
Ans:
The sufficient conditions for the function w =f (z) to be analytic in argion R are 1.ux
,uy ,vx and vy are existing to ux =vy and uy = -vx at every point of R.
In this example ,given : f (z) = z 3
U+iw =(x+iy)3=x 3+3x2 (iy)+3x(iy)2+ (iy)3
= (x3-3xy2) +I (3x2y-y3)
Part b.
1. Evaluate c 1/(z2+4) dz where c is |z|=3.
2. Obtain Taylor’s series to represent the function z2-1/(z+2)(z+3) in the region |z|<2.
3. Expand 1/ (z-1)(z-2) as a Laurent’s series valid in the regions 1<|z|<2 and |z|>2.
4. Find the residue o Z3/ (z-1)4(z-2)(z-3).
5. If c is the circle |z|=3, evaluate ctanz dz.
6. Evaluate cdz/ zsinz where c is |z|=1 using cauchy’s residue theorem.
7. Using contour integration evaluate 02 d/(13+5sin).
8. Using contour integration evaluate 02 cos2 d/(5-4cos).
9. Using contour integration evaluate 0 d/(a2+sin2). a>0.
10. Using contour integration evaluate - xdx/(x+1)(x2+1).
11. Using contour integration evaluate - (x2-x+2) dx /(x4+10x2+9).
12. Prove that 0 log(1+x2) / (1+x2) dx =  log2
Unit –IV-Complex Integration
1. What is the Value of
Soln.
3z 2  7 x  1
c z  1 dz if c is |x| = ½
3z 2  7 x  1
c z  1 dz
The pole is at z= -1 lies outside the circle |z| = ½
3z 2  7 x  1

dz = 0 [By Cauchy’s integral Theorem]
z 1
c
2. Find the residue at z = 0 of the function f(z) =
Soln. [Reside of f(z) at z = 0]
( z  0)(1  e z )
= lt
z 0 z cos z  sin z
0
=
0
By using L's Hospital Rule
1  ( z  1)e z
= lt
z 0  z sin z  cos z  cos z
1  ez
z cos z  sin z
2(1)
0 11
=1
=
3. State Cauchy’s integral Formula
Soln. If f(z) is analytic within and on a simple closed curve c and z0 is any point
inside c then
1
f ( z)
f(z0) =
dz

2i z  z 0
4. Find the Laurent’s Expansion of f(z) =
Soln. |z| > 1 |
1
in |z| > 1
z (1  z ) 2
1
| 1
z
2
1
1  1
1
=
= 3 1  
2
1
z  z
z (1  z )
z 3 (1  ) 2
z
2

1 
1 1
= 3 1  2   3   
z 
z z

1
2
3
= 3  4  5 
z
z
z
5. Define essential single point of f(z)
Soln. A point z =a is an essential singularity if
i)
z =a is singular point
ii)
z =a should not be a pole (or) removable singularity
6. State Cauchy’s residue theorem
if f(z) is analytic at all point inside and on a single closed curve C except at a finite
number of poise z1 , z2 , … ,zn within C then
 f ( z)dy  2ix sum of the residue of f (z) at z1,z2, ... ,zn
c
7. Find the zeroes of
z3 1
z3 1
Soln:
z3 1
=0
z3 1
z3 1  0
z3 1  0
z  (1)1 / 3
The zero are given by
The cube roots of unity are the roots.
dz
8. Evaluate 
,where C is 1z1=1
2z  1
c
Soln:
The poles is z =

c
1
lines inside the circle 1z1=1
2
dz
 ixR
2z  1
Where R = residue at z =
1
s
2
1
2 1
= lt
1
1
2
z
2 2( z  )
2
dz
1

 2i 
2z  1
2
z
dz
 2 z  1  i
9. The Laurent’s Series expansion of
1
valid in |z| > 1 is
z ( z  1)
1
1
1


z ( z  1) 2 z  1
1
1
1 1  1 


 1  
=
z
z z z
 1
z 1  
 z
ez
10. If C is the Circle x2 + y2 = 4 the value  2 dz of is
z
c
Soln.
1
 1;
z 
Soln.
ez
c z 2 dz where c is circle x2+y2=4
The pole is z = 0 is of order 2 and it is inside the circle |z| = 2
ez
c z 2 dz  2iR where R is Reside of z=0zz
d  z 2e  z 


R = lt
z 0 dz  z 2 



= lt  e  z
z 0

R= -1
ez
c z 2 dz  2i 1
ez
c z 2 dz  2i
11. If C is a circle with Z=4+i and radius 4, Then
 z
3

 z 2 dz  0 say true or false?
c
Soln :
True, f(z)= Z 3  Z 2 is analytic and C is a circle which is closed.
By cauchy’s integral theorem,the given integral is zero.

12. Expand cos z is a jaylor’s series at Z=
4
1
Soln. f ( z )  cos z;
f ( ) 
4
2
1
f ( z )   sin z;
f ( )  
4
2
1
f ( z )   cos z;
f ( )  
4
2
1
f ( z )  cos z;
f ( ) 
4
2
2
za
( z  a)
f ( z )  f (a) 
f (a) 
f (a)  
1!
2!






z  
z  
z  
1
4  1  
4  1  
4  1 
cos z   






2
1!  2 
2!
3!
 2
 2
2
3
 
     
z   
 z   z  
1 
4 
4
4 
cos z  1 



2
1!
2!
3! 




13. State Cauchy’s Integral theorem
Soln. If f (z ) is an analytic and f (z ) is continuous inside and on a simple closed curve C,
2
then
3
 f ( z)dz  0
c
14. Determine the poles of f(z)= and investigate their order
Soln. Z=1  Pole of Order 1
Z=2  Pole of Order 2
ez
dz where C is Circle |z|=2
15. Evaluate 
1

z
c
Soln. The pole is at z=-1 lies inside |z|=2
ez

dz  2i ( r1 ) where r1=residue at z=-1
1 z
c
Residue at z  1  lt z  1
z 1
ez
1 z
=e
z
e
dz  e
1 z
c

16. Expand the function
Soln.
f ( z )  sin z;
f ( z )  cos z
f ( z )  sin z
sin z
about z  
z 
sin(  )  0ie ( f ( ))
cos   1(ie., f ( ))
f ( )  0
f ( z )   cos z
sin z
1

z  z 
f ( )  1


(z   )2

f
(

)

(
z


)
f
(

)

f ( )  

2!


( z   )3


 
 ( z   ) 
3!

2
4
(z   )
(z   )
= 1 


3!
4
ez
17. Derive residue of f(z), where f ( z ) 
inside|z|=1
cos z
Soin :
ez
f ( z) 
inside the circle |z|=1
cos z
1
The pole ( z )  inside the unit circle
2
1
z 
=
18. If a=5, then
Soln.
dz
?
za
Z 2

dz
0
za
Z 2

1
is analytic and z = 5 lies outside |z|=2)
z 5
z4
19. Obtain the Poles 2
z  2z  5
z4
Soln. f ( z )  2
z  2z  5
The poles are given by z2+2z+5 =0
 2  4  20  2  4i
z

2
2
z = -1+2i
ez
20. The value of  2 dz is
z
Soln. Let us take c as unit circle z=0 is a double pole
ez
c z 2 dz  2i(r ) where r = residue at z=0
( f ( z ) 
d z
e
dz
 lt  e  z = -1
 lt
z 0
z 0
z
e
z
c
2
dz  2i
dz
z
21. The Value of
2
where c is |z| = 1 is
c
Soln.
e
z
z
2
dz the pole is z =0 (double) inside |z| =1
c
 2i (r ) where r is residue at z = 0
d 2 ez
z 2
z 0 dz
z
z
 lt  e
 lt
z 0
= -1
ez
  2 dz  2i (1)  2i
z
c
22. The value of
 f ( z)dz   when f (z ) is analytic inside the close curve
c
 f ( z)dz  0 [Cauchy’s Integral Theorem]
c
1
at z = 1 as a Taylor series
z2
1
1
1


z  2 ( z  1)  1 (1  ( z  1))
= -[1-(z-1)]-1
1
  1  ( z  1)  ( z  1) 2  ( z  1) 3  
z2
23. Expand
Soln.


1
1
in laurents expansion of
valid in |z-1| < 1 is equal to
z 1
z ( z  1) 2
Soln. Put z-1 =u, |u| < 1
1
1
1
 2 (1  u ) 1  2 (1  u  u 2  )
2
u (1  u ) u
u
1 1
 2  1 u
u
u
1
1


 1  ( z  1)
2
z 1
( z  1)
1
Co efficient of
is -1
z 1
cos 
dz where c is |z| = 1.5
25. Evaluate 
z

1
c
Soln. The pole is at z = 1, It is a simple pole and inside c
cos z

dz  2i(r )
z 1
c
Where r = residue at z = 1
cos(z )
 lt
( z  1)
z 1 z  1
24. Coefficient of
 cos 
= -1
cos z
dz  2i(1)
z

1
c
 2i
2
d
26. Express 
as a contour integral around the circle |z| = 1
0 1  a cos 
dz  e i id
z  e i ,
Soln. Put
dz
d 
dz  zid ,
iz
dz
2
d
2
dz
zi
c 1  a cos  c  1   i c az 2  2 z  a
z 
z
1  a
 2 





27. The Laurents expansion of
1
vaid in |z-1|<1 is
z2
Soln. Let u = z-1
 z  1  u | u | 1
1
 (1  u ) 2  1  2u  3u 2  4u 3  
2
(1  u )
 1  2( z  1)  3( z  1) 2  4( z  1) 3  
28. For a simple pole at z = a the residue of f(z) at z = a where f ( z ) 
Soln. The residue at z = a is
p( z )
is ___
q( z )
P(a )
Q (a )
29. The residue at the pole of the function
z
are equal
z 1
2
Soln. Since z  i are the poles of order 1
z
1
[resf ( z )]  lt ( z  i)

z i
( z  i)( z  i) 2
z i
z
1
[resf ( z )]  lt ( z  i)

z


i
( z  i)( z  i) 2
z i
2
z 1
dz where c is ellipse x2+y2=4, then the value of f (1) is ___
30. If f (a)  
z

a
c
Soln.
z 2 1
dz
z

1
c
The pole is at z=1 which is inside x2+4y2=4
 f (1)  2i ( R) where r is the residue at z = 1
f (1)  
( z 2  1)( z  1)
2
z 1
( z  1)
Res [ f ( z )] z 1  lt
 f (1)  2i(2)
f (1)  4i
Part b.
13. Evaluate c 1/(z2+4) dz where c is |z|=3.
14. Obtain Taylor’s series to represent the function z2-1/(z+2)(z+3) in the region |z|<2.
15. Expand 1/ (z-1)(z-2) as a Laurent’s series valid in the regions 1<|z|<2 and |z|>2.
16. Find the residue o Z3/ (z-1)4(z-2)(z-3).
17. If c is the circle |z|=3, evaluate ctanz dz.
18. Evaluate cdz/ zsinz where c is |z|=1 using cauchy’s residue theorem.
19. Using contour integration evaluate 02 d/(13+5sin).
20. Using contour integration evaluate 02 cos2 d/(5-4cos).
21. Using contour integration evaluate 0 d/(a2+sin2). a>0.
22. Using contour integration evaluate - xdx/(x+1)(x2+1).
23. Using contour integration evaluate - (x2-x+2) dx /(x4+10x2+9).
24. Prove that 0 log(1+x2) / (1+x2) dx =  log2
1.
Soln.
2.
Soln.
3.
Soln.
4.
Soln.
UNIT – V LAPLACE TRANSFORMATION
Find L[e  3e ]
L[e 2t  3e 5t ]  1L[e 2t ]  3L[e5t ]
1
3


s2 s5
Taking LCM & on simplification
4s  1
 2
s  3s  10
Find L[cos 2 3t ]
1  cos 2
Formula : cos 2  
2
1
 1 cos 6t  1
L[cos 2 3t ]  L  
 L[1]  L[cos 6t ]

2  2
2
2
1  cos 2
1 1
s 
=   2
Note : sin 2  

2
2  s s  36 
3
L[cos 3t ]
Formula: cos 3   4 cos 3   3 cos
1
 3 cos 3t  cos 9t  3
L[cos 3 3t ]  L 
 L[cos 3t ]  L[cos 9t ]

4
4

 4
3 s
s
 . 2

2
4 s  9 4( s  9 2 )
L[sin 3t cos t ]
Formula: 2 sin A cos B  sin( A  B)  sin( A  B)
1
L[sin 3t cos t ]  L(sin 4t )  L(sin 2t )
2
1 4
2 
2
1
  2
 2
 2
 2

2  s  16 s  4  s  16 s  4
Note:
2t
5t
2 cos A sin B  sin( A  B)  sin( A  B )
2 cos A cos B  cos( A  B)  cos( A  B)
2 sin A sin B  cos( A  B)  cos( A  B)
5. L[sin( wt   )]
Soln. Formula: sin( A  B)  sin A cos B  cos A sin B
L[sin( wt   )]  L[sin wt cos   cos wt sin  ]
(cos , sin  are constants)
 cos L[sin wt ]  sin L[cos wt ]
w
s
 cos  2
 sin  2
2
s w
s  w2
Note:
sin( A  B)  sin A cos B  cos A sin B
cos( A  B)  cos A cos B  sin A sin B
cos( A  b)  cos A cos B  sin A sin B
1
6. L[et (cosh 2t  sinh 2t )]
2
Soln. We know that L[e at f (t )]  F (s  a)
1
L[cosh 2t  sinh 2t ]  L[cosh 2t ]  L[sinh 2t ]
2
s 1
1
L[e t (cosh 2t  sinh 2t )] 

2
( s  1)  4 ( s  1) 2  4
7. Find L[t 2 e 2t ]
2!
2
 3
Soln. L[t 2 ]  2
s 1 s
e
L[t 2 e  2t ] 
( s  2) 3
8. L[e 3t sin 2 t ]
1
1  cos 2t  1
l sin 2t  l 
 l 1  l cos 2t 

Soln.
2
2

 2



1
s

2
2 s 2( s  4)


1 1
s 3
l e 3t sin 2t  

 4
2
2  s  3 (s  3)

9. lcosh t.sin 2t 
 e t  e  t 

 sin 2t 
Soln: l cosh t.sin 2t   l 
 2 

= 1 l e t sin 2t  e t sin 2t
2
1
2
1
2
2
l[sin 2t ]  2
l cosh t. sin 2t   
 
,
2
2 ( s  1)  4 2 ( s  1) 2  4
s 4
10. Soln:
d
F ( s)
We know that L[tf (t )]   F ' ( s)or
ds


2
s 4
d
2
d
L[t sin 2t ] 


2( s 2  4) 1
ds s 2  4 ds
2
= 2
 2s
( s  4) 2
4s
.
2
( s  4) 2
F ( s )  L[ f (t )]  L[sin 2t ] 
2
11. L[t 2 e 3t ]
Solution :
L t 2 f (t )  (1) 2 F " (s)
f(t)= e 3t


1
s3
1
1
2
F(s)=
; F ' (s) 
; F " (s) 
2
s3
( s  3)
( S  3) 3
2
.
 l (t 2 e 3t ) 
( s  3) 3
F(s)=L[f(t)]= L[e 3t ] 
t
sin t
dt
t
o
12. Find the Laplace transformation of e t 
Solution:
t
 1
L   f (t )dt   [ Lf (t )]
o
 s
 t sin t  1  sin t  1 
1
1
ds
L
dt   L 
sin
tds
=
=
2


 s
t
s
t
s
s

1

o

s
s
1
1
= [tan 1 ( s )]s = [tan 1   tan 1 s ]
s
s
1 
1
= [ 2  tan 1 s ] = [cot 1 s ]
s
s
t
 sin t 
1
L e t 
dt  
[cot 1 ( s  1)]
 o t
 s 1
 1 
13. INVERSE LAPLACE TRANSFORMFind L1 
2
 ( s  1) 
Soln:
t
 1 
t 1  1 
t
L1 

e
L

e
t.
2
 s 2 
 ( s  1) 


1
14. Find L1 

2
 (s  1)  1
Soln :


1
 1 
L1 
 e t L1  2   e t sin t

2
 s  1
 (s  1)  1
 s 3 
15. Find L1 
 e 3t L1 s ( s 2 4) = e 3t cos 2t.

2
 ( s  3)  4 




s
16. Find L1 

2
 (s  2)  1


 s22 
s
Soln : L1 
 L1 


2
2
 (s  2)  1
 ( s  2)  1
 s  2  1 

2
= L1 
-L 


2
2
 (s  2)  1
 (s  2)  1
  s 
 2 
= e 2t  L1  2   L1  2  
 s  1 
  s  1
2t
2t
= e cos t  e (2) sin t.


s
17. Find L1 

2
 ( s  b)  4 
Soln :


 s bb 
s
L1 
 L1 


2
2
 ( s  b)  4 
 (s  b)  4 
 s  b  1 

b
 L1 
L 


2
2
 ( s  b)  4 
 ( s  b)  4 
  s  1  b 
 e bt  L1 
 L  2

 s  4 
  s4
sin 2t
2
 s 
18. Find L1 
2
 (s  2) 
d
Soln. L1[ sf ( s )]  [ L1 F ( s )]
dt
d 
 1 
 e  2t L1  2  
dt 
 s 
 e bt cos 2t  be bt


d 2t
e t
dt
 e  2 t  t ( 2 ) e  2 t

 s 
19. Find L1 
2
 (s  2) 
 s 
 s  2  2
 L1 
Soln. L1 
2
2 
 ( s  2) 
 ( s  2) 
 s  2  1  2 
 L1 
L 
2
2
 ( s  2) 
 ( s  2) 
 s
 2 
 e 2t  L1  2   L1  2  
 s 
 s 
 e 2t [1  2t ]
 1 
20. L1 

 s ( s  3) 
t
 F (s) 
Soln. L1 

L1 F ( s )dt


0
 s 
t
1
  L1
dt
0
( s  3)
t
1
  e 3t L1 dt
0
s
t
  e 3t dt
0
t
 e  3t 


 3 0
 e 3t 1


3
3
1
 [1  e 3t ]
3


1
21. L1  2
2 
 s( s  a ) 
Soln.
t
 F (s) 
L1 

L1 F ( s )dt


0
 s 

 t 1 

1
1
L1  2
 L  2
dt
2 
2
 s( s  a )  0  ( s  a ) 
t sin at

dt
0
a
1 t
  sin atdt
a 0
1   cos at 
 
a  a  0
t
1   cos at cos 0 

a  a
a 
1
cos at

 2
a
a2



1
22. L1 

 ( s  1)( s  3) 


1
A
B
 ( s  1)( s  3)   ( s  1)  ( s  3)


A( s  3)  B( s  1)  1
Sub s = -3 B   1
2
1
Sub s = -1 A 
2
Subt.


1
1/ 2
1/ 2
 ( s  1)( s  3)   ( s  1)  ( s  3)


-1
Taking L on both sides

 1 1  1  1 1  1  1 t
1
 e  e  3t
L1 
 L 
 L 



 ( s  1)( s  3)  2  s  1 2  s  3  2
23. L e at cos(bt  c)
Soln.  by 1st Shifting theorem
L e at cos(bt  c)  Lcos(bt  c)ss a
Soln.





Lcos bt cos c  sin bt sin c ss a
cos cL[cos bt ]ss a  sin cL[sin bt ]ss a
cos c.( s  a)
b sin c

2
2
( s  a)  b
( s  a) 2  b 2
24. Evaluate
Soln.


0
te2t sin tdt
L[t sin t ]   e  st sin tdt
Hence the given integral is the value of L[t sint] when s = 2
d  1 
L[t sin t ] 
ds  s 2  1
Sub, s = 2
4
L[t sin t ] 
25
 2t
e  e 3t
dt
25. Evaluate 
t
0
Soln.
 e 2t  e 3t   st  e 2t  e 3t 
L
  e 
dt
t
t

 0


 e 2t  e 3t 
L
 , When s = 0
t


 e 2t  e 3t   1
1
L


ds

t
s

2
s

3

 0
 [log( s  2)  log( s  3)]s


 ( s  2) 
 1 2 / s 
 log
 log

( s  3)  s  1  3 / s  s


 log 1  log
s2
s3
Sub s = 0
s3
 log
s2

3
sin tdt  log( )
2
0
26. Intial value theorem
Soln. If L[ f (t )]  F ( s) , then lt f (t )  lt sF ( s )
 te
2t
t 0
s 
Proof:
We know that
L[ F (t )]  sF ( s )  f (0)
Taking limit on both side s  

lt L[ F (t )]  lt L[ F (t )]
s 
s 
 lt  e st f (t )dt  0
lt sF ( s)  f (0)  lt L[ F (t )] s 0
s 
s 
 F (s)  f (0)  0
lt  F ( s)  f (0)
(i.e)., lt  F (s)  lt f (t )
lt
s 
s 
s 
t 0
27. Find value theorem
If L f (t )  F (s);
lt f (t )  lt
t 
s 0
 f ( s)
Proof:We know that
L[ f (t )]   F ( s )  f (0)
Taking limit on both sides s  0
lt L[ f (t )]  lt  F ( s)  f (0)
s 0
lt
s 0
s 0
 F (s)  f (0)  lt
s 0
2
e
 st
f (t )dt
0
2
  f (t )dt
0
  f (t )0  f (2)  f (0)
2
lt
s 0
lt
s 0
 F (s)  f (0)  lt
 F (s)  lt f (t )
t 
f (t )  f (0)
t 
PART B
1.
2.
3.
Write the change of scale property for Laplace transform with proof and also
write inverse Laplace transform for the same.
Write the first shifting property for Laplace transform with proof and also write
inverse Laplace transform for the same.
Write the second shifting property for Laplace transform with proof and also
write inverse Laplace transform for the same.
4.
5.
6.
7.
8.
9.
10.
11.
12.
State and the convolution theorem.
Find the inverse Laplace transform of (s2 + s – 2)/s(s + 3)(s – 2).
Find the inverse Laplace transform of e-2s/s2(s2 + 1).
Find the inverse Laplace transform of s/(s2 + 1)(s2 + 4)(s2 +9).
Find the inverse Laplace transform of (14s + 10)/(49s2 + 28s + 13).
Find the inverse Laplace transform of (2s3 + 4s2 – s + 1)/(s2)(s2 – s + 2).
Find the inverse Laplace transform of 1/(s4 + 4).
Solve y” + y’ -2 y = 3 cost -11 sin3t ,y(0)=0 and y’(0)=6
Solve using Laplace transforms x’ – y = sint; y’ – x = -cos t given that x =2 , y=0
for t=0.