Sol. We have,

+ x sin 2y = x 3 cos 2 y

Dividing both sides by cos 2 y, we get

sec 2 y + x = x 3 sec

2 y + 2x tan y = x 3 Putting tan y = v and sec

2 y = , we get

+ (2x) v = x 3 .. (i)

This is a linear differential equation of the form

+ Pv = Q, where P = 2x and Q = x 3 Now, I.F. = = =

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

+ 2x v. = . x

3 Integrating both sides with respect to x, we get

v =

[Using: v (I.F.) = ]

⇒ v = + C, where t = x

2 ⇒ v =

⇒ v = (te

t –e t ) + C

⇒ v = (t –1) e t + C

⇒ v = (x 2 –1) + C [  t= x 2 ]

⇒ tan y = (x 2 –1) + C [  tan y = v]

This is the required solution of the given differential equation.