$y′′ + 2y′ + 2y = δ(t − \pi) + aδ(t − T)$ , $y(0) = y′(0) = 0$
$a$ and $T$ are positive numbers and $T > \pi$. I need to find values for $a$ and $T$ such that $y(t) = 0$ for all $t \ge T$?
I just started working with Laplace transforms. I am having hard time figuring out values for $a$ and $T$. I would be grateful for any guidance.
Because $y(0)=y'(0)=0$, the LT of the ODE yields
$$(s^2+2 s+2) Y(s) = e^{-\pi s} + a e^{-T s}$$
where $Y$ is the LT of $y$. Thus we have that
$$y(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{e^{(t-\pi) s} + a e^{(t-T) s}}{s^2+2 s+2} $$
which, by the residue theorem, is
$$e^{-(t-\pi)} \sin{(t-\pi)} \theta(t-\pi) + a e^{-(t-T)} \sin{(t-T)} \theta(t-T)$$
where $\theta$ is the Heaviside step function. Thus, for $t \ge T$, the step functions are both equal to one. Thus, for $t \ge T$, we have
$$y(t) = e^{-t} \left [ -e^{\pi} \sin{(t)} + a e^{T} \sin{(t-T)}\right ] $$
For $y(t)=0$ in this time domain, $T=k \pi$, where $k \in \mathbb{Z}$ and $k \gt 1$, and $a=(-1)^k e^{-(k-1) \pi}$.