Let $f$ be a twice-differentiable real-valued function satisfying $f(x)+f''(x)= -xg(x)f'(x)$, where $g(x) \geq 0$ for all real $x$. Prove that $|f(x)|$ is bound.
Honnestly I worked on this problem for a good while and I just don't know how to solve it.
Is anyone could give me the principal details how to solve it (hints)?
Consider the function $h(x)=f(x)^2+f'(x)^2$. Observe that $h(x)$ is a non-negative function. On the other hand, we see that $$ h'(x)=2f'(x)\left[f(x)+f''(x)\right]=-2xg(x)\ f'(x)^2. $$ Thus $h'(x)\leq 0$ when $x\geq 0$ and $h'(x)\geq 0$ when $x\leq 0$. Hence $h(x)$ attains its maximum at $x=0$. Since $|f(x)|$ is bounded by $\sqrt{h(x)}\leq \sqrt{h(0)}$, it follows that $f(x)$ is bounded.
