Intermediate value existence: $f''(\xi)-f(\xi)(1+2\cot^2\xi)=0$. 
Let $f$ be twice continuouly differentiable on $[0,\pi]$. $f(0)f(\pi)<0$. Show that there exists $\xi\in(0,\pi)$, such that $f''(\xi)-f(\xi)(1+2\cot^2\xi)=0$

I do not whether the problem is right, which is recalled by my classmate. Clearly, some zero of $f$ exists. $\cot\xi$ where to stand? Of course, is should not be $\pi/2$.
 A: Here is a similar question from IMC2013,It can help you solve your question.

A: 
Solve $y''=y(1+2\cot^2x)$

Solution: Let $\cos x=t$, we have $x=\arccos t\Rightarrow \frac{\mathrm dx}{\mathrm dt}=-\frac{1}{\sqrt{1-t^2}}$
$$\frac{\mathrm dt}{\mathrm dx}=-\sqrt{1-t^2},\quad \frac{\mathrm d^2 t}{\mathrm dx^2}=-t,\quad \cot^2x=\frac{\cos^2x}{\sin^2x}=\frac{t^2}{1-t^2}$$
so
$$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm dt}\frac{\mathrm dt}{\mathrm dx}=-\sqrt{1-t^2}\frac{\mathrm dy}{\mathrm dt}$$
$$\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{\mathrm d}{\mathrm dx}\bigg(\frac{\mathrm dy}{\mathrm dt}\cdot\frac{\mathrm dt}{\mathrm dx}\bigg)=\frac{\mathrm dy}{\mathrm dt}\cdot\frac{\mathrm d^2 t}{\mathrm dx^2}+\frac{\mathrm d^2y}{\mathrm dt^2}\cdot\frac{\mathrm dt}{\mathrm dx}\cdot\frac{\mathrm dt}{\mathrm dx}=-t\frac{\mathrm dy}{\mathrm dt}+(1-t^2)\frac{\mathrm d^2y}{\mathrm dt^2}$$
we get
$$(1-t^2)\frac{\mathrm d^2y}{\mathrm dt^2}-t\frac{\mathrm dy}{\mathrm dt}=y\bigg(1+\frac{2t^2}{1-t^2}\bigg)$$
$$\Rightarrow\frac{\mathrm d^2y}{\mathrm dt^2}-\frac{t}{1-t^2}\frac{\mathrm dy}{\mathrm dt}=\frac{1+t^2}{(1-t^2)^2}y$$
Since
$$\frac{\mathrm d}{\mathrm dt}\bigg(\frac{\mathrm dt}{1-t^2}\bigg)=\frac{1+t^2}{(1-t^2)^2}$$
so
$$\frac{\mathrm d^2y}{\mathrm dt^2}=\frac{1+t^2}{(1-t^2)^2}y+\frac{t}{1-t^2}\frac{\mathrm dy}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}\bigg(\frac{t}{1-t^2}y\bigg)$$
$$\Rightarrow \frac{\mathrm dy}{\mathrm dt}=\frac{t}{1-t^2}y+C_1$$
$$\Rightarrow \frac{\mathrm d}{\mathrm dt}\big(y\sqrt{1-t^2}\big)=-C_1\sqrt{1-t^2}$$
$$\Rightarrow y\sqrt{1-t^2}=\frac{1}{2}\big(-t\sqrt{1-t^2}+\arccos t\big)C_1+C_2$$
Finally,
$$y=C_1\bigg(\frac{x}{\sin x}-\cos x\bigg)+C_2\csc x$$

Let $f$ be twice continuouly differentiable on $[0,\pi]$. $f(0)f(\pi)<0$. Show that there exists $\xi\in(0,\pi)$, such that $f''(\xi)-f(\xi)(1+2\cot^2\xi)=0$

Proof:
$$f''(\xi)-f(\xi)(1+2\cot^2\xi)=0 \Rightarrow y''=y(1+2\cot^2x)$$
Derivation
$$y=C_1\bigg(\frac{x}{\sin x}-\cos x\bigg)+C_2\csc x$$
$$\Rightarrow y'=C_1\bigg(\frac{\sin x-x\cos x}{\sin^2x}+\sin x\bigg)-C_2\frac{\cos x}{\sin^2x}$$
Since
$$y\cos x=C_1\bigg(\frac{x\cos x}{\sin x}-\cos^2x\bigg)+C_2\cot x$$
$$y'\sin x=C_1\bigg(\frac{\sin x-x\cos x}{\sin x}+\sin^2x\bigg)-C_2\cot x$$
Add
$$y\cos x+y'\sin x=2C_1\sin^2x\Rightarrow \frac{y\cos x+y'\sin x}{\sin^2x}=2C_1$$
Now consider the function
$$h(x)=\frac{f(x)\cos x+f'(x)\sin x}{\sin^2x}$$
$f(0)f(\pi)<0\Rightarrow\exists\xi_0\in(0,\pi)$, s.t. $f(\xi_0)=0$. Let $g(x)=f(x)\sin x$,
$$g(0)=g(\xi_0)=g(\pi)=0$$
Rolle’s theorem, we have $g'(\xi_1)=g'(\xi_2)=0\Rightarrow h(\xi_1)=h(\xi_2)=0$, Rolle’s theorem,  there exist $\xi\in(\xi_1,\xi_2)$ for which
$$h'(\xi)=\frac{f''(\xi)-f(x)(1+2\cot^2\xi)}{\sin\xi}=0$$
