If $f''+f'=f$ then $f\equiv 0$ Let , $f:\mathbb R\to \mathbb R$ be a two times continuously differentiable function that satisfies the differential equation $f''+f'=f$ for all $x\in [0,1]$. If $f(0)=f(1)=0$ then , which is correct ? 
(A) There exists $(a,b)\subset [0,1]$such that $f(x)>0$ in $(a,b)$.
(B) There exists $(a,b)\subset [0,1]$such that $f(x)<0$ in $(a,b)$.
(C) $f\equiv 0$ in $[0,1]$.
(D) such function does not exists.
If $f(x)=0$ for all in $x\in[0,1]$ then $f$ satisfies the given conditions. So option (D) is false. I think option (C) is correct , but I can't prove this.

The function, which is continuous on $[0,1]$, must have a maximum point $x$. In the maximum point (if $x\in(0,1)$) you have: $f'(x)=0$ and $f''(x)\le 0$. So the equation says $f(x) = f''(x) \le 0$. With a similar reasoning you find that on the minimum you have $f(x)\ge 0$. Hence the function is identically zero.
