sorry for bad title.
Suppose $f:\mathbb{R} \rightarrow \mathbb R$ is differentiable twice and $f = f''$, $ f(0) = f'(0)=0$ and I want to prove that $f=0$.
So I started with proving that for $[-1,1]$ because it's differentiable it's continuous in $[-1,1]$, and there is $M\geq |f(x)| $ for any $x$ in $[-1,1]$, suppose $f(x_1)=M$.
We can develop a Taylor polynomial around zero and say that there is a $c$ between zero and $x$ such that: $$f(x) =f(0) + f'(0)x + \frac{f''(c)x^2}{2}= \frac{f(c)x^2}{2}$$ $$ f'(x) = (\frac{f(c)x^2}{2} )' =f(c)x$$
Now my question is:
Is it possible to derive this polynomial and treat $f(c)$ as a constant number?
Because then, since $M$ is maximum then $f'(x_1)=0$ and $f(c)x_1=0$ either way $M=0$.
Thanks for the help.
Things can be saw in three ways :
1-Uniqueness of solution with given initial conditions
Knowing that for a linear homogenous differential equation there exists an unique solution given initial conditions for $f$ and $f'$.
Because $0$ is solution you're function must be 0.
2-Direct Solving
Your solution is of the form :
$$\exists(A,B), \forall x, f(x)=A\cosh(x)+B\sinh(x)$$
So giving your conditions
$$f=0$$