, y(1) = 0

Given and y(1) = 0

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

[∵ m log a = log a^m]

∴ I.F = x^–1 [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let

By substituting this in the above integral, we get

We know

∴ y = –e^–x + cx

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

⇒ 0 = –e^–1 + c(1)

⇒ 0 = –e^–1 + c

∴ c = e^–1

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

y = –e^–x + (e^–1)x

∴ y = xe^–1 – e^–x

Thus, the solution of the given initial value problem is y = xe^–1 – e^–x