TITLE: Show that $f$ is bounded if $f(x)+f''(x)=-xg(x)f'(x)$ where $g\ge0$
QUESTION [5 upvotes]: Let $f$ be a twice differentiable real valued function such that 
$f(x)+f''(x)=-xg(x)f'(x)$
Where $g(x)\geq 0$ for all real $x$
Show that $|f(x) |$ is a bounded function.

REPLY [3 votes]: Multiplying both sides by $f'(x)$, one has
$$ \frac12[(f(x))^2+(f'(x))^2]'=-xg(x)(f'(x))^2\le 0 $$
for $x>0$. Then integrating both sides from $0$ to $x$, one has
$$ (f(x))^2+(f'(x))^2\le(f(0))^2+(f'(0))^2$$
from which one obtains 
$$ (f(x))^2\le (f(x))^2+(f'(x))^2\le (f(0))^2+(f'(0))^2. $$
So $|f(x)|$ is bounded. If $x<0$, one can use the same way to establish.