Answer
See below
Work Step by Step
Given: $y''-2ay'+(a^2+b^2)y=f(t)$
Since $y(0)=\alpha\\
y'(0)=\beta$
obtain: $L[y]=\frac{F(s)}{(s-a)^2+b^2}+\frac{\alpha(s-a)}{(s-a)^2+b^2}+\frac{\beta-a\alpha}{(s-a)^2+b^2}$
Using the Convolution Theorem
$y(t)=L^{-1}[\frac{F(s)}{(s-a)^2+b^2}]+L^{-1}[\frac{\alpha(s-a)}{(s-a)^2+b^2}]+L^{-1}[\frac{\beta-a\alpha}{(s-a)^2+b^2}]\\
=\alpha e^{at}\cos (bt)+\frac{\beta-a\alpha}{a-b}e^{at}\sin (bt)+\frac{1}{b}\int^t_0 e^{a\tau}\sin (b\tau)f(t-\tau)d \tau$
