Page 96 - Engineering Mathematics Workbook_Final

Differential Equations & Partial Differential Equations

65. The solution of the equation 68. The solution of the equation
dQ + Q = 1 with Q = at t = 0 is x dy + y = 0 passing through the
0
dt dx
point (1, 1) is
Q
(a) ( ) t = e − t − 1
2
(a) x (b) x
t −
Q t =
+
(b) ( ) 1 e
−
1
2
−
(c) x (d) x
Q t =
−
t
(c) ( ) 1 e [GATE-2018 (CE – Afternoon Session]
t −
Q t =
−
(d) ( ) 1 e 69. If y is the solution of the differential
dy
equation y 3 + x = 3 0, ( ) 0 = ,
1
y
[GATE-2017-CE-SECTION-1] dx
y −
66. For the initial value problem the value of ( ) 1 is
−
−
dx (a) 2 (b) 1
( ) ( ) 0 =
,
= sin t x 0
dt
(c) 0 (d) 1
the value of x at t =  , is [GATE-2018 (ME-Afternoon Session]
3
__________. 70. A curve passes through the point
( x = 1, y = ) 0 and satisfies the
[GATE-2017-(CH)]
differential equation
67. The solution of the differential dy x + y 2 y
2
equation = + .
dx 2y x
−
2
−
2
y 1 x dy + x 1 y dx = is
0
The equation that describes the curve
is
−
(a) 1 x = 2 c  y 2 


(a) ln 1+ 2    = x − 1

−
(b) 1 y = 2 c  x 
1   y 2  

−
2
c

(c) 1 x− 2 + 1 y = (b) ln 1+ 2   = x − 1
2  x 
2
+
c
(d) 1 x+ 2 + 1 y =   y  
(c) ln 1+    = x − 1

 x 
[ESE-2017 (EE)] 1  y 

(d) ln 1+     = x − 1

2  x 
[GATE-2018-(EC)]

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