MATHEMATICS OPTIONAL

ORDINARY DIFFERENTIAL
EQUATIONS (ASSIGNMENT)

By
Avinash Singh
(Ex IES, B.Tech IIT Roorkee)

DELHI CENTRE BHOPAL CENTRE

27-B, Pusa Road, Metro Pillar No. 118, Plot No. 46 Zone - 2, M.P Nagar,
Near Karol Bagh Metro, New Delhi-110060 Bhopal - 462011

Phone: 8081300200, 8827664612 | E-mail : info@nextias.com | Web: www.nextias.com

 2 Mathematics Optional

Tutorial Sheet - I

(1) Solve : y  x  dy dx =a  y 2


dy

.  x  a  1  ay   cy 
dx 

(2) Solve :  1  x 2  y 2  x2 y 2   xy  dy 
0
dx 

 
Ans. logx-log 1  (1  x 2 ) 1 2  (1  x 2 ) 1 2  (1  y 2 )1 2  c

(3) ( x  y )(dx  dy )  dx  dy ;  x  y  c  log( x  y )

(4) Solve ( x  2 y  1) dx  ( x  2 y  1)dy

Ans. 3x  6 y  1  ce 3( x  y )/2

(5)  y x   ydx  xdy   y sin  y x   xdy  ydx

x cos

Ans. xy cos  x   c
y

(6) (4 y  3x)dy  ( y  2 x )dx  0

1
2
 ( 3  1)x  2 y 
2
2 3
Ans. c( x  2 xy  2 y )   
 ( 3  1)x  2 y 

(7) ( x 2  4 xy  2 y 2 )dx  ( y 2  4 xy  2 x 2 )dy  0

Ans. y 3  6 xy 2  6 x 2 y  x 3  c

dy
dx  ( x  2 y  3) /(2 x  y  3)
(8)

(9) dy/dx = (x+y+4)/(x-y-6)

 ( y  5) 
Ans: ( x  1)2  ( y  5)2  ce 2 tan 1  
 ( x  1) 

(10) (2 x 2  3 y 2  7) xdx - (3x 2  2 y 2  8) ydy=0

Ans: ( x 2  y 2  1)5  c( x 2  y 2  3)

(11) ( x  y )2 dx  ( y 2  2 xy  x 2 )dy  0

Ans: x 3  y 3  3xy( x  y )  c

x /y
(12) Solve (1  e )dx  e
x /y
  
1  x y dy  0

Ans: x  ye x/y  c

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
3
xdy  ydx
(13) Solve xdx  ydy  0
x2  y 2

Ans: x 2  y 2  2 tan 1 ( x / y )  c

(14) y sin 2 xdx  (1  y 2  cos 2 x )dy  0

Ans: 3 y cos 2 x  6 y  2 y 3  c
(15) Show(4x+3y+1)dx+(3x+2y+1)dy=0 is a family of hyperbolas with a common axis and tangent at
the vertex.
(16) Find the values of constant  such that (2xey+3y2) (dy/dx) +(3x2+ey)=0 is exact. further for this
value of , Solve the equation.

Ans: = 2 , x 3  2 e x  y 3  c

(17) Solve: ( x 3 y 3  x 2 y 2  xy  1)ydx  ( x 3 y 3  x 2 y 2  xy  1)xdy  0

Ans: xy  ( 1 xy )  2 log y  c

(18) Solve (2 ydx  3xdy )  2 xy(3 ydx  4xdy )  0

Ans: x2y3+2x3y4=c

(19) (2 x 2 y 2  y )dx  ( x 3 y  3 x )dy  0

10 5 4 12
Ans: 4x 7 y 7  5x 7 y 7 c

(20) Sinx dy dx +3y=cosx


Ans:  y 

1
3
 
 tan x 2  2 tan x 2  x  c
3
 
(21) ( x  2 y 3 ) dy
dx 
y 
Ans: x y  y  c
2

(22) (1  y 2 )dx  (tan 1 y  x )dy

1
Ans: x  tan 1 y  1  ce tan y

(23)
dy
dx
 y cos x  1  2 sin 2x
Ans: y  ce  sin x  sin x  1

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 4 Mathematics Optional

dz z z
(24) Solve  log z  2 (log z)2
dx x x

1 1
Ans: x(log z)  2  c
2x

(25) ( x 2  2 x  2 y 2 )dx  2 xydy  0

2x3 x4
Ans: y 2 x 2   c
3 4

1
(26) ( xy 2  e x 3 )dx  x 2 ydy 0

 3 e
1
y2
Ans: 2  2 x3 c
x

(27) Solve x( dy / dx )  y  y 2 log x

1
Ans: y  log x  1  cx

(28) ( x 3 y 2  xy )dx  dy

x2
Ans: y 1  (2  x 2 )  ce  2

(29) ( x 2 y 3  xy ) dy dx  1

y2
1 
Ans:  2  y 2  ce 2
x
(30) Find the curve for which the position of y axis cut off between the origin and tangent varies as
the cube of the abscissa of the point of contact.

Ans: 2y   kx 3  cx
(31) Find the Cartesian equation of the curve in which the perpendicular from the foot of the ordinate
on the tangent is of constant length.
Ans: y = k cosh {(x+c)/ k}

(32) Find the family of curves whose tangent form an angle  4 will the hyperbola xy=c

Ans: y  x  2 c tan
1 x
c
 c  
(33) Show that the Curve in which the angle between the tangent and the radius vector at any
point is half of the vectorial angle is a cardioid.

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
5
Ans: kr – 1 = rce
(34) If the population of country doubles in 50 years in how many years will treble under the
assumption that rate of increase is proportional to the number of inhabitants.
Ans: 78.25 yrs
(35) A metal bar at temperature of 100°F is placed at a constant temp of 0°F. If after 20 Minutes the
temperature of the bar is half , find an expression for the temperature of the bar at any time.
Ans: T = 100 e(-0.035)t
(36) Show that the only curves having constant curvatures are circles and straight lines.
(37) Find the curve for which sum of the reciprocal of the radius vector and the polar subtangent is
costant.

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 6 Mathematics Optional

Tutorial Sheet - II
(38) Find the orthogonal trajectories of the family of curves 3xy=x3-a3 a being parameter of the family.
2 2y
 
Ans. x  y  1 2 e  c

39) Find the orthogonal trajectories of x2+y2=2ax.

Ans. x 2  y 2  cy , c being parameter.

(40) Find the orthogonal trajectories of the family of circles x 2  y 2  2 fy  1  0 , where f is parameter

Ans) x 2  y 2  2 gx  1  0

(41) Find the orthogonal trajectories of family of parabolas y 2  4 a( x  a ) , where a is parameter.

   
2
Ans. y  2 x dy y
dy
dx dx

2 y2
(42) Find the orthogonal trajectories of the family of curves x 2   1 , where  is a
(a   ) (b 2   )
parameter.

(Ans) { x  y dy dx} x  y dx dy   a 2

 b2

(43) Find the orthogonal trajectories of r=a(1+cosn  )

2
Ans. r n  b(1  cos n )

(44) Find the equation of the family of oblique trajectories which cut the family of concentric circles at
30°.

(45) p 3  2 xp 2  y 2 p 2  2 xy 2 p  0

2 1
 
Ans. ( y  c )( y  x  c )[ x  y  c ]  0

(46) p( p  y )  x( x  y )

(47) x 2 p 2  2 xyp  2 y 2  x 2

 sin
1 y x 
Ans. x  ce

(48) p 2  2 py cot x  y 2


Ans. ( y  c sec 2 x / 2) y  c cos ec 2 x / 2  0 

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
7
(49) x 2 p 2  2 xyp  y 2  x 2 y 2  x 4  0

y  x sinh(c  x )
Ans.  0
 y  x sinh(c  x )

(50) y  2 px  p 2 y

Ans. 2 xc  y 2  c  0

(51) y 2 log y  xpy  p 2

Ans. log y  cx  c 2

(52) x= y+a log p

y  c  a log(1  p )
Ans.  

 x  c  a log(1  p )  a log p 
(53) x = 4(p + p3)

 y  2 p2  3 p 4  c 
Ans.  3 
x  4( p  p ) 

(54) y  px  x 4 p 2

Ans. xy + c = c2x
(55) y = yp2 + 2px
Ans. c2y2 = 4(1 + x)
(56) y = 2 px + f(xp2)
Ans. y = 2cx1/2 + f(c2)
(57) y = 2 px – p2
Ans. x = (2/3)p + cp-2, y = (1/3)p2 + 2cp–1
(58) y = 3px + 4p2
Ans. x = –(8/5)p + cp-3/2
y = 3cp–1/2 – (4/5)p2

(59) p  tan( px  y )

(60) sin px cos y  cos px.sin y  p

(61) Solve p 2 x( x  2)  p(2 y  2 xy  x  2)  y 2  y  0

(62) x 2 ( y  px )  yp 2

(63) ( px  y )( py  x )  h 2 p

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 8 Mathematics Optional

2 4 2 1 1
(64) y ( y  xp)  x p ( x  ,y )
u v

(65) y  2 px  y 2 p 3

(66) xp 2  2 yp  x  2 y  0. ( y  x  v & x 2  u )
(67) (px2 + y2)(px + y) = (p + 1)2
(68) ( x 2  y 2 )(1  p )2  2( x  y )(1  p )( x  yp )  ( x  yp )2  0

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
9
Tutorial Sheet - III
69. Find the differential equation of the family of circles x 2  y 2  2cx  2c 2  1  0 (c is arbitrary)
Determine singular solution of the differential equation.

Ans. 2 y 2 p 2  2 xyp  x 2  y 2  1  0, x 2  2 y 2  2  0

dy
70. Find the solution of the differential equation y  2 xp  yp 2 where p  also find the singular
dx
solution.
xy0
Ans. y 2  2cx  c 2 ; x  y  0

71. Find general and singular solutions of 3 xy  2 px 2  2 p 2 or y  (2 x / 3) p  (2 / 3 x )p 2

Ans. (3y  2c )2  4cx 3 , x 3  6 y  0

72. Solve the differential equation y  x  2 ap  ap 2 . Find the singular solution and interpret it
geometrically.
Ans. 4a(y – c) = x2 + c2 – 2xc; y – x + a = 0
73. Find the complete primitive (general solution) and singular solution of (xp – y)2 = p2 – 1.
Ans. x2 – y2 = 1

74. Reduce the equation xyp2 – p(x2 + y2 – 1) + xy = 0 to clairaut's form by the substitutions x2 = u and
y2 = v. Hence, show that the equation represents a family of conics touching the four sides of a
square.
2 2 c
Ans. y  cx 
c1
x  y  1  0, x  y  1  0, x  y  1  0, x  y  1  0
75. Solve (D2 + a2)y = cot ax

 1   ax 
Ans. c1 cos ax  c 2 sin ax   2  sin ax log tan  
a   2 
76. (4D2 + 12D + 9)y = 144e-3x
2x
2
77. (9 D  12 D  4)y  e3

2x 2x
1
Ans.  c1  c 2 x  e3  ( )x 2 e 3
18
78. ( D  1)(D2  2 D  2)y  e x

Ans. e x (c 1  c 2 cos x  c 3 sin x  x )

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 10 Mathematics Optional

79. ( D 4  2 D3  5D 2  8D  4)y  e x
x
Ans. P.I. = x 2 e 10
80. ( D 2  6D  8)y  ( e 2 x  1)2

2x 4x 1 4x 2x
Ans. c1 e  c 2 e    (4 xc  8xe  1)
8
 d4y  4
81.  4   m y  sin mx
 dx 
 d2 y  dy
82. Find the solution of  2   4 y  8 cos 2 x , given that y  0 and  0 , when x = 0
 dx  dx

83. ( D 3  3D 2  2 D )y  x 2

84. ( D 3  D 2  6D )y  x 2  1

2
Ans. c1  c2 e3x  c 3 e 2 x  ( 1 18)( x 3  x 2  25 x 6 )

85. ( D 2  6 D  9)y  x 2 e 3 x

Ans.  c 1  c 2 x  e 3 x   1 12  x 4 e 3 x

86. Solve ( D3  3D  2)y  540 x 3 e  x

87. ( D 2  1)y  cosh x.cos x

Ans. c1 e x  c 2 e  x   2 5  sin x sinh x   2 5  cos x cosh x.

x
88. (D4  D2  1)y  e 2 cos( x 3
2)

89. Solve ( D 2  4D  4)y  8 x 2 e 2 x sin 2 x

Ans. (c 1  c 2 x )e 2 x  e 2 x (3sin 2 x  4x cos 2 x  2 x 2 sin 2 x )

d2 y
90.  5 dy / dx  6 y  e 4 x ( x 2  9)
dx 2

Ans. c1 e 2 x  c 2 e 3 x  (1 / 4)e 4 x (2 x 2  6 x  25)

91. ( D 4  2 D2  1)y  x 2 cos x

4 3x 2
Ans. (c 1  c 2 x )cos x  (c 3  c 4 x )sin x  (1 / 4)[( x / 12  )cos x  ( x 3 / 3)sin x ]
4
92. Solve (D2  1)2  24 x cos x given y  Dy  D 2 y  0 and D3 y  12 . When x  0 '.

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
11
 5 21  /2
Ans. y = c1 x 2  c 2 x 5 21  /2 3
x x 5

97. Find the values of  for which all solution of x 2  d2 y

dx 2   3x  dy
dx   y  0 tends to zero as x  
Ans. 1    0

98. ( x 2 D2  4 xD  2)y  e x

Ans. y  c1 x 2  c 2 x 1  x 2 e x

99. Solve ( x 3 D3  2 x 2 D2  2)y  10  x  1 x 

Ans. c1 x 1  x[c 2 cos log x  c 3 sin log x ]  5x  2 x 1 log x.

100.  x  1 4
D3  2(x  1)3 D2  ( x  1)2 D  ( x  1) y   1
( x  1)

1  1  2
Ans. y  c1  c 2 log(1  x )(1  x )  c 3 (1  x )     1  x 
9
 

2 2d2 y dy 2
101.  1  2 x  2
 6  1  2x   16 y  8  1  2 x 
dx dx

Ans. y( x )  (1  2 x )2 log(1  2 x )[1  log(1  2 x )]

102. (3  x )y " (9  4x )y   (6  3x )y  0

Ans. c 2 e x  ( 1 8 )c 1 e 3 x (4 x 3  42 x 2  150 x  183)

103. Solve: x 2 y   xy   y  0 , given that  x  1 x  is one integral by using the method of reduction of
order.
Ans. y  c 2 ( x  1 x )  c1 ( 1 x )

104. sin 2 x  d2y

dx 2   2y ,given that y  cot x is a solution.
Ans. cot x[c 2 ]  c 1 (1  x cot x )

105.  d2 y
dx 2   (1  x) dy
dx  xy  x

2
Ans. c1 e x  e  x  x /2
dx  c 2 e x  1 (Left in integral form).

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 12 Mathematics Optional

106. Make use of transformation y( x )  V ( x )sec x to obtain the solution of y   2 y  tan x  5y  0 , y(0)=0,

y  (0)= 6

Ans. y  sec x sin  x 6 

2
107. Solve y   4 xy   (4 x 2  1)y  3e x sin 2 x

2
Ans. y  e x (c1 cos x  c 2 sin x  sin 2 x )

108. Solve y   2 xy   ( x 2  1)y  x 3  3x

2
Ans. x  e  x 2 (c1 x  c 2 )

d2 y L dy  1 1 6 
109. 2
 1/3
 2  4  2 y  0
dx x dx  4 x 3 6 x 3 x 

 3  2/3
  x
Ans. y  e 4
(c 1x
3
 c 2 x 2 )

110. Solve y   (4 cos ec 2 x )y   (2 tan 2 x )y  e x cot x by changing the dependent variable.

Ans. cot x(c1 e x 2

 c2 e  x 2
 ex )

111. Solve sin 2 xy   sin x cos x y   4 y  0

  x   x
Ans. c1 cos  2 log tan     c 2 sin  2 log tan 
  2   2

112. Solve (1  x 2 )2 y   2 x(1  x 2 )y  4 y  0

Ans. (1  x 2 )y  c 1 (1  x 2 )  2c 2 x

113. Solve xy   y   4 x 3 y  x 5

2
Ans. c1 cos x 2  c 2 sin x 2  x 4

114. cos x y   y sin x  2 y cos 3 x  2 cos 5 x

2 sin x  2 sin x
Ans. c1 e  c2 e  sin 2 x

d2 y  dy 
115. If y  y1 ( x ) and y  y 2 ( x) are the two solutions of the equation 2
 p( x )    ( x )y  0 where
dx  dx 
p(x),Q(x) are continuous function of x ,

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
13
dy 2 dy
Prove that y1  y 2 1  ce   pdx , c arbitrary cont.
dx dx
116. Apply the method of variation of parameters to solve the equation

( x  2)y 2  (2 x  5)y 1  2 y  ( x  1)ex

Ans. c1 (2 x  5)  c 2 e 2 x  e x

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 14 Mathematics Optional

Tutorial Sheet - IV
d2 y
117. Solve by the method of variation of parameters  (1  cot x ) dy dx  y cot x  sin 2 x.
dx 2

Ans. y  c 1 (sin x  cos x )  c 2 e  x  ( 1 10 )(sin 2 x  2 cos 2 x )

118. Solve by the method of variation of parameters x 2 y   2 x(1  x )y   2( x  1)y  x 3

Ans. y  c1 x  c 2 xe 2 x   x 2    x 4
2

119. y 2  n 2 y  sec nx (use of vop)

Ans. y  c 1 cos nx  c 2 sin nx   1 n  cos nx log cos nx  ( x n)sin nx

2

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
15
Laplace Transform
120. Find out the Laplace transform of

(a) cos2at (b) sin32t

1 sin t, 0  t  
(c) (d) 0
t  t

0 0  t  1
x / a , 0  x  a 
(e) f (x)   (f) f(x) =  t 1  t  2
1 , xa
0 t2


 cos t   1 4 s
(g) Find L. sin t (h) L.   e
 t  5
1

e 4s
Ans. 3
2s 2

(i) (t  3)2 et

9s 2  12s  s
Ans.
(s  1)3

1 3 6 6
Show L (1  te )  
t 3
(j)  2
 3

s (s  1) (s  2) (s  3)4
   
sin  t   t
G(t )    3  3
 
t
0 3

 t 1
  3/2 , show that 1/ 2  L 
1 1 
(l) If L 2 .
  s s  t 
(m) Find L f (t), if f (t )  3 f (t )  2 f (t )  0 , f (0)  1, f (0)  2 .

(n) L(t sin at ), L((sin 2t / t )

(o) Find the replace transform of (sin at)/t. Does the replace transform of (cos at)/t exist ?

 cos at  cos bt  1 s 2  b2
(p) Prove that L    log 2
 t  2 s  a2

(q) sin 2 t 
0  2
dt 
t 2

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 16 Mathematics Optional

 sin x 
(r) Evaluation L  0  t dx 
 x 
cot 1 s
Ans.
s

sin t 0  t  
(s) Evaluate L(f(t), if f(t) has a period 2 and is given by f (t )  
0   t  2

1
Ans. (1  e  s )
1  s2
(t) Find the replace of square wave given by

E 0  t  T / 2
f (t )   and f(t + T) = f(t)
 E T 2  t  T

E ST
Ans. tanh
S 4

(u) 0   te2t cos tdt  3 25

(v) 0   e t sin t dt   4
t

(w) To Prove that  

L erf ( t ) 
1
s s 1
121. Evaluate L–1 of.

6 3  4S 8  6S
(a)  
2S  3 9S 2  16 16S 2  9

1 4t 4 4t 2 3t 3 3t
Ans. 3e 3t 2  sinh  cosh  sin  cos
4 3 9 3 3 4 8 4

1 1 t2 t4 t6
(b) cos  1  2
 2
  ....
s s (2!) (4!) (6!)2


1
 cos x dx  2 (
2 1
(c) 
2 ) 2



e
 x2
(d) dx 
0
2

 2s 2  6s  5 
(e) L1  3 t 2 t
 = e  2e  3e
3t

 s  6 s  11s  6 

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 Ordinary Differential Equations (Assignment)
17
(f) s 1
L1  2 (cos bt  cos at )
( s  a )( s  b ) a  b 2
2 2 2 2

sa 1
(g) (h)
(s  a)2  b2 s ( s  1) 2
2

 s  1
(i) L1  4 4 
 2 sinh at . sin at
 s  4a  2 a

 t t 1 2 t 
4
 1 t3 t2 t 1
(j) L1  5   e       e 
 (s  1) (s  2)   72 54 54 81 243 243 

 e 4  3s 
(k) L1  
 ( s  4) 2 
5

  2 s 
se 3 
(l) L1  2
 s  9 

 e  s  cos 2 t
1
1
(m) If L  
 s1/2  t

 e  s   cos 2 ta 
a

Find L1   

 s1/2   t 

 s 
(m) L1  2 2 2
 (s  a ) 

  1  s  1  et
(n) L1 log   Ans.
  s  t
  1  2  1  cosh t 
(o) L1 log  1  2   Ans.
  s  t

(p) 
L1 tan 1  1 s   1
Ans.   sin t
t

(q) L1  1
s
log
s2
s1 
1
122. Apply convolution theorem to show that 0  t sin u.cos(t  u)du  t sin t
2

By Avinash Singh, Ex IES, B.Tech IIT Roorkee

 18 Mathematics Optional

 s 
123. L1  2 2
 (s  2 s) 

124. Solve y (t )  y(t )  t with y(0)  1 , y( )  0

Ans. y   cos t  t

125. Solve ( D2  n 2 )y  a sin(nt   ) if y  Dy  0 when t=0

a
Ans. [cos  sin nt  nt cos(  nt )]
2n 2

126. Solve [tD2  (1  2t )D  2]y  0 , y(0)  1 , y(0)  2

Ans. y  e 2t

127. Solve y   ty   y  1 , if y(0)  1 , y(0)  2

Ans. y(t )  1  2t

By Avinash Singh, Ex IES, B.Tech IIT Roorkee