Answer
$y(t)=\cos 3t+\frac{4}{3}t\sin 3t$
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]+9Y(s)=\frac{8s}{s^2+9}\\
s(Y)(s^2+9)=\frac{8s}{s^2+9}+s$
which implies that:
$Y(s)=\frac{8s}{s^2+9}+\frac{s}{s^2+9}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{8s}{s^2+9}]+L^{-1}[\frac{s}{s^2+9}]$
We finally obtain
$y(t)=\cos 3t+\frac{4}{3}t\sin 3t$
