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:
$[s^2(Y)-sY(0)-y'(0)]-4[sY(s)-y(0)]+13Y(s)=e^{-\frac{\pi s}{4}}\\
s(Y)(s^2-4s+13)-3s+12=e^{-\frac{\pi s}{4}}$
which implies that:
$Y(s)=\frac{e^{-\frac{\pi s}{4}}}{(s-2)^2+9}+\frac{3s}{(s-2)^2+9}-\frac{12}{(s-2)^2+9}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{e^{-\frac{\pi s}{4}}}{(s-2)^2+9}+\frac{3(s-2)}{(s-2)^2+9}-\frac{6}{(s-2)^2+9}]$
We finally obtain
$y(t)=\frac{1}{3}e^{2(t-\frac{\pi}{4})}\sin 3(t-\frac{\pi}{4})u_{\frac{\pi}{4}}(t)+3\cos 3t e^{2t}-2\sin 3te^{2t}$
