Let $f(t) = f(t$) be the 2pi periodic("sawtooth wave"), f(t) = t for $0 \leq t \leq 2\pi$ and consider the equation $$y^{\prime \prime} + a^2y = f$$
For which values of $a$ (here $a$ >0) does this have a bounded solution? I would be grateful for any help. I am not sure how to solve this.
If $t$ is restricted to the domain $0 \leq t \leq |x|$ for some $x \in \Bbb{R}$ (in particular, for $x = 2\pi$, the equation will always have a bounded solution, and in fact, given any finite initial conditions, all solutions will be bounded.
Are you sure you were not thinking of a related problem like $$ f(t) = t \mbox{ for } 0 \leq t < 2 \pi \\ f(t) = f(t-2\pi) \mbox{ for } t\geq 2\pi $$ Becuase that sort of problem is much more interesting, and is unbounded for a countably infinite set of values of $a$.