Page 89 - Engineering Mathematics Workbook_Final

Differential Equations & Partial Differential Equations

+
(c) e x y is an integrating factor of the (a) y 2 cos x + x sin y = 0
differential equation 
+
(b) y 2 sin x x cos y =
(d) A suitable substitution transforms 2
the differentiable equation to the (c) y 2 sin x + x sin y = 0
variables separable form 
(d) y 2 cos x + x cos y =
[JAM MA 2010] 2
30. Consider the differential equation
11
27. If y is an integrating factor of the dy
=

differential equation dx − 2x  ( ) x , x R , satisfying
2xy dx − (3x − y 2 ) dy = , then the  0, x  0
2
0


0
value of a is y ( ) 0 = , where ( ) x =  1, x  0


(a) -4 (b) 4 . This initial value problem
(c) -1 (d) 1 (a) has a continuous solution which is
0
[JAM MA 2011] not differentiable at x =
28. The solution of the differential (b) has a continuous solution which is
equation differentiable at x =
0
dy = − ( x x + 2 y − 2 10 ) , y ( ) 0 = 1 is (c) has a continuous solution which is
dx ( y x + 2 y + 2 ) 5 differentiable on R

(a) (d) does not have a continuous
4
=
2
2
2
x − 2x y − y − 20x − 10y + 11 0 solution on R.
2
4
(b)
2
4
2
=
x + 2x y + y + 20x + 10y − 11 0 31. If ( ) x satisfies dy + 2y = 2 e − x 2
4
2
2
+
y
(c) dx
( )
y
0
x + 2x y − y + 20x − 10y + 11 0 with ( ) 0 = , then lim y x equals
=
2
4
4
2
2
2
x→ 
(d)
=
x + 2x y + y − 20x + 10y − 11 0 (a) 0 (b) 1
2
4
4
2
2
2

(c) 2 (d) -1
[JAM MS 2008]
[JAM GP 2008]
29. The solution of the differential
equation 32. The solution of the initial value
dy = y 2 cos x + cos y ; y      = 0 problem dy = sin x , y ( ) 0 = 0 is
y
dx x sin y − 2 sin x   2     dx y + 2
is (a) ( y y + ) 2 = ( 4 1 cos x− )

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