,

Given and

This is a first order linear differential equation of the form

Here, and

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

We have

∴ I.F = x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ xy = x sin x + c

However, when, we have y = 1

∴ c = 0

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

∴ y = sin x

Thus, the solution of the given initial value problem is y = sin x