TITLE: If $f(x)+f'(x)-\frac{1}{x+1}\int_{0}^{x}f(t)dt=0$ and $f(0)=0$, then what is $f'(x)$?
QUESTION [6 upvotes]: $f\in C^{1}[0,\infty)$, $f(0)=0$ and 
$$
f(x)+f'(x)-\frac{1}{x+1}\int_{0}^{x}f(t)dt=0
$$
then $f'(x)=$ ?
I'v tried in the following ways. First, let $F(x)=\int_{0}^{x}f(t)dt$, then we are left to solve a second order ODE with initial condition $F(0)=0$, and $F'(0)=0$, but the problem is it seems to me that it's not that easy to solve it.
I hope there are some other ways to handle it that a one year students can understand. (I have tried to write $g(x)=f(x)e^x$ to rewrite the equation, but it doesn't make it easier)

REPLY [2 votes]: Multiply the both sides by (x+1). Then differentiate. You'll get
$$(x+1) f''(x)+(x+2) f'(x)=0$$
Solve this equation. The solution is
$$f(x)=C_1 \text{Ei}(-x-1)+C_2$$
Now find derivative:
$$f'(x)=\frac{C_1 e^{-x}}{x+1}$$
From your condition that $f(0)=0$, it follows that $f'(0)=0$ and as such, $C_1=0$. So $f(x)$ is a constant zero.