, y(0) = 1

Given and y(0) = 1

This is a first order linear differential equation of the form

Here, P = tan x and Q = 2x + x²tan x

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

We have

∴ I.F = sec x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ y sec x = x²sec x + c

∴ y = x² + c cos x

However, when x = 0, we have y = 1

⇒ 1 = 0² + c(cos 0)

⇒ 1 = 0 + c(1)

∴ c = 1

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

y = x² + (1)cos x

∴ y = x² + cos x

Thus, the solution of the given initial value problem is y = x² + cos x