If $f(x)=f'(x)+f''(x)$ then show that $f(x)=0$ 
A real-valued function $f$ which is infinitely differentiable on $[a.b]$ has the following properties:
  
  
*
  
*$f(a)=f(b)=0$
  
*$f(x)=f'(x)+f''(x)$ $\forall x \in [a,b]$
  
  
  Show that $f(x)=0$ $\forall x\in [a.b]$

I tried using the Rolle's Theorem, but it only tells me that there exists a $c \in [a.b]$ for which $f'(c)=0$. 
All I get is:


*

*$f'(a)=-f''(a)$

*$f'(b)=-f''(b)$

*$f(c)=f''(c)$


Somehow none of these direct me to the solution.

Hint: Let $\alpha$ and $\beta$ be the roots of $X^2+X-1$. 
(a) Check that $f$ satisfies
$$
\left(\frac{d}{dx}-\alpha\right)\left(\frac{d}{dx}-\beta\right)f=0.
$$
(b) Solve the above equation by solving two ODE of the form $y'-cy=g(x)$. (If you don't know how to solve $y'-cy=g(x)$, I'll be happy to give another hint.)
(c) Conclude.
