Answer
See below
Work Step by Step
Given: $y''-(a+b)y'+aby=f(t)$
Since $y(0)=\alpha\\
y'(0)=\beta$
obtain: $L[y]=\frac{F(s)}{(s-a)(s-b)}+\frac{\alpha}{s-b}+\frac{\beta-b\alpha}{(s-a)(s-b)}$
Using the Convolution Theorem
$y(t)=L^{-1}[\frac{F(s)}{(s-a)(s-b)}]+L^{-1}[\frac{\alpha}{s-b}]+L^{-1}[\frac{\beta-b\alpha}{(s-a)(s-b)}]\\
=\alpha e^{bt}+\frac{\beta-b\alpha}{a-b}(e^{at}-e^{bt})+\frac{1}{a-b}\int^t_0 (e^{a\tau}-e^{b\tau})f(t-\tau)d \tau$
