The problem states that:
If $f$ is a real valued, twice differentiable function and satisfies the equation: $$f(x) + f''(x) = -xg(x)f'(x),$$
where $g(x)\geq 0$ for all real values of $x$. Prove that $|f(x)|$ is bounded.
My approach till now : adding $f'(x)$ on both sides of equation, we get $$f(x) + f''(x) + f '(x) = f '(x)[1-xg(x)]$$
Now consider $h(x) = 1-xg(x)$. If $x<0$, $h(x)>1$ [since $g(x)$ is always non negative] and if $x>0$, $h(x)<1$.
So let us consider 2 cases:
(I) If $x>0$ then $$f(x) + f''(x) + f'(x) >f'(x)$$ on solving this equation, we get that $f(x) \geq -f''(x)$.
(II) If $x<0$ then $$f(x) + f'(x) + f''(x) <f '(x)$$ on solving we get, $f(x)\leq -f''(x)$.
From results of both cases it comes out that , $$-f''(x)\leq f(x)\leq -f''(x)$$ which means that $f(x) = -f''(x)$, which is not what we had to prove.
Please help me with this problem.
Multiply with $2f'(x)$ and integrate $$ f'(x)^2+f(x)^2=f'(0)^2+f(0)^2-2x^2\int_0^1sg(sx)f'(sx)^2ds $$ The last term is always negative, so that $f'(x)^2+f(x)^2$ is bounded which implies the claim.