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:
$[s^2(Y)-sY(0)-y'(0)]+6[sY(s)-y(0)]+13Y(s)=e^{-\frac{\pi s}{4}}\\
s(Y)(s^2+6s+13)-5s-35=e^{-\frac{\pi s}{4}}$
which implies that:
$Y(s)=\frac{e^{-\frac{\pi s}{4}}}{(s+3)^2+4}+\frac{5s}{(s+3)^2+4}+\frac{35}{(s+3)^2+4}\\
=\frac{e^{-\frac{\pi s}{4}}}{(s+3)^2+4}+\frac{5(s+3)}{(s+3)^2+4}-\frac{50}{(s+3)^2+4}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{e^{-\frac{\pi s}{4}}}{(s+3)^2+4}+\frac{5(s+3)}{(s+3)^2+4}-\frac{50}{(s+3)^2+4}]$
We finally obtain
$y(t)=\frac{1}{2}e^{-3(t-\frac{\pi}{4})}\sin 2(t-\frac{\pi}{4})u_{\frac{\pi}{4}}(t)+5e^{-3t}\cos 2t-25e^{-3t}\sin 2t$
