I've been struggling with the following question:
$\ddot x + 2 \dot x +(1+e^{-t})x=t^{2}cos(t)$, show that all solution are bounded for $t>0$.
My problem: Im probably missing some theorem or fact. If we had, for example $\ddot x + 2 \dot x +(1+e^{-t})x=0$. Well that would be pretty easy to prove that all solutions tend to $0$ when $t$ tends to $\infty$. We only have to use a very well known theorem of perturbed systems. But for that one I cant find any theorem or anything that would help me to solve it.
Thanks so much. :)
Regarding the homogeneous, transforming Laplace we have
$$ (s+1)^2X_h(s) + X_h(s+1) = x_0s+x'_0 $$
or
$$ X_h(s) + \frac{X_h(s+1)}{(s+1)^2}=\frac{x_0s+x'_0}{(s+1)^2} $$
now if ${\cal{L}}^{-1}[X_h(s)]$ is bounded then ${\cal{L}}^{-1}[X_h(s+1)]$ is also bounded due to pole translation. The final value condition establishes that
$$ \lim_{s\to 0}s\left(X_h(s)+\frac{X_h(s+1)}{(s+1)^2}\right)= \lim_{s\to 0}s\frac{x_0s+x'_0}{(s+1)^2}= 0 $$
Regarding the complete DE we have
$$ X(s) + \frac{X(s+1)}{(s+1)^2}=\frac{x_0s+x'_0}{(s+1)^2}+\frac{1}{(s+1)^2}\frac{2s(s^2-3)}{(s^2+1)^3} $$
which is unbounded because $\frac{1}{(s+1)^2}\frac{2s(s^2-3)}{(s^2+1)^3}$ is unbounded.
NOTE
If $X(s+1)$ is bounded then $\frac{X(s+1)}{(s+1)^2}$ is also bounded because $|f * g | \le |f||g|$