Sol. We have,

3x (1 –x 2 ) y 2 + (2x 2 –1) y 3 = ax 3 Putting y

3 = v and 3y 2 = we get

3x (1 –x 2 ) + (2x 2 –1) v = ax 3 ⇒ x (1 –x

2 ) + (2x 2 –1) v = ax 3 ⇒ + v = ... (i)

This is a linear differential equation of the form

+ Pv = Q, where P =

and Q =

Now,

I.F. =

=

= [Using Partial fractions]

=

=

=

= =

Multiplying both sides of (i) by I.F. = , we get

+ v =

Integrating both sides w.r.t. x, we get

v. = dx + C

⇒ v. = dt + C, where t = 1 –x

2 ⇒ v. = – + C

⇒ v. = + C

⇒ v. = + C [  t = 1 –x

2 ]

⇒ y 3 = ax + C. x [  v = y 3 ]

This is the required solution of the given differential equation.