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.
 A: Since you know how to solve $y'' = y$ (and so I presume $y'' = ky$) here is a way to get your equation into that form.
We will use the identity
$$ (fg)'' = f'' g + 2f'g' + fg''$$
Now setting $g = e^{kx}$ gives us
$$ (fe^{kx})'' = e^{kx} (f'' + 2kf' + k^2f)$$
In order to eliminate $f''$ and $f'$, we set $k=\frac{1}{2}$, to get 
$$ (f e^{x/2})'' = e^{x/2} (f'' + f' + f/4) = e^{x/2} (5f/4)$$
(using the given $f = f' + f''$)
You can now set $y = f(x)e^{x/2}$ to get
$$ y'' = \frac{5}{4} y$$
A: Hint $f$ can't have a positive maximum at $c$ since then $f(c)>0, f'(c)=0, f''(c) \le 0$ implies that $f''(c)+f'(c)-f(c) < 0$. Similarly $f$ can't have a negative minimum. Hence $f = 0$.
A: Let $x=c$ be the $x$ coordinate of absolute max of $f(x)$ on $[a,b]$.  (This point exists by the extreme value theorem).  I will show that $f(c) = 0$.  Since $f(a) = 0$ and $c$ is the absolute max, $f(c)\geq 0$.  By Fermat's theorem, we know $f'(c) = 0$.  Hence, we learn that $f(c) = f''(c)\geq 0$.
Now, assume for a contradiction that $f(c) > 0$, so $f''(c) > 0$ and $c\neq a$ and $c\neq b$.  I claim that for $x$ close enough to $c$, but bigger than it, that $f(x) > f(c)$, contradicting maximality of $f(c)$.
Since $f''$ is continuous, for $x$ close enough to $c$ say, within $\delta$, we have $f''(x) > 0$.  On the interval where $c<x<c+\delta$, $f'(x) \geq 0$ with equality only at $x=c$.  This follows from the Mean value theorem applied to $f'$, because if $f'(x)\leq 0$ for a point $x\in(c,c+\delta)$, then by the MVT, $f''(d) \leq 0$ for some $d\in(c,c+\delta)$, giving a contradiction.
From this, it follows that $f(x)>f(c)$ for $x\in(c,c+\delta)$, because, again by the MVT, we have $\frac{f(x)-f(c)}{x-c} = f'(d) > 0$ for some $d\in(c,c+\delta)$, so, $f(x) - f(c) > 0$.
Thus, we contradict maximality of $f(c)$.  From this contradiction, we deduce $f(c) = 0$ is the maximum of the function.  Now, repeat a similar argument to $-f$ (changing the interval $(c,c+\delta)$ to $(c-\delta, c)$) to deduce the minimum value of $f$ is $0$.  From this it follows that $f$ is identically $0$.
A: 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.
