Page 110 - Engineering Mathematics Workbook_Final

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

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