Let $I$ a non-trivial interval and $f \in D^{2}(I)$ such that $f''(x) = f(x)$ for all $x \in I$. Show that, if there exists some $a \in I$ such that $f(a) = f'(a) = 0$, then $f(x) = 0$ for all $x \in I$.
My attempt:
A reasonable guess I've been pondering is that $f$ could be written as
$$f(x) = be^{x}+ce^{-x},\ \ b, c \in \mathbb{R}.$$
Because $f'(x) = be^{x}-ce^{-x}$ and so
$$ f''(x) = be^{x}+ce^{-x} = f(x).$$
And also, that would imply that
$$ be^{a}+ce^{-a} = 0\\ be^{a}-ce^{-a} = 0 $$
which gives $b=c=0$ and so $f$ is the zero map. The question is how are $b$ and $c$ suppossed to look like? Because if I treat $b$ and $c$ as unkowns in a $2x2$ system I get
$$b = \left(\frac{f(x)+f'(x)}{2}\right)e^{-x}\\ c= \left(\frac{f(x)-f'(x)}{2}\right)e^{x} $$ which doesn´t make any sense to me. Any hint someone could give me? Is my original guess a reasonable one?