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:
$[s^2(Y)-sY(0)-y'(0)]+2[sY(s)-y(0)]+5Y(s)=e^{-\frac{\pi}{2}s}\\
s(Y)(s^2+2s+5)-2=e^{-\frac{\pi}{2}s}$
which implies that:
$Y(s)=\frac{e^{-\frac{\pi}{2}s}}{(s+1)^2+4}+\frac{2}{(s+1)^2+4}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{e^{-\frac{\pi}{2}s}}{(s+1)^2+4}+\frac{2}{(s+1)^2+4}]$
Since $L^{-1}\{e^{-3s}(\frac{1}{s-2}-\frac{1}{s+2})\}=u_3(t)[e^{2(t-3)}-e^{-2(t-3)}]$ we finally obtain
$y(t)=\frac{1}{2}e^{-(t-\frac{\pi}{2})}\sin 2(t-\frac{\pi}{2})u_{\frac{\pi}{2}}(t)+2e^{-t}\sin 2t$
