Sol. We have,

= x 3 y 3 –xy

⇒ + xy = x 3 y 3 This is the form of Bernouli's differential equation.

Dividing both sides by y

3 , we get

+ x = x 3 Let = v. Then,

=

⇒ = –

Substituting these in the given differential equations we get

– + x. v = x

3 ⇒ + (–2x) v = –2x

3 ... (i)

This is a linear differential equation of the form,

+Pv = Q, where P = –2x and Q = –2x 3 We have, I.F. = = =

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

+ (–2x) v = –2x

3

Integrating both sides with respect to x, we get

v = –2x 3 dx + C

⇒ + C, where t = –x 2 ⇒

⇒ [te

t –e t ] + C

⇒ e t (t –1) + C

⇒ (–x 2 –1) + C [  t = – 2 ]

⇒ (x 2 + 1) + C

⇒ = (x 2 + 1) + C [  v = 1/y 2 ]

This is the required solution of the given differential equation.