Find the solution of the initial value problem.
$y'' +2y' +2y = \delta(t - \pi)$ with initial conditions $y(0) =1, y'(0) =0$.
What I did was take the Laplace and got:
$(s^2Y(s) - s) + 2(sY(s) -2) +2Y(s) = e^{-\pi s}$, then simplying and solving for $Y(s)$
$Y(s) = \frac{e^{-\pi s} +s + 2}{s^2 + 2s +2}$
Now taking the inverse I got:
$(e^{-\pi} + s +2) \dot\ e^{-t}sint$
But the answer is:
$y = e^{-t}cost + e^{-t}sint + u_\pi(t)e^{-(t-\pi )}sin(t-\pi )$
How did they get that?