Find the particular solution of the differential equation y ≠ 0 given that x = 0 when

Given

This is a first order linear differential equation of the form

Here, P = cot y and Q = 2y + y²cot y

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin y [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ x sin y = y² sin y + c

∴ x = y² + c cosec y

However, when, we have x = 0.

By substituting the value of c in the equation for x, we get

Thus, the solution of the given differential equation is