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WBJEE 2016 Math Solved Paper
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© examsnet.com
Question : 56
Total: 75
General solution of
y
d
y
d
x
+
b
y
2
= a cos x, 0 < x < 1
Here c is any arbitrary constant
y
2
= 2a(2b sinx + cosx) +
c
e
−
2
b
x
(4
b
2
+ 1)
y
2
= 2a (sinx + 2bcosx) +
c
e
−
2
b
x
(4
b
2
+ 1)
y
2
= 2a (sinx + 2bcosx) +
c
e
2
b
x
y
2
= 2a(2b sinx + cosx) -
c
e
−
2
b
x
Validate
Solution:
Let
y
2
= z
y
d
y
d
x
=
1
2
d
z
d
x
d
z
d
x
+ 2bz = 2a cosx
IF =
e
2
b
∫
d
x
=
e
2
b
x
z
.
e
2
b
x
=
∫
2
a
c
o
s
x
.
e
2
b
x
dx
y
2
e
2
b
x
=
2
a
4
b
2
+
1
(sin x + 2b cos x)
e
2
b
x
+ c
(4
b
2
+ 1)
y
2
= 2a (sinx + 2bcosx) +
c
e
−
2
b
x
© examsnet.com
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