The question is :
Let $f:\mathbb{R} \to \mathbb{R}$ be twice differentiable function satisfying
$f(x)+f''(x)=-xg(x)f'(x), x\in \mathbb{R} $ where $g(x) \ge 0, \forall x\in \mathbb{R}$
Which of the following is\are true?
$(1)$ If $f(0)=f'(0)=1$ , then $f(3)\lt 3$
$(2)$ If $f(0)=f'(0)=2$ , then $f(4)\lt 4$
$(3)$ If $f(0)=f'(0)=3$ , then $f(3)=5$
$(4)$ If $f(0)=f'(0)=3$ , then $f(3)=6$
My thoughts:-
I will first discuss about $(3)$ and $(4)$
Let $g(x)=0$
Then with some computation , we can show
$f(x)=3(\sin x+\cos x)$ as a suitable candidate to discard $(3)$ and $(4)$
Here , for option $(3)$
$f(3)=5$
$\Rightarrow \sin 3+\cos 3=\frac 53$
On squaring both sides
$1+\sin 6=\frac{25}9$
$\sin 6=\frac {16}9 \gt 1$, a contradiction
Similarly $f(3)= 6$ will give the contradiction
$\sin 3+\cos 3=2$ ( implying $\sin 3=\cos 3=1$ which is an impossibility) .
Thus we are left with $(1)$ and $(2)$
Note:A slight variant of the above example satisfies the condition in $(1)$ and $(2)$
I tried with simple examples like $g(x)=1 $ and $f(x)=x$ or like quadratics but couldn't reach conclusions .
Please help with the options $(1)$ and $(2)$ . Thanks for your time.
Best Answer
Consider the energy function $E=f(x)^2+f'(x)^2$. Then $$ \frac{d}{dx}E=2f'(x)(f''(x)+f(x))=-2xg(x)f'(x)^2 $$ so that $E$ is falling along solutions. As far as I can see this implies that 1) and 2) are true.
In 1) $f(x)\le\sqrt{E(x)}\le\sqrt{E(0)}=\sqrt2<3$ and similarly in 2) $f(x)\le\sqrt8<4$. The same way you get in 3) and 4) $f(x)\le\sqrt{18}<5$, so that the given values can never be reached.