TITLE: $f''(x)+e^xf(x)=0$ , prove $f(x)$ is bounded
QUESTION [7 upvotes]: Differentiable function in $\mathbb{R}$ for which $f''(x) + e^x f(x)=0$ for every $x$. Prove that $f(x)$ is bounded as $x \rightarrow +\infty$
I have tried a few stuff but they didnt work out, for example i noticed that the function has infinite max and min as $x \rightarrow +\infty$ but thats still not enough to prove it, any ideas?

REPLY [4 votes]: The general solution of your differential equation is
$$ f(x) = A J_0(2 e^{x/2}) + B Y_0(2 e^{x/2})$$
where $J_0$ and $Y_0$ are Bessel functions of the first and second kinds.
Their asymptotics are known.
EDIT: if $R(x) = f'(x)^2 + e^x f(x)^2$, show that $R'(x) = e^x f(x)^2 \le R(x)$,
and thus $\dfrac{d}{dx} \log R(x) \le 0$.