The question is to find the conditions (about constant $a, b$) for $f(x)$ to be a bounded function. $f(x)$ is a solution of the IVP $$f''+f=\sin(bx), ~~ f(0)=a, ~f'(0) = 0$$
I solved the function (using Laplace transform) and I get the solution: $$f(x) = \left\{ \begin{array}{cl} -\frac{1}{2}x\cos x+\frac{1}{2}\sin x+a\cos x,& b=1 \\ \frac{1}{2}x\cos x-\frac{1}{2}\sin x+a\cos x,& b=-1 \\ \dfrac{b\sin x-\sin(bx)}{b^2-1}+a\cos x,& \text{others}\\ \end{array} \right. $$ Then I found that when $b = 1$ or $b = -1$, the function is not bounded.
Is that all? I did not understand the intent of the question. Is there any limitation in Laplace transform that influences the solution to be unbounded?