VITEEE 2012 Paper

dy

dx

+

2yx

1+x2

=

1

(1+x2)2

which is a linear differential equation
Here, P =

2x

1+x2

, Q =

1

(1+x2)2

Now, IF = e∫Pdx
= e∫

2x

1+x2

dx = elog(1+x2) = (1+x2)
∴ Solution of differential equation is
y . (1+x2) = ∫

1

(1+x2)2

. (1 + x2) dx + C
⇒ y (1 + x2) = ∫

1

1+x2

dx + C
⇒ y (1 + x2) = C + tan−1 x