,

Given and

This is a first order linear differential equation of the form

Here, P = cot x and Q = 4x cosec x

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

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ y sin x = 2x² + c

∴ y = (2x² + c) cosec x

However, when, we have y = 0

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

Thus, the solution of the given initial value problem is