TITLE: Twice differentiable function related problem
QUESTION [5 upvotes]: Let $f(x)$ be a real valued twice differentiable function on interval $[1,5]$ such that $f(1) = f(5) = 0$ and $f(x) = f'(x) + f''(x)$ , $\forall x \in \left[ {1,5} \right]$, then $(f(2)+f(4)–f'(3))$ is equal to___________________.
I am not able to approach this problem. Initially I thought of using Rolle's Theorem but function needs to be continuous in the closed interval $[1,5]$ and differentiable in the open interval $(1,5)$.
How do I perform the mathematical operation on these numbers

REPLY [2 votes]: The linear ODE
$$f=f'+f''$$
has the particular solution $f\equiv0$. Moreover using the Ansatz $f=ce^{ax}$ you obtain the equation
$$1=a+a^2\iff a_{\pm}=\frac{1\pm\sqrt{5}}{2}\in\mathbb{R}.$$
The general solution is thus given by the linear combination
$$f=c_{+}e^{a_+x}+c_{-}e^{a_-x}.$$
Inserting the boundary conditions, we obtain the linear system of equations
$$\begin{cases}c_+e^{a_+}+c_-e^{a_-}=0\\[12pt]
c_+e^{5a_+}+c_-e^{5a_-}=0
\end{cases}\begin{cases}c_+=-c_-e^{a_-}/e^{a_+}\\[10pt]
\underbrace{\left(e^{4a_-}-e^{4a_+}\right)e^{a_-}}_{\neq 0}\cdot c_-=0
\end{cases}$$
whose only solution is $c_{\pm}=0$. Thus $f\equiv 0$ which confirms Albus Dumbledore's result.