Twice diffrentiable function $f: \mathbb R \to \mathbb R$ such that $f''(x)+e^xf(x)=0, \forall x \in \mathbb R$

Question

Let $f: \mathbb R \to \mathbb R$ be a twice differentiable function such that $f''(x)+e^xf(x)=0, \forall x \in \mathbb R$. Then is it true that $f(x)$ is bounded in $[0,\infty)$ i.e. does there exist $M >0$ such that $|f(x)| <M, \forall x >0$ ?

Answer 1

The explicit solution of $f''(x) = -e^x f(x)$ is given by $$f(x) = a_1 J_0(2e^{x/2}) + a_2 Y_0(2e^{x/2}),$$ where $J_0$ denote the Bessel function of the first kind and $Y_0$ the Bessel function of second kind. Since $\lim_{x \rightarrow - \infty} J_0(x) =1$ and $\lim_{x \rightarrow \infty} J_0(x) =0$, resp. $\lim_{x \rightarrow \pm \infty} Y_0(x) =0$, $f(x)$ is bounded.