Answer
See below
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]+9Y(s)=\frac{30}{s^2+4}+e^{-\frac{\pi s}{4}}\\
s(Y)(s^2+9)-=\frac{30}{s^2+4}+e^{-\frac{\pi s}{4}}$
which implies that:
$Y(s)=\frac{30}{(s^2+4)(s^2+9)}+\frac{e^{-\frac{\pi s}{4}}}{(s^2+4)(s^2+9)}\\
=6(\frac{1}{s^2+4}-\frac{1}{s^2+9})+\frac{e^{-\frac{\pi s}{4}}\times3}{3(s^2+9)}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[6(\frac{1}{s^2+4}-\frac{1}{s^2+9})+\frac{e^{-\frac{\pi s}{4}}\times3}{3(s^2+9)}]$
We finally obtain
$y(t)=3\sin 2t-3\sin 3t+\frac{\sin 3(t-\frac{\pi}{4})}{3}u_{\frac{\pi}{4}}(t)$
