, y(0) = 0

Given and y(0) = 0

This is a first order linear differential equation of the form

Here, P = 2 and Q = e^–2xsin x

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

We have

∴ I.F = e^2x

Hence, the solution of the differential equation is,

Recall

⇒ ye^2x = –cos x + c

⇒ ye^2x × e^–2x = (–cos x + c) × e^–2x

∴ y = (–cos x + c)e^–2x

However, when x = 0, we have y = 0

⇒ 0 = (–cos 0 + c)e⁰

⇒ 0 = (–1 + c) × 1

⇒ 0 = –1 + c

∴ c = 1

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

y = (–cos x + 1)e^–2x

∴ y = (1 – cos x)e^–2x

Thus, the solution of the given initial value problem is y = (1 – cos x)e^–2x