Differential Equations & Partial Differential Equations
98. The differential equation representing (a) 4 (b) 3
the family of circles touching y axis
at the origin is _______ (c) 2 (d) 1
(a) Linear differential equation of [IISC 2008]
first order
102. Consider the differential equation
(b) Linear differential equation of 2 2
( ) dy =
( ) dx xy
second order 2cos y − sin y 0.
Then ______
(c) Nonlinear differential equation of
first order
x
(a) e is an integrating factor
(d) Nonlinear differential equation
of second order (b) 3x is an integrating factor
−
x
[JAM 2006] (c) e is an integrating factor
99. If k is a constant such that + = (d) x is an integrating factor
3
( −1) 2
2 satisfies the differential [JAM 2009]
equation
dy 103. Consider the differential equation
2
x
x
x = ( x − − ) 1 y + − 1, then k
dx dy = ay by , where a, b > 0 and
−
2
is equal to ________ dx
y ( ) 0 = y as x → , the solution y(x)
(a) 1 (b) 0 0
tends to _________
−
−
(c) 1 (d) 2
a
[JAM 2007] (a) 0 (b)
b
100. One of the integrating factors of the (c) b (d) y
differential equation a 0
( y − 3xy ) dx + ( x − xy ) dy = 0 is [JAM 2010]
2
2
_____ 104. The solution of
2
d y dy
0 y
4
1 1 dx 2 + 4 dx + 4y = , ( ) 0 = ,
(a) (b)
2
2
x y 2 x y y 1 ( ) 0 = is __________
8
1 1 2x
(c) (d) (a) 4e
xy 2 xy
(b) (16x + ) 4 e − 2x
[JAM 2007]
(c) 4e − 2x + 16x
101. The number of independent solutions
2x
of the differential equation 1 − (d) 4e − 2x + 16xe [JAM 2011]
2 11 + = 0 is _______
99
Differential Equations & Partial Differential Equations
y
a
105. If y is a integrating factor of the 109. Let ( ) x be the solution of the ODE
differential equation d y dy
2
2xy dx − (3x − y 2 ) dy = then the dx 2 + 2 dx + By = 0 where
2
0
value of a is _________ 0 B 1. Then
lim ( )_____________
−
(a) 4 (b) 4 →∞
−
(c) 1 (d) 1 (a) 0 (b)
B
(c) − (d)
[JAM 2011] 2
[IISC 2011]
106. The differential equation 110. Consider y + by + cy = 0 where b,
11
1
(1 x y+ 2 3 + x y 2 ) dx + c are real constants. If It is given that
2
2x
(2 x y + x 3 ) y dy = is exact if y = e is a solution. Then,
+
3
2
0
=_________ 2 2
(a) b + 4c 0 (b) b + 4c 0
1 3
(a) (b) (c) b − 2 4c 0 (d) b − 2 4c 0
2 2
[IISC 2007]
(c) 2 (b) 3
111. Consider the second order differential
[JAM 2012]
11
1
equation y + 3y + 2y = . Then
0
107. An integrating factor of lim ( ) is ________
(2xy + 3x y + 6y 3 ) dx + ( x + 6y 2 ) dy = 0 →∞
2
2
(a) a non-zero finite number
is ________
(b) 0
3
3
(a) x (b) y
(c) −
3y
3x
(c) e (d) e
(d) [IISC 2007]
[JAM 2012]
112. The general solution of the first order
2
d y dy differential equation
108. Consider + b + cy = 0 where 2
1
2
dx 2 dx xy + 2x y − xe − x = 0 is ________
b and c real constants. If y = xe − 5x is
)
+
a solution, then _______ (a) ( ) x = y e x 2 (x c
(a) both b and c are positive 2
)
+
(b) ( ) x = y e − x (x c
(b) b is positive but c is negative
(c) b is negative, but c is positive (c) ( ) x = y xe + x 2 c
(d) both b and c are negative
+
y
(d) ( ) x = x c [IISC 2006]
[IISC 2010]
100
Differential Equations & Partial Differential Equations
113. The general solution of first order (a) y 2 cos x + x sin y = 0
differential equation
xy + x y − = 0 is _________ 2
2
1
y
+
(b) y sin x x cos y =
2
(a) ( ) x = y xe − x 2 /2 + c
(c) y 2 sin x + x sin y = 0
(b) ( ) x = y e x 2 /2 (x c+ )
(d) y 2 cos x + x cos y = 2
(c) ( ) x = y e x 2 /2 − cx
[MS 2009]
y
(d) ( ) x = c xe − x 2 /2 [IISC 2006]
116. The differential equation
2
114. The solution of the differential d y + cos (x + y =
) sin x is
2
dy ( x x + 2 y − 2 10 ) dx
equation = , __________
dx ( y x + 2 y + 2 ) 5
(a) Linear, homogeneous
1
y ( ) 0 =
(b) Nonlinear, nonhomogeneous
(a)
2
2
=
4
2
x − 2x y − y − 20x − 10y + 11 0 (c) Linear, Non homogeneous
2
4
(b) (d) Nonlinear, homogeneous
2
4
2
=
4
x + 2x y + y + 20x + 10y − 11 0
2
2
(c) [IISC 2005]
=
x + 2x y − y + 20x − 10y + 11 0 117. Which of the following pair of
4
2
4
2
2
2
(d) functions is not a linearly
x + 2x y + y − 20x + 10y − 11 0 independent pair of solutions of
=
2
2
4
4
2
2
y + 11 9y = 0
[MS 2008] (a) sin3x, sin3x − cos3x
3
115. The solution of the differential (b) sin3x + cos3x , 3sin x − 4sin x
equation
dy = y 2 cos x + cos y , y = 0 (c) sin3x, sin3x cos3x
y
dx x sin y − 2 sin x 2 (d) sin3x + cos3x ,
3
is ________ 4cos x − 3cosx
[IISC 2005]
101
Differential Equations & Partial Differential Equations
118. A solution of the differential equation (a) 1, 2 (b) 2, 8
dy = y + 1 is _________ (c) 3, 8 (d) 4, 8
dx
122. The degree of the differential
(a) y = e − x 1 (b) y = e + x 1 2 2 1/4 3
equation y + x d y = d y
2
(c) y = e + x x (d) y = e x− 1 dx dx 3
[IISC 2002] is
119. The differential equation (a) 2 (b) 3
2
d y + 4 dy + 4y = has solution (c) 4 (d) 1
0
dx 2 dx
_________ 123. The differential equation of the
family of circles of radius ‘r’ and
+
A
(a) cos2x B sin2x whose centre lies on ‘x’ axis
(b) Ae − 2x + Bxe − 2x 2 dy 2
2
(a) r 1+ dx = x
2x
2x
(c) Ae + Bxe dy 2
2
2x
(d) Ae − 2x + Be [IISC 2002] (b) y 2 1+ dx = r
120. The solution of first order ODE (c) x 2 1+ dy 2 = 2
1
xy = xy + + + 1 is ________ dx r
y
x
=
(a) y cx e − ( x ) 1 (d) y 2 1 − dy 2 = r
2
dx
(b) y = cxe − x 1 124. The differential equation of the
family of curves of the form
+
(c) y = ce − x x y = Ax Bx is
2
(d) y ce= x − x − 1 [IISC 2005] (a) x y − 2xy + 2y =
1
2 11
0
121. The order and degree of the (b) x y + 2xy + 2y =
2 11
1
0
differential equation
3 2 4/3 (c) 2x y − xy + 0
2 11
1
d y d y 2y =
2
1 + 3 = 2 are
2 11
dx dx (d) x y − 2xy − 2y =
1
0
102
Differential Equations & Partial Differential Equations
dy 4 + y 2 log3 log2
2
y
125. Consider = , if ( ) 1 = (a) 2 (b) 2
+
dx 1 x 2 log2 log3
y
then ( ) 2 = log3 log2
(c) (d)
(a) 0 (b) 5 log2 log3
(c) 14 (d) 21 129. The rate at which a body cools is
proportional to the difference
dy x (2log x + ) 1 between the temperature of the body
126. The solution of =
dx sin y + y cos y and that of the surrounding air. If a
0
is body in air at 25 C cools from
0
0
100 C to 75 C in one minute then
+
(a) sin y = y x 2 log x c
the temperature at the end of three
− +
y
(b) cos y = x 2 log x x c minutes is
0
0
(a) 47.22 C (b) 42.22 C
−
2
c
y
(c) sin y = x 2 log x x +
0
0
(c) 37.22 C (d) 39.22 C
− +
y
(d) sin y = x logx x c
130. A radium decomposes at a rate
dy proportional to the amount of radium
−
2 −
y
127. The solution of = e x y + x e is
dx present at that time. If 5% grams of
original amount disappears after 50
x 3 years then the amount disappears
−
x
y
(a) e = e + + c
3 after 50 years then the amount will be
remain after 1000 years is
x 3
−
x
(b) e − y = e + + c
3 (a) 95.95% of the original amount
(b) 95% of the original amount
x 3
x
y
(c) e = e + + c
3 (c) 90% of the original amount
x 3 (d) 90.25% of the original amount
−
x
(d) e − y = e + + c
3 dy 2
y
131. The solution of = (4x + + ) 1 is
128. The rate of which bacteria multiply is dx
proportional to the instantaneous 1 4x + 4y + 1
number present, if the original (a) tan − 1 = c
number doubles in 2 hours then it will 2 2
be triple in
103
Differential Equations & Partial Differential Equations
1 4x + 4y + 1 134. The solution which transforms the
+
(b) tan − 1 = x c
2 2 non-homogenous differential
dy 2x + y + 6
equation = to
1 dx − x + − 3
y
+
(c) tan − 1 (4x + 4y + ) 1 = x c
2 homogenous form is
1 (a) x = X + 3, y Y + 1
=
c
(d) tan − 1 (4x + 4y + ) 1 =
2
=
(b) x = X + 3, y Y − 1
dy
132. The solution of = Sin (x + ) y is (c) x = X − 3, y Y +
=
dx 0
=
( +
) sec x y =
(a) tan x y − ( + ) x 2 + c (d) x = X − 3, y Y + 1
2
135. Which of the following differential
) tan x y =
(b) sec x y − ( + ) x 2 + c equation is linear
( +
2 dy
(a) + x y = 2 sin y
(
(c) tan x + y − ( ) y = + dx
) cos x +
x c
(
(d) tan x + y − ( ) y = + (b) dy − x y = 2 sin x
) cot x +
x c
dx
133. The solution of
y
y
dy = + tan is (c) (1 y ) dy + sin x =
+
0
x
x
dx dx
y (d) dy + ( y y + ) x = x
2
(a) sin = xc dx
x
dy
=
−
y 136. The solution of + 2xy ex with
2
(b) tan = xc dx
x y ( ) 0 = is
1
y
(c) Cosec = xc (a) (1 x e+ ) x 2 (b) (1 x e+ ) − x 2
x
)
y (c) (1 x e− ) x 2 (d) (1 x e − x 2
−
(d) Cot = xc
x
104
Differential Equations & Partial Differential Equations
+
137. If y − 1 x 0 then the solution of the (c) cos y = (x c )sin x
differential equation
+
(d) sec y = (x c )cos x
1
y 1 ( y + ) y = ( x x + ) y with
2
y ( ) 0 = is 141. The general solution of the
differential equation
x
+
−
(a) 1 x e− + − x (b) 1 x e dy + x sin2y = x 3 cos y is
2
dx
(c) 1 x e+ + − x (d) 1 x e+ + x
1 − 2
2
c e
dy 2log x (a) Tan y = ( x − ) 1 + x
138. If x 2 + 2xy = and 2
dx x
y
0
y ( ) 1 = then ( ) e = (b) Tan y = 1 ( x − ) 2 + − x 2
2
c e
2
(a) e (b) 1
2
c e
(c) Tan y = ( x − ) 1 + − x 2
1 1
(c) (d)
e e 2 1 2
2
(d) Cot y = ( x − ) 1 + x
c e
dy 2
139. If + 2 Tan x = y Sin x and
dx 142. The general solution of
dx
y = 0 then the maximum value ( x y + 3 2 xy ) = 1 is
3 dy
of ‘y’ is − 1
2
2 c e
(a) = x − + − x 2 /2
1 1 y
(a) (b)
4 8
+
2
(b) 1 = x + 2 c e − x 2 /2
1 y
(c) 2 (d)
2
1 2 x 2 /2
2 c e
140. The general solution of the (c) y = x + +
differential equation
dy + Tan x Tan y = Cos x Sec y (d) 1 = x + 1 c e − x 2 /2
2
+
dx y
is
143. The differential equation
(a) 2sin y = (x c+ − sincos x )sec x ( xy 3 + y cos ) x dx +
(b) sin y = (x c+ )cos x ( x y + 2 2 sin ) x dy = 0 is exact for
105
Differential Equations & Partial Differential Equations
3 3 (a) ( x − x 2 y 2 ) c
=
e
(a) , = = 1 (b) 1, = =
2 2
e
(b) ( x + x 2 y 2 ) c=
2 2
(c) , = = 1 (d) 1, = =
3 3
=
−
(c) ( x + x 2 y 2 ) c
e
144. The differential equation
−
=
e
( 27x + ky cos ) x dx + (2sin x − 27y 3 ) dy = 0 (d) ( x − x 2 y 2 ) c
2
is exact for k =
148. If the integrating factor of
(a) 2 (b) 3 7 2 3
( x y + 3y ) dx + (3x y − ) x dy =
0
(c) 4 (d) 5
is x y then
145. The integrating factor of (a) 7, = − = 1
)
(Cos sin 2x dx + (cos y − cos 2 ) x dy = 0
2
y
is (b) 1, = = − 7
(a) sec y + 2 sec y tan y (c) = = 0
(b) tan y + 2 sec y tan y (d) = = 1
1 149. The solution of
(c) 2 2
0
sec y + 2 sec y tan y ( y − xy ) dx − ( x + x ) y dy = is
1 x x
(d) (a) ln − = c
tan y + 2 sec y tan y y y
146. Consider the differential equation x x
dy = (3x + 2x )e y x + 1, if ( ) 1y = 1 (b) ln + = c
−
2
dx y y
y
then ( ) 0 = x
=
(c) ln + xy c
−
3
(a) log (b) 0 y
e
x
3
=
(c) 1 (d) log (d) ln − xy c
e
y
147. The solution of the differential
equation 150. The orthogonal trajectory of the
( x + y + 2x ) dx + 2ydy = is family of circles x + 2 y = 2 2cx is
2
2
0
described by the differential equation
106
Differential Equations & Partial Differential Equations
y
(a) ( x + 2 y 2 ) dy = 2xy 153. Let ( ) x is solution of
dx y 111 − y + 4y − 4y = , ( ) 0 = ,
1
11
2
0 y
0
(b) ( x − 2 y 2 ) dy = 2xy y 11 ( ) 0 = then the value of
dx
y 2 =
(c) ( y − 2 x 2 ) dy = 2xy
dx 1
(a) (4e /2 − ) 6
(d) ( y − 2 x 2 ) dy = xy 5
dx 1
(b) (6e /2 − ) 4
151. The orthogonal trajectory of family of 5
straight lines y = ( k x − ) 1 , k 1
R
are given by (c) (8e /2 − ) 2
5
2
2
2
(a) (x − ) 1 + ( y − ) 1 = c 1
(d) (8e /2 + ) 2
5
2
(b) x + 2 y = 2 c
154. The set of linearly independent
2
2
2
(c) x + ( y − ) 1 = c solutions of the differential equation
4
2
d y − d y =
2
2
2
(d) (x − ) 1 + y = c dx 4 dx 2 0 is
152. The orthogonal trajectories of family (a) 1, , ,e − x
x
x e
)
of centroids r = a + (1 cos is
x
x e
(b) 1, , ,e − x ,xe − x
(a) r = c (1 sin− )
x
x e
(c) 1, , ,xe x
(b) r = c (1 sin+ )
x
x e
)
(c) r = c − (1 cos (d) 1, , ,xe − x
)
(d) r = c + (1 cos 155. The differential equation whose
linearly independent solutions are
−
x
cos2x, sin2x and e is
3
2
(a) ( D + D + 4D + ) 4 y = 0
3
2
0
(b) ( D − D + 4D − ) 4 y =
107
Differential Equations & Partial Differential Equations
3
2
(c) ( D − D − 4D − ) 4 y = (a) c e + 2x c e + 3x e x
0
1
2
2
3
2
(d) ( D − D − 4D + ) 4 y = x
0
(b) c e + 1 2x c e − 3x + e
2
2
d y dy 2
156. Consider + b + cy = 0 where
dx 2 dx e x
b & c are real constants. If (c) c e + 1 2x c e + 2 3x 4
y = x e − 5x is a solution then
e x
(d) c e + 2x c e − 3x +
(a) both b and c are positive 1 2 4
(b) b is positive and c is negative 160. The particular solution of
x
1
11
(c) b is negative but c is positive y 111 − y − y = − e is a constant
multiple of
(d) both b and c are negative
−
x
x
−
−
x
x
157. If e and xe are two independent (a) xe (b) xe
2 −
2 x
x
2
d y dy (c) x e (d) x e
solutions of + + y = 0 then
dx 2 dx
the value of = 161. The solution of the differential
2
d y dy
2x
equation − − 2y = 3 e ,
(a) 1 (b) -1 dx 2 dx
1
0 y
(c) -2 (d) 2 y ( ) 0 = , ( ) 0 = − 2 is
158. The particular integral of (a) y e − e + xe
−
2x
x
=
2x
2
d y − 2 dy + 2y = log2 is
=
x
2x
dx 2 dx (b) y e − e − 2x + xe
e 2 log2 x e
2 2x
2x
(a) (b) (c) y e= − x + e +
2 2 2
log2 log2 x
2x
(c) (d) (d) y e= x − e − 2x − e
4 8 2
159. The general solution of the 162. Consider y − 11 y = 2e if ( ) 0y = 0 ,
x
differential equation y 1 ( ) 0 = then ( ) 1y
0
2
d y − 5 dy + 6y e is =
=
x
dx 2 dx
108
Differential Equations & Partial Differential Equations
( )
(a) e + sinh ( ) 1 (b) cosh 1 166. The particular integral of
( D − 4D + ) 4 y = x is
2
2
( )
(c) sinh 1 (d) cosh 1 +
( ) 1
1 3
(a) x + 2 2x +
163. Suppose ( ) x = x cos2x is a 4 2
y
p
particular integral of 1 x + 2 2x − 3
y + y = − 4sin2x, then the (b) 4 2
11
constant is 1 3
(c) x − 2 2x +
(a) -4 (b) -2 4 2
(c) 2 (d) 4 (d) 1 x − 2 2x − 3
4 2
164. The particular integral of the
differential equation 167. The particular integral of the
( D − 4D + ) 3 y = cos x is differential equation
2
( D − 4D + ) 4 y = 2x
2
x e is
cos x − 2sin x
(a) xe 2x x e
2 2x
10 (a) (b)
6 6
cos x + 2sin x
3 2x
3 2x
(b) x e x e
10 (c) (d)
24 6
2cos x − 4sin x
(c) 168. The particular integral of
5 ( D − 5D + ) 6 y = e 2x cos x is
2
2cos x + 4sin x 2x
(d) (a) − e sin
5 (cos x + ) x
2
165. The particular integral of e 2x
( D − ) 9 y = e + sin2x is (b) − 2 (cos x − sin ) x
3x
2
e 3x sin2x e 2x
(a) + (c) (cos x + sin ) x
6 13 2
e 3x 1 e 2x
(b) − sin2x (d) (cos x − sin ) x
6 32 2
e 3x 1 169. The particular integral of the
(c) − sin2x
6 13 differential equation
2
x
xe 3x 1 ( D + ) 9 y = sin2x is
(d) − sin2x
6 13
109
Differential Equations & Partial Differential Equations
sin2x 4 (b)
(a) x + cos2x
5 25 y = c 1 + c cos (log x + sin (log x
))
) c
x ( 2 3
x sin2x 4
(b) − cos2x (c)
5 25
y = c + x c cos (log x + 3 sin (log x ))
) c
( 2
1
x 2 sin2x 4 (d)
(c) + cos2x
5 25 y = c x + x c cos (log x + 3 sin (log x ))
2
) c
( 2
1
(d) None
170. The particular integral of 173. The differential equation for which x
2
x
( D + 2 ) 4 y = x cos2x is , ln x and x are independent
solutions is
x 2 sin2x x 2 sin2x
3 111
2 11
1
(a) (b) (a) x y + x y − 3xy + 3y =
0
4 8
2 11
3 111
1
0
x 2 cos2x x 2 cos2x (b) x y − 2x y + 3xy − 6y =
(c) (d)
4 8
2 11
1
3 111
0
(c) x y − x y + 2xy − 2y =
171. The particular integral of
11
1
0
( D + 2 ) 9 y = sec3x is (d) y 111 − y + 2y − 3y =
x cos3x 174. Consider the differential equation
)
(a) sin3x + log (cos x x y − 3xy + 4y = then the two
2 11
1
0
3 9
linearly independent solutions of the
x cos3x differential equation are given by
)
(b) sin3x + log (cos3x
3 9
3
,
2
2
,
(a) x x (b) x x 2 / nx
x sin3x
)
(c) cos3x + log (sec3x x
3 3 (c) v (d) , x xe
x sin3x 175. The partial differential equation
)
(d) cos3x + log (cos3x 2 2 2
3 3 y u u
2 + 4 + 3 = 2 is
x 2 x y y 2
172. The general solution of the
differential equation (a) Elliptic (b) Hyperbolic
3
( x D + 2x D + ) 2 y = is
3
2
2
0
(c) Parabolic (d) None
x
+
(a) y c x= 1 + ( cosc 2 x c 3 sin ) x e
110
Differential Equations & Partial Differential Equations
176. The partial differential equation 181. The degree of the differential
2 u 2 u 2 2 1/4 3
2
xy − ( x − y 2 ) − d y d y
x 2 equation y + x dx 2 = dx 3
x y
2 u u u
2
xy + y − x = ( 2 x − y 2 ) is.
y 2 x y
is 182. Match each of the items A, B, C with
an appropriate item from 1, 2, 3, 4
(a) Elliptic (b) Hyperbolic and 5
(c) Parabolic (d) None d y dy
2
(A) a 1 a y a y = a
3
2
4
177. The partial differential equation dx 2 dx
2 u + 4 2 u + 4 2 u = d y
3
x 2 y 2 0 is (B) a 1 dx 3 a y = 2 a
x y
3
(a) Elliptic (b) Hyperbolic d y dy
3
2
(C) a 1 2 a x + a x y = 0
2
3
(c) Parabolic (d) None dx dx
178. The partial differential equation (1) non-linear differential equation
2 u + 2 u − 2 u = (2) linear differential equation with
x 2 y 2 z 2 0 is constant coefficients
(3) linear homogeneous differential
(a) Elliptic (b) Hyperbolic equation
(c) Parabolic (d) None (4) non-linear homogeneous
differential equation
179. One dimensional heat equation
u = 2 u is (5) non-linear first order differential
t x 2 equation
(a) A – 1, B – 2, C – 3
(a) Elliptic (b) Hyperbolic
(b) A – 3, B – 4, C – 2
(c) Parabolic (d) None
(c) A – 2, B – 4, C – 3
180. The two dimensional heal equation (d) A – 3, B – 1, C – 2
2 u + 2 u =
x 2 y 2 0 is [GATE-1994 (EC)]
183. The differential equation
(a) Elliptic (b) Hyperbolic 5
)
3
11
y + ( y 3 sin x y + = cos x is
1
y
(c) Parabolic (d) None
111
Differential Equations & Partial Differential Equations
(a) homogeneous (c) third order linear ordinary
differential equation
(b) nonlinear
(c) second order linear (d) mixed order nonlinear ordinary
differential equation
(d) non homogeneous with constant
coefficients [GATE-1995] [GATE-2010-ME]
2
d y dy 187. The order and degree of the
8
2
184. + ( x + 4x ) + y = x − 8. differential equation
dx 2 dx
3
The above equation is a d y + 4 dy 3 + y = 0 are
2
dx 3 dx
(a) partial differential equation
respectively.
(b) nonlinear differential equation
(a) 3 and 3 (b) 2 and 3
(c) non-homogeneous differential
equation (c) 3 and 3 (d) 3 and 1
(d) ordinary differential equation [GATE-2010-CE]
[GATE-1999] 188. The partial differential equation
u u 2 u
185. The degree of the differential t + u x = x 2 is a
2
d x
equation + 2x = 3 0 is
dt 2 (a) linear equation of order 2
(a) zero (b) 1 (b) non-linear equation of order 1
(c) 2 (d) 3 (c) linear equation of order 1
[GATE-2007-CE] (d) non-linear equation of order 2
186. The Blasius equation, [GATE-2013-ME]
2
3
d t + fd t = LINEARLY DEPENDENT AND
d 3 2d 2 0, is a INDEPENDENT SOLUTIONS
(a) second order nonlinear ordinary 189. Choose the CORRECT set of
differential equation functions, which are linearly
dependent.
(b) third order nonlinear ordinary
2
differential equation (a) sin , sinx 2 x and cos x
(b) cos x, sin x and tan x
112
Differential Equations & Partial Differential Equations
2
2
x
(c) cos2 , sin x and cos x number doubles in 2 hours then it will
be triple in
(d) cos 2x, sin x and cos x
log3 log2
(a) 2 (b) 2
[GATE-2013 (ME)] log2 log3
190. Consider the following statements log3 log2
about the linear dependence of the (c) (d)
real valued functions y = 1, y = log2 log3
x
1
2
and y = x , over the field of real 192. A radium decomposes at a rate
2
3
numbers. proportional to the amount of radius
present at that time. If 5% grams of
(I) , y y and y are linearly original amount disappears after 50
1
3
2
independent on 1 x 0 years then the amount will be remain
−
after 100 years is
(II) , y y and y are linearly
3
2
1
dependent on 0 x 1 (a) 95.95% of the original amount
(b) 95% of the original amount
(III) , y y and y are linearly
1
3
2
dependent on 0 x 1 (c) 90% of the original amount
(IV) , y y and y are linearly (d) 90.25% of the original amount
3
1
2
independent on 1 x 1 193. The orthogonal trajectory of family of
−
R
Which on among the following is straight lines y = ( k x − ) 1 , k
correct? are given by
(a) Both I and IV are true (a) (x − ) 1 + ( y − ) 1 = c
2
2
2
(b) Both I and III are true
(b) x + 2 y = 2 c
2
(c) Both II and IV are true
2
2
2
(c) x + ( y − ) 1 = c
(d) Both III and IV are true
2
2
2
[GATE-2017 EC SESSION-1] (d) (x − ) 1 + y = c [GATE]
FORMATION OF DIFFERENTIAL 194. The differential equations of the
EQUATION
family of circles of radius ‘r’ and
191. The rate of which bacteria multiply is whose centre lies on ‘x’ axis.
proportional to the instantaneous
number present, if the original
113
Differential Equations & Partial Differential Equations
dy 2 SOULUTION OF DIFFERENTIAL
2
2
(a) r 1+ = x EQUATIONS
dx
197. A solution of the first order
2 differential equation
2
(b) y 1+ dy = r
2
dx sin (x y )
+
)
+
x
y 'cos (x y + = e − cos (x y )
+
x
dy 2 is
2
2
(c) x 1+ = r
dx ( x
(a) sin x + ) y − e = constant
2 x
2 dy (b) e tan (x + y = ) constant
(d) y 1− = r
2
dx
x x
(
x
(c) cos x + ) y − e + e = constant
195. A spherical naphthalene ball exposed
(
x
x
to the atmosphere loses volume at a (d) sin x + ) y − e + e = constant
x
rate proportional to its instantaneous
surface area due to evaporation. If the [GATE]
initial diameter of the ball is 2 cm and
the diameter reduces to 1 cm after 3 198. Consider the following differential
months, the ball completely equation :
evaporates in
y y
+
(a) 6 months (b) 9 months ( x ydx xdy )cos x = ( y xdy − ydx )sin
x
(c) 12 months (d) infinite time Which of the following is the solution
[GATE-2006] of the above equation (c is an
arbitrary constant?
0
196. A body originally at 60 C cools x y x y
0
down to 40 C in 15 min when kept (a) cos x = c (b) sin x = c
y
y
0
in air at a temperature of 25 C .
What will be the temperature of the y y
body at the end of 30 min? (c) xy cos x = c (d) xy sin x = c
0
0
(a) 35.2 C (b) 31.5 C [GATE-2015-CE-SET-II]
0
0
(c) 28.7 C (d) 15 C 199. The solution of the differential
equation
[GATE-2007-CE]
( x + y + 2x dx + 2y dy = is
)
2
2
0
114
Differential Equations & Partial Differential Equations
x
e x −
(a) ( 2 y 2 ) = c 203. The differential equation
2
( xy 3 + y cos ) x dx + ( x y + sin ) x dy = 0
2
(b) ( 2 y 2 ) = c is exact for
e x +
x
−
x
e
(c) ( x + 2 y 2 ) = c (a) 3 , = = 1
2
(d) ( x − 2 y 2 ) = c (b) 1, = = 3
−
x
e
2
200. The solution of 2
( y xy dx − ( x x y dy = 0 is (c) 3 , = = 1
)
)
2
+
2
−
x x (d) 1, = = 2
(a) ln − = c 3 [GATE]
y y
204. For the differential equation,
x x
(b) ln + = c f ( , x y ) dy + g ( , x y = ) 0 to be
y y dx
exact,
x
=
(c) ln + xy c f g f g
y (a) = (b) =
y x x y
x
=
(d) ln − xy c 2 f 2 g
y (c) f = g (d) x 2 = y 2
EXACT DIFFERENTIAL EQUATION
[GATE-1997-CE]
201. The differential equation
VARIABLE SEPARABLE
( 27x + ky cos ) x dx + ( 2sin x − 27y 3 ) dy = 0
2
is exact for k = ____. 205. The solution of dy = y with initial
2
dx
[GATE]
1
y
value ( ) 0 = bounded in the
202. If the integrating factor of interval
)
( x y + 3y dx + ( 3x y x dy = (a) − (b) − 1
)
7
2
−
8
0
x
x
is x y then = ____ and (c) x 1, x 1 (d) 2 x−
2
= ____ .
[GATE-2007-ME]
[GATE]
115
Differential Equations & Partial Differential Equations
206. The solution of the differential 209. The solution of
equation dy = + tan
y
y
dx x is
x
2
−
−
2
0
y 1 x dy + x 1 y dx = is
y
−
(a) 1 x = 2 c (a) sin x = xc
y
(b) 1 y− 2 = c (b) tan = xc
x
−
2
−
2
(c) 1 x + 1 y = y
c
(c) cosec = xc
+
2
(d) 1 x + 1 y = x
2
+
c
(d) cot y = xc
[ESE-2017 (EE)] x
207. The figure shows the plot of y as a 210. Solve the differential equation,
function of x. The function shown is 2 dy 3 3
the solution of the differential xy dx = x + y
equation (assuming all initial
conditions to be zero) is [GATE-1994-ME]
2
d y dy 211. A curve passes through the point (x =
(a) = 1 (b) = + x 1, y = 0) and satisfies the differential
dx 2 dx
2
dy dy dy = x + y 2 + y
(c) = − x (d) = x equation dx 2y x . The
dx dx
equation that describes the curve is
[GATE-2014 (IN-SET 1)]
y 2
HOMOGENEOUS DIFFERENTIAL (a) ln 1+ 2 = x − 1
EQUATION x
dy 1 y 2
208. The solution of = sin (x + ) y is (b) ln 1+ = x − 1
dx 2 x 2
(
) sec x +
x c
(a) tan x + y − ( ) y = + y
(c) ln 1+ x = x − 1
2
(
(b) sec x + y ) tan x− ( + ) y = x + c
2
1 y
(
+
(c) tan x + y ) cos x− ( + ) y = x c (d) ln 1+ x = x − 1
2
(
+
(d) tan x + y ) cot x− ( + ) y = x c
[GATE 2018 (EC)]
116
Differential Equations & Partial Differential Equations
LINEAR DIFFERENTIAL EQUATION dv + − ) − )
ST
OF 1 ORDER (a) dt (1 n pv = (1 n q
211. The general solution of the dv
)
)
+
−
differential equation (b) dt + (1 n pv = (1 n q
dy + tan x tan y = cos x sec y is
dx dv
)
)
+
−
(c) + (1 n pv = (1 n q
dt
+
(a) 2sin y = (x c − sin cos x )sec x
)
)
(b) sin y = (x c )cos x (d) dv + (1 n pv = (1 n q
+
+
+
dt
+
(c) cos y = (x c )sin x [GATE-2005-CE]
(d) sec y = (x c )cos x [GATE] 214. A system described by a linear,
+
constant coefficient, ordinary, first
212. The general solution of order differential equation has an
( x y + 3 2 xy ) dx = 1 is exact solution given by y(t) for y > 0,
dy when the forcing function is x(t) and
the initial condition is y(0). If one
− 1 2
2
2 c e
(a) = x − + − x /2 wishes to modify the system so that
y the solution becomes -2y(t) for t > 0,
we need to
1 − 2
2
2 c e
(b) = x + + x /2
y (a) change the initial condition to -
y(0) and the forcing function to 2x(t)
1 2
2
2 c e
(c) = x + + x /2 (b) change the initial condition to
y
2y(0) and the forcing function to (t)
1 − 2 (c) change the initial condition to
2
1 c e
(d) = x + + x /2
y j 2y ( ) 0 and the forcing function to
213. Transformation to linear form by j 2x ( ) t
−
substituting v = y 1 n of the equation
(d) change the initial condition to -
dy + p ( ) t y = q ( ) t y ; n > 0 will be 2y(0) and the forcing function to -
n
dt
2x(t)
[GATE-2013 (EC)]
117
Differential Equations & Partial Differential Equations
215. Which ONE of the following is a (a) sec y + 2 sec y tan y
linear non-homogeneous differential
equation, where x and y are the (b) tan y + 2 sec y tan y
independent and dependent variables
respectively? 1
(c) sec y + 2 sec y
dy − dy tan y
x
(a) + xy = e (b) + xy = 0
dx dx 1
(d) 2 [GATE]
dy − dy − tan y + sec y tan y
y
(c) + xy = e (d) + e y = 0
dx dx 219. The solution of the differential
equation
[GATE-2014-EC-SET 3] ( x + y + 2x dx + 2ydy =
)
2
2
0
216. The integrating factor for the
differential equation (a) ( 2 y 2 ) = c
e x −
x
dp + K p = K L e − K t
1
dt 2 1 0 is (b) ( 2 y 2 ) = c
e x +
x
(a) e − K t (b) e − K t (c) ( 2 2 )
2
1
−
x
e x + y = c
(c) e K t (d) e K t
2
1
(d) ( x − 2 y 2 ) = c
−
x
e
[GATE-2014-CE-SET 2]
2
217. Consider the differential equation 220. Consider d y + b dy + cy = 0 where
( t − 81 ) dy + 5ty = sin ( ) t with dx 2 dx
2
dt b & c are real constants. If
y ( ) 1 = 2 . There exists a unique y = x e − 5x is a solution then
solution for this differential equation (a) both b and c are positive
when t belongs to the interval (b) b is positive, and c is negative
(a) (-2, 2) (b) (-10, 10) (c) b is negative but c is positive
(c) (-10, 2) (d) (0, 10) (d) both b and c are negative
[GATE-2017 EE SESSION-1] 221. The differential equation y + 11 y = 0
is subjected to the conditions
218. The integrating factor of
y ( ) 0 = , ( ) 0y = . In order that
0
2
(cos sin 2y ) x dx + ( cos y − cos 2 ) x dy = 0
the equation has non-trivial solutions
is
the general value of is
118
Differential Equations & Partial Differential Equations
[GATE-1993 (ME)] The form of non-zero solutions y
(where m varies over all integers) are
222. The Solution to the differential
equation (a) y = A sin m x
x +
x +
( ) 0
f 11 ( ) 4 f 1 ( ) 4 f x = is m m a
f
(a) ( ) x = e − 2x (b) y = A m cos m x
1
(b) ( ) x = f 1 e 2x , f 2 ( ) x = xe − 2x m a
(c) ( ) x = f 1 e − 2x , f 2 ( ) x = xe − 2x (c) y = A x m
m
a
m
−
x
(d) ( ) x = f 1 e 2x , f 2 ( ) x = xe
(d) y = A e m x
[GATE-1995-ME] m m a
223. The complete solution of the ordinary [GATE-2006-EC]
differential equation
2
d y + p dy + qy = 0 is 226. The homogeneous part of the
differential equation
dx 2 dx d y dy
2
y c e + 1 − x c e − 3x . Then, p and q are dx 2 + p dx + qy = (p, q, r are
=
r
2
(a) p = 3, q = 3 (b) p = 3, q = 4 constants) has real distinct roots if
(c) p = 4, q = 3 (d) p = 4, q = 4 2 2
(a) p 4q 0 (b) p 4q 0
[GATE-2005-ME]
=
(c) p 2 4q = 0 (d) p 2 4q r
224. For the equation
t +
1
x 11 ( ) 3x t + ( ) 5 [GATE-2009 (PI)]
( ) 2x t = , the
solution x(t) approaches to the 227. A function n(x) satisfies the
following value as t → differential equation
2
(a) 0 (b) 5/2 d n ( ) x − n ( ) x = 0 where L is a
dx 2 L 2
(c) 5 (d) 10
constant. The boundary conditions
[GATE-2005-EE] are: n(0) = K and ( ) 0n = . The
225. For the differential equation solution to this equation is
2
d y + k y = 2 0 the boundary (a) ( ) x = K exp ( /x L
)
n
dx 2
)
conditions are
(b) ( ) x = n K exp − ( / x L
(i) y = 0 for x = 0 and
(ii) y = 0 for x = a (c) ( ) x = n K 2 exp ( x L− / )
119
Differential Equations & Partial Differential Equations
)
(d) ( ) x = n K exp − ( x L dy
/
boundary conditions = 1 and
dx x=
[GATE-2010-EC] 0
dy
228. If the characteristic equation of the dx x= = − 1 has
differential equation 2
2
d y + 2 dy + y = 0 has two equal (a) no solution
dx 2 dx
roots, then the values of are (b) exactly two solutions
(a) 1 (b) 0, 0 (c) exactly one solution
(c) j (d) 1/ 2 (d) infinitely many solutions
[GATE 2017]
[GATE-2014-EC-SET 2]
−
−
x
x
231. If e and xe are two independent
229. Consider two solution ( ) t = x 1 ( ) t d y dy
x
2
and ( ) t = x 2 ( ) t of the differential solutions of dx 2 + dx + y = 0 then
x
2
d x ( ) t the value of = ____ .
( ) 0, t ,
equation + x t = 0
dt 2 PARTICULAR INTEGRAL
dx ( ) t
such that ( ) 0 = 1, 1 = 0, 232. For initial value problem
x
1
dt ) ) x
t= 0 y + 2y + (101 y = (104 e , y(0) =
dx ( ) t 1.1 and y(0) = -0.9. Various solutions
x = 0, 2 = 1. The are written in the following groups.
2
dt
t= 0 Match the type of solution with the
x 1 ( ) t x 2 ( ) t correct expression.
Wronskian ( ) t = dx 1 ( ) t dx 2 ( ) t Group-1
W
dt dt P. General solution of homogeneous
at t / 2 is equations
=
(a) 1 (b) -1 Q. Particular integral
(c) 0 (d) / 2 S. Total solution satisfying boundary
conditions
[GATE-2014-ME-SET 3]
Group-II
230. The differential equation x
2
d y + 16y = 0 for y(x) with the two (1) 0.1e
+
dx 2 (2) e − x cos 10A x B sin10x
120
Differential Equations & Partial Differential Equations
x
x
(3) e − x cos10x + 0.1e 235. Consider y − 11 y = 2e if y(0) = 0,
0
y 1 ( ) 0 = then y(1) =
(a) P – 2, Q – 1, R – 3
(b) P – 1, Q – 3, R – 2 (a) e + sin h (1) (b) cos h(1)
(c) P – 1, Q – 2, R – 3 (c) sin h(1) (d) cos h(1) + 1
[CSIR]
(d) P – 3, Q – 2, R – 1
236. Suppose ( ) x = x cos2x is a
y
[GATE-2006 (IN)] p
particular integral of
11
233. The solution of the differential y + y = − 4sin2x , then the
2
d y constant is _____
equation k 2 = y − y under the
dx 2 2 [GATE]
boundary conditions
CAUCHY HOMOGENEOUS LINEAR
(i) y = y at x = 0 and DIFFERENTIAL EQUATIONS
1
(ii) y = y at x = , where k, y and 237. The general solution of the
2
1
y are constants, is differential equation
2
2
d y − x dy + y =
2
x
exp −
(a) y = ( y − y 2 ) ( / x k 2 ) + y dx 2 dx 0 is:
1
2
2
(b) y = ( y − y 1 )exp − / ) y (a) Ax + Bx (A, B are constants)
( x k +
1
2
+
)
( /
(c) y = ( y − y 2 )sinh x k + y (b) Ax B log x (A, B are constants)
1
1
+
( x k +
(d) y = ( y − y 2 )exp − / ) y (c) Ax Bx 2 log x (A, B are
1
2
constants)
[GATE-2007-EC] (d) Ax Bx logx (A, B are
+
234. Consider the following second-order constants) [GATE-1998]
differential equation: 238. The radial displacement in a rotating
2
1
11
y − 4y + 3y = 2t − 3t . The disc is governed by the differential
particular solution of the differential d u 1 du u
2
equation is equation 2 + − 2 = 8x
dx x dx x
−
−
2
− −
(a) 2 2t t − 2 (b) 2t t where u is the displacement and x is
the radius. If u = 0 and x = 0, and u =
2
2
(c) 2t − 3t (d) 2 2t− − − 3t
[GATE-2017 CE SESSION-II]
121
Differential Equations & Partial Differential Equations
2 at x = 1, calculate the displacement 2 dy + 2xy = 2log x
1 242. If x dx x and y(1) = 0
at x = [GATE-1998]
2 then y(e) = 0
2 11
1
239. The solution to x y + xy − = 0 (a) e (b) 1
y
1
is (c) (d) 1
e e 2
−
=
3
(a) y C x + 1 2 C x
2
[GATE]
−
2
(b) y C= 1 + C x u u
2
243. The solution of = 4 ,
C x y
=
(c) y C x + 2 − 3y
1
) 8e
x u (0, y = is _________
4
+
=
u
) 8e
(d) y C x C x (a) ( , x y = − 12x− 3y
1
2
u
) 8e
[GATE-2015 (PI)] (b) ( , x y = − 3x− 12 y
240. The differential equation for which x, (c) ( , x y = − 3y− 4x
) 8e
u
2
x ln x and x are independent − 3x− 3y
) 8e
u
solutions is (d) ( , x y =
3 11
2 11
1
0
(a) x y + x y − 3xy + 3y = 244. The solution of p + q = 1 is
_____
1
(b) x y − 2x y + 3xy − 6y =
2 11
3 11
0
+
=
+
(a) z ax by c
3 11
1
(c) x y − x y + 2xy − 2y = ( ) 2
2 11
0
+
(b) z = ax + 1− a y c
1
11
0
(d) y 111 − y + 2y − 3y =
)
+
(c) z = ax + ( 1− a y c
[CSIR]
+
(d) z = ax − ay c
241. Consider the differential equation
p +
x y − 3xy + 4y = then the two 245. The solution of (1 q = ) qz is
2 11
1
0
linearly independent solutions of the ______
differential equations are given by
+
+
=
(a) az − 1 e x ay c
3
2
2
,
,
(a) x x (b) x x 2 ln x (b) z e= x ay c
+
+
x
(c) v (d) ,x xe
(c) z ax by c= + +
)
+
[CSIR] (d) z = ax by + f ( ,a b
122
Differential Equations & Partial Differential Equations
246. The solution of p + q = + is 249. The partial differential equation that
2
2
y
x
can be formed from z = ax + by + ab
____
has the form
+
=
+
(a) z ax by c z z
with p = and q =
2 3/ 2 2 3/ 2 x y
+
−
(b) z = (a x ) + ( y a ) + b
3 3
(a) z = px + qy
3 2/3 3 2/3
−
(c) z = (a + ) x + ( y a ) + b (b) z = px + pq
2 2
−
(d) z = (a + ) x 1/2 + ( y a ) 1/2 + b (c) z = px + qy + pq
(d) z = py + pq
247. The general integral of the partial [GATE-2010-CE]
differential equation
y p xyq = ( x z − 2y ) is 250. The complete integral of
2
−
(
)
2
3
2
−
(z − px qy ) = pq + 1 p + q is
)
2
2
(a) ( x + y 2 , y − yz = 0
(
)
+
2
2
2
(b) ( x − y 2 , y + yz = 0 (a) z = ax by + 3 pq + 2 p + q ) 2
(
2
+
) 0
,
© ( xy yz = (b) z = ax by + 3 ab + 2 a + b ) 2
−
(d) (x + y ,ln x z = 2
) 0
2
+
(c) z = ax by + 3 ab + 3 ( 2 a + ) b
+
+
=
[ESE 2018 (EE)] (d) z ax by c
248. The solution at x = 1, t = 1 of the [ESE 2017 (COMMON PAPER)]
partial differential equation
2 u = 25 2 u subject to initial 251. The solution of the following partial
2
2
x 2 t 2 differential equation u = 9 u is
2
2
conditions of u(0) = 3x and x y
u 3 (a) sin (3x − ) y (b) 3x + y
2
2
t ( ) 0 = is ______
2
(a) 1 (b) 2 (c) sin (3x − 3y ) (d) ( 3y − x 2 )
© 4 (d) 6 [ESE 2017 (COMMON PAPER)]
[GATE-2018 (CE-MORNING SESSION)]
123
Differential Equations & Partial Differential Equations
CLASSIFICATION OF P.D.E ( n t cosn )
2
−
u
, x
(b) ( ) t = A e x with
n
252. Consider the following partial n= 0
2
differential equation: A = f x
( )cosnx dx
n
2 2 2 0
0
3 + B + 3 + 4 = (c)
x 2 x y y 2
− 2n+ 1 2
For this equation to be classified as u ( ) t = A e 2 sin 2n + 1 x
, x
n
2
2
parabolic, the value of B must be with n= 0
_____ [GATE 2017] 2 2n + 1
A = f x x dx
( )sin
253. The type of partial differential n 0 2
equation
2 P + 2 P + 3 2 P + 2 P − P = 0 (d)
x 2 y 2 x y u ( ) t = , x A n exp − ( n t ) sin nx
2
x y
( )
is n= 1
2
nx
(a) elliptic (b) parabolic with A = 0 f ( )sinx ( )dx
n
(c) hyperbolic (d) none of these
255. If u = (x, t) is such that
[GATE-2016-CE-SET 1] 2 u = 4 2 u , 0 x , t ,
0
t 2 x 2
254. The solution of the initial boundary
u 2 u u ( ) u ) 0
( ,t =
0,t =
value problem = 0 x
t x 2
, t 0 with boundary and initial u x ) 0; u ( ,0 = x ) sin x then
( ,0 =
conditions t
u ( ) 0 u ), 0
0,t = =
,
x ( ,t t and u 3 6 is ________
( ,0 =
u x ) f x
( ), 0 x is ____
3 3
(a) (b)
(a) 4 8
− − 2n+ 1 2 t
)
u ( , x t = A e 2 cos 2n + 1 x (c) 3 (d) 3
n
n= 0 2 4 8
with
2 2n + 1
A = f ( )cosx x dx
n
0 2
124
Differential Equations & Partial Differential Equations
256. The number of boundary conditions (d) The solutions are not dependent
required to solve the differential on the boundary conditions.
2 2
equation + is [GATE-2016, 2 MARKS]
x 2 y 2
259. Solution of Laplace’s equation
(a) 2 (b) 0 having continuous second-order
partial derivatives are called
(c) 4 (d) 1
(a) biharmonic functions
[GATE-2001 (CE)]
(b) harmonic functions
257. The solution of the partial differential
u 2 u (c) conjugate harmonic functions
equation = is of the form
t x 2
(d) error functions
( k / ) x − ( k / ) x
( ) C e
(a) cos kt 1 + C e [GATE-2016; 2 MARKS]
C
2
260. Solution of the differential equation
dy
−
( k / ) x − ( k / ) x = e x y (e − x e y ) is
(b) Ce kt C e + C e dx
1
2
x
y e
x
(a) e e = e e x (e − ) 1 + C
(c)
kt ( ) ( ) y x x e x
−
+
Ce C 1 cos / k x C 2 sin − / k x (b) e = e − e + C
x
y
(d) (c) e = (e − ) 1 + ce − e x + C
(
)
(d) ye = e e − 1 +
x
x
ex
e
+
C sin kt ( ) C 1 cos ( / k ) x C 2 sin − ( / k ) x C
[GATE-2016-CE-SET 1] 261. Solution of the equation
dy 1 1
y
258. Which one of the following is a + tan y = 2 tan sin y is
property of the solutions to the dx x x
Laplace equation: 2 f = 0 ? (a) 2x = sin y (1 2cx+ 2 )
(a) The solutions have neither 2
maxima nor minima anywhere except (b) 2x = sin y (1 cx+ )
at the boundaries.
(c) 2x + sin y (1 cx+ 2 ) 0=
(b) The solutions are not separable in
the coordinates. (d) x + 2sin y (1 2cx+ 2 ) 0=
(c) The solutions are not continuous.
125
Differential Equations & Partial Differential Equations
262. The curve for which the area of the 266. In which of the following differential
triangle formed by the x-axis, the equation degree is not defined?
tangent lie at any point P and line OP 2
2
2
2
is equal to a is (a) d y + 3 dy = x log d y
dx 2 dx dx 2
a 2
2
−
(a) x cy= + (b) y = x cx 2 2 2
y d y dy 2 d y
(b) dx 2 + dx = x sin dx 2
a 2 a 2
=
=
(c) y cx + (d) x cy −
x y dy
(c) x = sin dx − 2y , x 1
263. For the differential equation whose
)
2
2
solution is (x h + ( y k ) = a dy
−
−
2
(d) x − 2y = log
(a is a constant), its dx
(a) order is 2 (b) order is 3 267. y = ae − 1/ x + b is a solution of
(c) degree is 2 (d) degree is 3 dy = y , then
dx x 2
+
dy ax h
264. The solution of =
+
dx by k (a) a R − 0
represents a parabola when
0
(b) b =
(a) a = 0, b 0 (b) a 0, b 0
(c) b = 1
(c) b = 0, a 0 (d) a = 0, b R
(d) a takes finite number of values
265. The solution of the differential 268. Which of the following equation(s) is
equation ( x y − ) 1 dy + 2xy dx = / are linear?
2
2
3
0
is dy y
(a) + = log x
(a) 1 x y+ 2 2 = cx dx x
dy
+
(b) 1 x y = 2 2 cy (b) y dx + 4x = 0
0
(c) y = dy
(c) (2x + y 3 ) = 3y
1 dx
(d) y = −
x 2 (d) None of these
126
Differential Equations & Partial Differential Equations
: →
269. The equation of the curve satisfying 272. Let y R R be a solution of the
the differential equation d y
2
=
−
x
QDE, − y e , x R ,
dy 2 dy dx 2
y + (x − ) y − x = 0 can be y ( ) 0 = y 1 ( ) 0 = 0 . Then which of the
dx dx
a following are true?
(a) y attains its minimum on R.
(a) Circle (b) Straight line
(b) y is bounded on R.
(c) Parabola (d) Ellipse 1
( )
−
x
(c) lime y x = .
270. For the boundary value problem, x→ 4
y + 11 y = 0 , (d) lime y x = 1
( )
x
)
)
( ), y −
1
( y = y 1 ( = y x→− 4
−
( ) .
273. Consider the Lagrange equation
To each eigen value , there z z
corresponds x 2 + y 2 = (x + ) y z . Then the
x y
(a) only one eigen function general solution of the given equation
is
(b) two eigen functions xy x − y
(a) F , = 0 for an
(c) two linearly independent eigen z z
functions arbitrary differentiable function F.
(d) two orthogonal eigen functions x − y 1 1
(b) F , − = 0 for an
2
d y z x y
271. Let − q ( ) x y = 0, 0 x , arbitrary differentiable function F.
dx 2
dy 1 1
y ( ) 0 = 1, ( ) 0 = 1, where q(x) is (c) z = f − for an arbitrary
dx x y
monotonically increasing continuous differentiable function f.
function. Then, 1 1
(d) z = xy f − for an
y
(a) ( ) x → as x → x y
arbitrary differentiable function f.
dy
(b) → as x →
dx 274. Which of the following are complete
integral of the partial differential
(c) y(x) has finitely many zeros in 2
)
0, equation pqx + yq = 1.
(d) y(x) has infinitely many zeros in
)
0, .
127
Differential Equations & Partial Differential Equations
x ay (a) The value of u(2, 2) = -1
(a) z = + + b
a x
(b) The value of u(2, 2) = 1
x ay
(b) z = + + b 1 1 1
b x (c) The value of u , =
2 2 2
2
b
(c) z = ( 4 ax + ) y + 1 1 1
(d) The value of u , =
2
−
(d) (z b ) = ( 4 ax + ) y 2 2 2
Partial Differential Equations
275. For an arbitrary continuously
differentiable function f, which of the 277. The one dimensional heat conduction
following is a general solution of partial differential equation
)
−
( z px qy = y − x 2
2
2
T = T , is
2
2
2
(a) x + y + z = f xy t x 2
( )
(a) parabolic (b) hyperbolic
(b) (x + ) y 2 + z = f xy
2
( )
(c) elliptic (d) mixed
2
2
2
(c) x + y + z = ( f y x− )
[GATE – 1996]
(d) x + y + z = f ( ( x + ) y 2 + z 2 ) 278. The one dimensional heat conduction
2
2
2
partial differential equation
, x
276. Let ( ) t be the solution of initial T = 2 T is
t x 2
boundary value problem
2 u 2 u (a) parabolic (b) hyperbolic
0
, t
= , 0 x ;
t 2 x 2 (c) elliptic (d) mixed
x [GATE – 1996 (ME)]
u ( ,0x ) cos= , 0 x ;
2 279. The number of boundary conditions
required to solve the differential
u
x
( ,0 = 2 2
) 0, 0 x ,
t equation + is
x 2 y 2
u ( ) 0t = , t 0.
0,
x (a) 2 (b) 0
(c) 4 (d) 1
128
Differential Equations & Partial Differential Equations
[GATE – 2001 (ME)] u u 2 u
+ u = 2 is a
280. The partial differential equation t x x
2 + 2 + + = 0 has (a) linear equation of order 2
x 2 y 2 x y
(b) non-linear equation of order 1
(a) degree 1 and order 2
(c) linear equation of order 1
(b) degree 1 and order 1
(d) non-linear equation of order 2
(c) degree 1 and order 1
[GATE – 2013 - ME]
(d) degree 2 and order 2
284. The type of partial differential
[GATE – 2007 (ME)] equation
2 P 2 P 2 P P P
281. The partial differential equation that 2 + 2 + 3 + 2 − = 0
+
+
=
can be formed from z ax by ab x y xdy x y
is
has the form
(a) elliptic (b) parabolic
z z
with p = and q =
x y (c) hyperbolic (d) none of these
+
(a) z = px qy [GATE – 2016 – CE – SET-1]
(b) z = px + pq 285. The solution of the partial differential
u 2 u
+
(c) z = px qy + pq equation t = x 2 is of the form
=
(d) z qy + pq ( k / ) x − ( k / ) x
kt
(a) cosC ( ) C e + C e
1
2
[GATE – 2010 - CE]
282. The type of the partial differential (b) Ce kt ( k / ) x − ( k / ) x
f 2 f C e + C e
1
2
equation = is
t x 2
(c)
(a) parabolic (b) elliptic Ce kt C 1 cos ( / k ) x C 2 sin − ( / k ) x
+
C
( )[C
(c) hyperbolic (d) nonlinear (d) sin kt 1 cos ( / k ) x +
[GATE – 2013 (IN)] ( )
C 2 sin − / k ] x
283. The partial differential equation
[GATE – 2016 – CE – SET-1]
129
Differential Equations & Partial Differential Equations
286. Which one of the following is a 289. Consider the following partial
property of the solutions to the differential equation u(x, y) with the
Laplace equation: 2 f = 0 ? constant c > 1:
u u
(a) The solutions have neither y + c x = 0
maxima nor minima anywhere except
at the boundaries. Solution of this equation is
(b) The solutions are not separable in (a) ( ,u x y = ) ( f x cy+ )
the coordinates.
)
−
u
(b) ( , x y = ) ( f x cy
(c) The solutions are not continuous.
u
(c) ( , x y = ) ( f cx + ) y
(d) The solutions are not dependent
on the boundary conditions. (d) ( , x y = ) ( f cx − ) y
u
[GATE – 2016, 2 MARKS]
[GATE – 2017 – ME – SESSION-1]
287. Solution of Laplace’s equation 290. The complete integral of
having continuous second-order 3 2
−
2
partial derivatives are called (z − px qy ) = pq + ( 2 p + ) q is
(a) biharmonic functions 2
+
2
(a) z = ax by + 3 pq + ( 2 p + ) q
(b) harmonic functions
+
(c) conjugate harmonic functions (b) z = ax by + 3 ab + ( 2 a + ) b 2
2
(d) error functions
(c) z = ax by + 3 ab + 3 ( 2 a + ) b 2
+
2
[GATE – 2016; 2 MARKS]
=
+
+
288. Consider the following partial (d) z ax by c
differential equation
[ESE – 2017 (Common Paper)]
2 2 2
0
3 + B + 3 + 4 = 291. The solution of the following partial
x 2 x y y 2 2 u 2 u
differential equation 2 = 9 2 is
For the equation to be classified as x y
2
parabolic, the value of B must be (a) sin (3x − (b) 3x + 2
2
____________. ) y y
2
[GATE – 2017 – CE – SESSION-1] (c) sin (3x − 3y ) (d) (3y − x 2 )
[ESE – 2017 (Common Paper)]
130
Differential Equations & Partial Differential Equations
292. The general integral of the partial 295. If ‘a’ and ‘b’ are arbitrary constants
differential equation then the partial differential equation
2
−
y p xyq = ( x z − 2y ) is corresponding to the equation
+
z = ax by a + b is _________
+
2
2
2
2
(a) ( x + y 2 , y − yz =
) 0
296. If ‘a’ and ‘b’ are arbitrary constants
2
(b) ( x 2 − y 2 , y + yz ) 0= then the partial differential equation
corresponding to the equation
,
) 0
(c) ( xy yz = z = xy + y ( x − a 2 ) + b is
2
2
−
) 0
(d) (x + y ,ln x z =
297. Form a partial differential equation
[ESE – 2018 (EE)] by eliminating the arbitrary function
from the relation
293. The solution at x = 1, t = 1 of the 2
partial differential equation z = y + 2 f (1/ x + log ) y .
2 u = 25 2 u
x 2 t 2 subject to initial 298. Form a partial differential equation
conditions of ( ) 0 = 3x and by eliminating the arbitrary function
u
from the relation
u ( ) 0 = is _____________. ( f x + y 2 ,z − xy =
2
) 0 .
3
t
−
(a) 1 (b) 2 299. The solution of p q = log (x + ) y .
(c) 4 (d) 6 300. The solution of
)
−
y x
[GATE – 2018 – CE – MORNING (z − ) y p + (x z q = −
SESSION]
301. The solution of q = 3p is
2
294. Consider a function it which depends
on position x and time t. The partial 302. The solution of q = 2 2 2 − 2 )
differential equation z p (1 p is
u 2 u 303. The solution of p + q = + y is
2
x
2
= is known as the
t x 2
304. The solution of
)(
(a) Wave equation ( p q z − px qy =
−
−
) 1 is
(b) Heat equation
u 2 u
(c) Laplace’s equation 305. The solution of the PDE =
t x 2
(d) Elasticity equation is of the form
[GATE – 2018 – ME – AFTERNOON
SESSION]
131
Differential Equations & Partial Differential Equations
k x − k x 2 y 2 y
(a) cos kt C ( ) C e + C e (d) t 2 = c 2 x 2
2
1
309. The number of arbitrary constants in
k x − k x the general solution of one
(b) Ce kt C e + C e
1 2 dimensional wave equation is
310. Which of the following differential
(c) equations is parabolic?
kt k k
+
Ce C cos x C sin − x
1 2 (a) one dimensional wave equation
(b) one dimensional heat equation
(d) (c) Laplace equation
k k
+
C sin kt C cos x C sin − x (d) Poisson’s equation
1 2
311. The solution of the equation
2 y 2 y
u u = 4 , representing the
306. The solution of 3 + 2 = 0, t 2 x 2
x y vibrations of a string of length 5
( ,0 =
x
−
u x ) 4e subject to the conditions,
y
0,t =
5,t =
2 u 2 u y ( ) 0; y ( ) 0; t = 0
)
307. The equation + = f ( ,x y
x 2 y 2 when t = 0 and
( ,0 =
is y x ) sin2 x − 2sin5 x is
(a) Elliptic (b) Parabolic 2 u 2 u
312. The solution of = with
(c) Hyperbolic (d) circular t 2 x 2
0,t =
1,t =
u x
u ( ) u ( ) 0; ( ,0 = and
) 0
308. The one dimensional wave equation
is u ( ,0 =
x
) u (u is a constant) is
t 0 0
2 u 2 u
(a) + = 0
x 2 y 2 313. A tightly stretched string with fixed
end points x = 0 and x l = is initially
2 u 2 u in a position given by
)
(b) + = f ( ,x y
x 2 y 2 x
y = y 0 sin 3 l . If it is released
u 2 u
(c) = c 2
t x 2
132
Differential Equations & Partial Differential Equations
from rest from this position then the
y
, x
displacement ( ) t is.
u 2 u
314. The solution of = c 2 ,
t x 2
0 x , t 0 under the boundary
u
0,t =
) 0
u
conditions ( ) 0, ( ,t =
and ( ,0 =
u x
) sin x . Where
0 x is
, x
u
315. What is the temperature ( ) t in a
laterally insulated copper bar 80 cm
long if the initial temperature is
x 0
100sin C and the ends are
80
0
kept at 0 C ?
2 u 2 u
316. The solution of + = 0 with
x 2 y 2
the conditions
)
u (0, y = y ( , l y = ( ,0 = and
) u x
) 0
n x
u ( , x a = is.
) sin
l
2 u 2 u
317. The solution of + = 0 with
x 2 y 2
) 0
u
the boundary conditions (0, y = ,
=
) 0 u
u ( , l y = , ( , x ) 0 and
u ( ,0x ) u= 0 is (where 0 x ls ).
133
5
Complex
Variables
Complex Variables
1. Let denote the boundary of the e 2
square whose sides lie along x = 1 4. The value of (z + ) 1 4 dz is
z =
and y = 1, where is described in 2
the positive sense. Then the value of (a) 2 ie (b) 8 i e
−
−
1
2
z 2 dz is 3
2z + 3 2 i
−
2
(c) e (d) 0
i 3
(a) (b) 2 i
4
5. The conjugate (also called
−
+
(c) 0 (d) 2 i symmetric) point of 1 i with respect
to the circle z − 1 = 2 is
→
2. Let : f C C be given by
( ) 2 (a) 1 i − (b) 1 4i
+
f ( ) z = z / z when Z 0; .
+ (d) 1 i −
−
0 when Z = 0 (c) 1 2i
Then f 6. The harmonic conjugate of
)
2
2
u ( , x y = x − y + xy is
(a) is not continuous at Z = 0
(b) is differentiable but not analytic at (a) x − 2 y + 2 xy
Z = 0
(b) x − 2 y − 2 xy
(c) is analytic at Z = 0
1
(d) satisfied the Cauchy – Riemann (c) 2xy + ( y − 2 x 2 )
equations at Z = 0 2
3. The function (d) xy + 1 ( 2 y − 2 x 2 )
( ) 2 2 2
f ( ) z = z / z if z 0
7. The singularity of e sin z at z = is
0 if z = 0
(a) a pole
(a) satisfied the Cauchy – Riemann
equations at z = 0 (b) a removable singularity
(b) is not continuous at z = 0 (c) nonisolated essential singularity
(c) is differentiable at z = 0 (d) isolated essential singularity
(d) is analytic at z = 1
135
Complex Variables
8. The value of the integral 12. The fixed points of ( ) z = 2iz + 5
f
2
sin z + cos z 2 dz where C is z − 2i
C (z − 4 )(z − ) 2 are
the circle z = 3 traced anti- (a) 1 i (b) 1 2i
clockwise, is
(c) 2i 1 (d) i 1
−
(a) 2 i (b) i
2
f
13. The function ( ) z = z is
−
(c) i (d) 2i
(a) differentiable everywhere
z − sin z
f
9. For the function ( ) z = , (b) differentiable only at the origin
z 3
the point z = 0 is (c) not differentiable anywhere
(a) a pole of order 3 (d) differentiable on real x – axis
(b) a pole of order 2 14. The function ( ) z = z maps the
2
f
(c) an essential singularity first quadrant onto
(d) a removable singularity (a) itself
−
1 e − z (b) upper half plane
f
10. For the function ( ) z = , the
z (c) third quadrant
point z = 0 is
(d) right half plane
(a) an essential singularity
15. The radius of convergence of the
(b) a pole of order zero power series of the function
(c) a pole of order one f ( ) z = 1 about z = 1 is
−
(d) a removable 1 z 4
singularity 1
(a) 1 (b) 4
11. The value of the integral dz ,
C z − 1 3
2
4
C : z = is equal to (c) (d) 0
4
(a) i (b) 0
(c) − i (d) 2 i
136
Complex Variables
16. Let T be any circle enclosing the 20. Let ( ) z be an analytic function
f
origin and oriented counter- with a simple pole at z = 1 and a
clockwise. Then the value of the double pole at z = 2 with residues 1
integral cos z dz is and -2 respectively. Further if
z 2 3
f ( ) 0 = , ( ) 3 = − and f is
0 f
(a) 2 i (b) 0 4
bounded as z → , then f(z) must be
−
(c) 2 i (d) undefined
1
(a) ( z z − ) 3 − + 1 − 2 + 1
1 4 z − 1 z − 1 (z − ) 2 2
f z =
17. For the function ( ) sin , z = 0
z
1
is a (b) − + 1 − 2 + 1
4 z − 1 z − 2 (z − ) 2 2
(a) removable singularity
1 2 5
(b) simple pole (c) − +
z − 1 z − 2 ( z − ) 2 2
(c) branch point
15 + 1 + 2 − 7
(d) essential singularity (d) 4 z − 1 z − 2 (z − ) 2 2
2
f
18. A z = 0, the function ( ) z = z z 21. An example of a function with a non-
isolated essential singularity at z = 2
(a) does not satisfy Cauchy – is
Riemann equations
1 1
(b) satisfied Cauchy – Reimann (a) tan (b) sin
equations but is not differentiable z − 2 z − 2
z − 2
(c) is differentiable (c) e − (z− ) 2 (d) tan
z
(d) is analytic
)
22. Let ( ) z = f u ( ,x y ) i+ ( ,x y be an
19. The bilinear transformation , which
maps the points 0, 1, in the z-plane entire function having Taylor’s series
onto the points , , i − 1 in the − expansion as a z . If
n
n
plane is n= 0
f ( ) x = u x )
( ,0 and
−
z − 1 z i
)
( ) i=
(a) (b) f iy (0, y then
+
z i z + 1
0
(a) a = for all n
+
z i z + 1 2n
(c) (d)
−
z − 1 z i (b) a = a = a = a = 0, a
0
4
1
2
0
3
137
Complex Variables
(c) a 2n+ 1 = 0 for all n 26. In the Laurent series expansion of
1 1
0
0
(d) a but a = f ( ) z = z − 1 − z − 2 valid in the
0
2
1
2
23. Let I I = cot ( 2 ) z dz , where C is region z , the coefficient of z 2
c (z i − ) is
the contour 4x + 2 y = 2 2 (counter
(a) -1 (b) 0
clock-wise). Then I is equal to
(c) 1 (d) 2
(a) 0
27. Let = f ( ) z be the bilinear
−
(b) 2 i
transformation that maps -1, 0 and 1
1 to -i, 1 and i respectively. Then
(c) 2 i 2 − f (1 i − ) equals
sinh
+
−
2
2 i (a) 1 2i (b) 2i
(d) −
−
sinh (c) 2 i + (d) 1 i
+
−
2
24. The real part of the principal value of 28. For the positively oriented unit circle,
−
4 4 i is 2Re ( ) z
dz =
(a) 256 cos (ln 4) z = 1 z + 2
(b) 64 cos (ln 4) (a) 0 (b) i
(c) 16 cos (ln 4) (c) 2 i (d) 4 i
(d) 4 cos (ln 4) 29. The number of zeroes, counting
multiplicities, of the polynomial
5
3
2
25. If sin z = a n (z − / ) 4 n , then a z + 3z + z + 1 inside the circle
6
n= 0 z = 2 is
equals
1 (a) 0 (b) 2
(a) 0 (b)
720 (c) 3 (d) 5
1 − 1 30. f = u + i and g = i + be non-
(c) (d)
720 2 720 2 zero analytic functions on z 1.
Then it follows that
(a) ' 0f
138
Complex Variables
(b) f is conformal on z 1 (a) 0, i (b) 0, ,2 i i
(c) f kg for some real k (c) 0, i , 2 i (d) 0
(d) f is one to one 2
i
35. The value of exp (e − i ) d
1 0
31. The principal value of log i 4 is equals
(a) 2 i (b) 2
i
(a) i (b) (c) (d) i
2
36. The sum of the residues at all the
i i cot z
f
(c) (d) poles of ( ) z = 2 , where a is
+
4 8 (z a )
a constant, (a 0, 1, 2,...... ) is
32. Consider the functions
2
f ( ) z = x + 2 iy and 1 1
2
(a) − cosec a
+
2
2
g ( ) z = x + y + ixy . At z = 0, n=− (n a ) 2
(a) f is analytic but not g 1 1
2
(b) − 2 + cosec a
+
(b) g is analytic but not f n=− (n a )
(c) both f and g are analytic 1 1 2
(c) − − cosec a
n=− (n a ) 2
+
(d) neither f nor g is analytic
1 (d) 1 1 + cosec a
2
33. The coefficient of in the expansion ) 2
z n=− (n a+
z
of Log , valid in z 1, is 37. Which of the following is not the real
z − 1 part of an analytic function?
(a) -1 (b) 1 (a) x − y
2
2
1 1
(c) − (d) 1
2 2 (b)
1 x + 2 y 2
+
34. Let be a simple closed curve in the
(c) cos coshx y
complex. Then the set of all possible
values of dz is (d) x + x
2
z (1 z− 2 ) x + y 2
139
Complex Variables
→
38. The radius of convergence of 42. Define : f C C by
1 n 2 0, if Re z = ( ) 0 or Im z = ( ) 0
1+ f ( ) z =
n − − − − − z is , z otherwise
n
n= 0 n 3 . Then the set of points where f is
(a) e (b) 1/e analytic is
(c) 1 (d) (a) :Rez ( ) 0z and Im ( ) 0z
z
( ) 0
n
39. It is given that a z converges at (b) :Re z
n
n= 0
z = 3 i + 4. Then the radius of (c) :Re z z ( ) 0 or Im z ( ) 0
convergence of the power series
a z is (d) :Im z
z
( ) 0
n
n= 0 n
43. Let S be the positively oriented circle
(a) 5 (b) 5
given by z − 3i = 2. Then the value
(c) <5 (d) > 5 dz
of 2 is
f
40. Let ( ) z be an analytic function. s z + 4
Then the value of 2 f e it ( ) (a) − (b)
( )cos t dt
0 2 2
equals − i i
(c) (d)
(a) 0 (b) 2 f ( ) 0 2 2
(c) 2 f 1 ( ) 0 (d) f 1 ( ) 0 44. Let ( ) cos z − sin z for non-
f z =
z
1 zero z C and ( ) 0 = . Also, let
0
f
f
41. Let ( ) z = . Then the
z − 2 3z + 2 g z =
( ) sinh z for z C .
1
coefficient of in the Laurent
z 3 (i) Then f(z) has a zero at z = 0 of
f
2
series expansion of ( ) z for z order
is (a) 0 (b) 1
(a) 0 (b) 1 (c) 2 (d) greater than 2
(c) 3 (d) 5
140
Complex Variables
g ( ) z − 1
(ii) Then has a pole at z = 0 of (c) cos (1) (d) sin ( ) 1
zf ( ) z 2
order 48. Consider the function
(a) 1 (b) 2 f ( ) z = e iz
2
( z z + ) 1
(c) 3 (d) greater than 3
45. For the function (i) The residue of f at the isolated
1 singular point in the upper half plane
f z = , the point z z = + : y 0 is
( ) sin
x iy C
cos (1/ z )
= 0 is − 1 − 1
(a) (b)
(a) a removable singularity 2e e
(b) a pole (c) e (d) 1
2
(c) an essential singularity
(ii) The Cauchy Principal Value of
(d) a non-isolated singularity sin xdx
the integral 2 is
15
n
f
46. Let ( ) z = z for z C . If − ( x x + ) 1
n= 0 − −
+
−
−
C : z i − = 2 then f ( ) z dz = (a) 2 (1 2e 1 ) (b) (1 e 1 )
C (z i − ) 15
)
+
−
+
(c) 2 (1 e (d) (1 e − 1 )
)
+
(a) 2 i (1 15i
)
−
u
49. Let ( , x y = ) 2x (1 y for all real x
)
(b) 2 i (1 15i− ) and y. then a function ( , x y , so
)
(c) 4 i (1 15i+ ) that ( ) z = f u ( , x y + ) i ( , x y is
analytic, is
(d) 2 i
2
2
(a) x − 2 ( y − ) 1 (b) ( x − ) 1 − 2 y
n
47. Let a n (z + ) 1 be the Laurent
2
n=− (c) ( x − ) 1 + 2 y (d) x + 2 ( y − ) 1
2
series expansion of
z
f
f ( ) sinz = . Then a = 50. Let ( ) z be analytic o n
−
z + 1 2 D = z C : z − 1 1 such that
(a) 1 (b) 0 f ( ) 1 = 1. If ( ) z = f z 2
f
( ) for all
141
Complex Variables
z D, then which one of the 53. Which of the following is the
following statements is NOT correct? imaginary part of a possible value of
( )
i ?
f
(a) ( ) z = f ( ) z 2 for all z D
z 1
(b) f = f ( ) z for all z D (a) (b) 2
2 2
3
f z
(c) ( ) = f ( ) z 3 for all z D (c) (d)
8
4
→
(d) ( ) 1f 1 = 0 54. Let : f C C be analytic except for
a simple pole at z = 0 and let
( ) z
f
C
51. Let I = dz , where : g C → be analytic. Then, the
C (z − 1 )(z − ) 2 Re ( ) ( )
z z z= 0 f z g z
f z = ( ) sin + cos and C is the value of is
2 2 Re f ( ) z
z=
0
curve z = 3 oriented anti-clockwise.
Then the value of I is (a) g(0)
(a) 4 i (b) 0 (b) g’(0)
( )
(c) 2 i− (d) 4 i− (c) lim z f z
z→ 0
n
( ) ( )
52. Let b z be the Laurent series (d) lim z f z g z
n
n=− z→ 0
1
expansion of the function , z
a sinh z 55. Let ( ) z = , z = x iy
+
f
−
0 z . Then which one of the 8 z 3
following is correct?
( )
(i) Res f z is
1 7 z= 2
(a) b = 1, b = − , b =
−
0
2
6 2 360 1 1
(a) − (b)
1 7 8 8
(b) b = 1, b = − , b =
−
−
3
1
6 1 360 1 1
(c) − (d)
1 7 6 6
(c) b = 0, b = − , b =
−
0
2
6 2 360 56. The Cauchy principal value of
1 7 f ( ) x dx
(d) b = 1, b = − , b =
0 2 4 −
6 360
142
Complex Variables
The residue of ( ) z at its pole is
f
(a) − 3 (b) − 3
6 8 equal to 1. Then the value of is
−
(c) 3 (d) 3 (a) -1 (b) 1
(c) 2 (d) 3
0
:
57. The straight lines L x = ,
1
L 2 : y = and L x + : y = 1 are 60. For the value of obtained in above
0
g
problem, the function ( ) z is not
3
mapped by the transformation = z conformal at a point
2
into the curves C , C and C (1 3i ) (3 i + )
+
3
1
2
respectively. The angle of (a) 6 (b) 6
intersection between the curves at
= 0 is (c) 2 (d) i
3 2
(a) 0 (b) 61. The coefficient of (z ) in the
2
−
4
Taylor series expansion of
sin z
(c) (d) if z
z
2 f ( ) z = − around
=
58. Let − 1 if z
2 is
1 − ( − ) 2 + 4 dz = 4 1 − 1
(z − ) 2 4 z (a) 2 (b) 2
, where the close curve C is the 1 1
triangle having vertices at (c) (d) −
− i i − i i − 6 6
,
, i , the integral f
2 2 62. Let ( ) z be an entire function on C
being taken in anti-clockwise such that ( ) z 100log z for each
f
direction. Then one value of is
f i =
2
z with z . If ( ) 2i , then f(1)
(a) 1 i+ (b) 2 i+ must be
(c) 3 i+ (d) 4 i+ (a) 2
59. Consider the functions (b) 2i
z + z (c) i
2
f ( ) z = 2 ,
(z + ) 1 (d) cannot be determined
g ( ) sinhz = z − , 0.
2
143
Complex Variables
+
−
63. Let C be the contour z = oriented initial point 1 2i and final point
2
+
in the anti-clockwise direction. The 1 2i. The value of C 1 2z dz =
+
+
3/z
value of the integral C ze dz = 1 z
1
+
(a) 3 i (b) 5 i (a) 4 − ln2 i
2 4
(c) 7 i (d) 9 i 1
−
+
→
64. Let : f C / 3i C defined by (b) 4 + 2 ln2 i 4
z i −
f ( ) z = 1
iz + 3 (c) 4 + ln2 i −
2 4
Which of the following statements 1
+
about f is FALSE? (d) 4 − ln2 i
2 2
(a) f is formal on C/3i
68. If a C with a 1, then the value
(b) f maps circles in C/3i onto circles 2
−
in C of 1 a dz dz where is
z a 2
+
(c) All the fixed points of f in the
region z C :Im ( ) 0z the simple closed curve z = 1 taken
with the positive orientation is
(d) There is no straight line in C/3i
which is mapped onto a straight line 69. Let D = z C : z 1 . Then there
in C by f. exists a non-constant analytic
function f on D such that for all
2
f
65. The function ( ) z = z + iz + 1 is n 2
differentiable at
− 1
(a) i (b) 1 (a) f n = 0
(c) -i (d) no point in C
1
66. The radius of convergence of the (b) f n = 0
power series 4 ( ) 1 n z− n 2n is
n= 0 1
(c) f 1 − = 0
( ) 0 and let
67. Let = z C :Im z n
C be a smooth curve lying in with 1 1
(d) f − = 0
2 n
144
Complex Variables
n
70. Let a z be the Laurent series (d) Both f and g are analytic
n=−
n
1 n
expansion of ( ) z = 76. If a (z − ) 2 is the Laurent
f
2z − 2 13z + 15 − n
3 a series of the function
in the annulus z 5. Then 1 z + 4 z + 3 z 2
2 a f ( ) z = for z C / 2 .
2 (z − ) 2 3
is equal to
Then a equals.
−
2
i dz
71. The value of dz =
−
4 z = 4 z cos z 77. For n Z define
1 ( i n i− )x
72. Let C = z C : z = 2 . Then c = 2 − e dx , then
n
1 z 7 cos 1 c 2 .
2 i C z 2 dz n Z n
73. Let (a) cosh (b) sinh
)
3
2
+
2
3
u ( , x y = x + ax y bxy + 2y be (c) cosh 2 (d) sinh 2
)
a harmonic function and ( , x y its ( 2 / i )
−
−
harmonic conjugate. If (0,0 ) 1= , 78. The principal value of ( ) 1 is
+
+
then a b ( ) (a) e (b) e
1,1
2i
2
−
74. Let C be a simple positively oriented (c) e − 2i (d) e
2
circle of radius 2 centered at origin in
the complex plane. Then 79. In the Laurent series expansion of
1
2 C ze + 1/z tan + z 1 dz f ( ) z = valid for z − 1 1,
i 2 (z − 1 )(z − ) 3 2 ( z z − ) 1
1
the coefficient of is
z − 1
f z
75. Let ( ) ( x= 2 + y 2 ) 2ixy+ and (a) -2 (b) -1
2
g ( ) 2z = xy + ( i x − y 2 ) for z C . (c) 0 (d) 1
Then in the complex plane C
80. Let :f C C→ be an entire function
(a) f is analytic but g is not analytic with f(0) = 1, f(1) = 2 and f’(0) = 0. If
there exists M > 0 such that
(b) g is analytic but f is not analytic
f 11 ( ) z M for all z C , then f(2)
(c) neither f nor g is analytic
=
145
Complex Variables
(a) 2 (b) 5 the greatest integer less than or equal
to is
(c) 2 + 5i (d) 5 + 2i
→
81. Let be the circle given by z = 4e i 85. Let : f C C be non-zero and
, where varies from 0 to 2 . Then analytic at all points in Z. If
( ) z =
( )cot
f z
F
( ) z for
e
z − z 2z dz = z C / Z , then the residue of F at
2
n Z is
)
(
−
i
2
2
(a) 2 i e − ) 1 (b) (1 e
( )
f
(a) f n (b) ( ) n
2
−
(c) ( i e − ) 1 (d) 2 i (1 e 2 ) f ( ) n
' f
(c) (d) ( ) z z n
=
2
3 z
f
82. Let ( ) z = z e for z C and let
i
=
be the circle z e , where 86. Which of the following is the
varies from 0 to 4 . Then imaginary part of the possible value
i
1 f 1 ( ) z of ln ( )
2 i f ( ) z dz
(a) (b)
→
83. Let : f C C (the set of all complex 2
numbers) be defined by
)
2
3
3
f ( , x y = x + 3xy + ( i y + 3x 2 ) y . (c) 4 (d) 8
' f
Let ( ) z denote the derivative of f
f z =
+
87. If ( ) 11 iv is analytic then, the
with respect to z. Then which one of
the following statements is TRUE? harmonic conjugate of
u = x − y + xy is
2
2
(a) ( ' 1f + ) i exist and 2 2
f ( ' 1 i+ ) = 3 5 (a) x − y − xy
(b) x + 2 y − 2 xy
(b) f is analytic at the origin
(c) f is not differentiable at I (c) 2xy + 1 ( y − 2 x 2 )
2
(d) f is differentiable at 1
xy + ( 2 y − 2 x 2 )
(d)
i z
84. Let = C 2z − 2 e 5z + 2 dz , 2
+
C :cost i sint , 0 t 2 . Then
146
Complex Variables
−
−
f
88. An analytic function ( ) z is such (d) 1,1 2 ,1 2 − 2
=
z
Re
that ( f 1 ( )) 2y and
4
2
2 0
93. The roots of z − z − 2z + = are
f (1 i+ ) 2= then imaginary part of
(a) 1
f ( ) z =
(b) 1 i−
2
2
(a) 2xy− (b) x − y
(c) both (a) & (b)
2
2
(c) 2xy (d) y − x (d) neither (a) nor (b)
2
f
89. The function ( ) z = z maps first 94. Consider the functions
quadrant onto ___ f ( ) z = x + 2 iy and
2
(a) itself (b) upper half
plane (c) third quadrant (d) right g ( ) z = x + 2 ixy at z = 0
half plane
)
−
90. Let u = 2x (1 y for real x and y (a) f is analytic, but not g
)
v
then a function ( , x y so that
(b) g is analytic but not f
+
f ( ) z = u iv is analytic
2
2
(a) x − 2 ( y − ) 1 (b) ( x − ) 1 + 2 y (c) both f and g are analytic
2
2
(c) ( x − ) 1 − 2 y (d) x + 2 ( y − ) 1 (d) neither f nor g is analytic
− −
91. If z − 1 = 2 , then zz z z = z
95. Lt is
z→ 0 z
(a) 1 (b) 2
(a) 0
(c) 3 (d) 4
(b) 1
2
92. If 1, , are cube roots of units,
3
then the roots of (x − ) 1 + = are (c) 1
8 0
2
(a) 1, 1, 1− − −
(d) does not exists
(b) 1, ,2
(c) 1 1 2 ,1 2− + + + 2
147
Complex Variables
96. Which of the following is not c 1
harmonic 100. The value of z a dz where ‘c’ is
−
−
(a) u = sinh y cos y the circle z a = r is
(a) 0 (b) 2 i
1
(b) u = log ( x + 2 y 2 ) (c) 2 (d) i
2
101. The value of c zdz from z = 0 to
2
(c) u = x + 2 y z = 4 2i along the curve ‘c’ given
+
=
by z t + 2 it
2
(d) u = x − 2 y 8i 8
(a) 10 − (b) 10i +
97. The function ( ) secf z = z is 3 3
8
(a) analytic for all ‘z’ (c) 10 − (d) 0
3i
=
(b) analytic for z sin z
102. The residue of ( ) z = at z = 0
f
z 8
(c) not analytic at z = is
2
(a) 0 (b) − 1
(d) None 7!
1
+
f
98. If z = , ( ) z = u iv is analytic (c) (d) None
2 7!
sin2x f z =
and u = then 103. The residue of ( ) cot z at any
cosh2y + cos2x one its poles is
f ( ) z =
(a) 0 (b) 1
(a) tan z c+ (b) secz c+
(c) 3 (d) none
+
+
(c) cot z c (d) sin z c
104. Let = e i /10 , then residue of
−
2
99. Let U V = (x − y )( x + 4xy + y 2 ) 1
=
f ( ) z = at z is
+
+
and ( ) z = u iv is analytic then 1 z 10
f
f ( ) z in terms of z is
(a) − (b)
10 10
2
c
(a) iz− 3 + c (b) z +
i − i
2
(c) z + 1 (d) iz (c) 5 (d) 5
148
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