Let $c>0$ be a constant and suppose that $f$ is a $C^2$ function of $x+ct$ plus a $C^2$ function of $x−ct$. That is, $f(x,t)=g(x+ct)+h(x−ct)$ where $g:R→R$ and $h:R→R$ are $C^2$ functions.
Show that such an $f$ is always a solution of the wave equation $c^2f_{xx}=f_{tt}$ and show that any $C^2$ solution of the wave equation $c^2f_{xx}=f_{tt}$ must be of the form $f(x,y)=g(x+ct)+h(x−ct)$.
If anyone can prove this would much appreciated it!
Compute $f_{xx}$ and $f_{tt}$ using the chain rule:
$$f_{xx} = g_{xx} + h_{xx},\\ = g''(x+ct) + h''(x-ct) $$
and
$$f_{tt} = g_{tt} + h_{tt},\\ =c^2g''(x+ct) + c^2h''(x-ct) $$
Now substitute into the wave equation:
$$c^2 f_{xx} = f_{tt}, \implies c^2 \left(g''(x+ct) + h''(x-ct)\right) = c^2g''(x+ct) + c^2h''(x-ct)$$
and we can see both sides are equal.
For the second part of the question, define new variables, $$\xi = x+ct, \eta = x-ct$$ and find expressions for the operators $\frac{\partial}{\partial t}, \frac{\partial}{\partial x}$ in terms of these variables (use the chain rule). If you get stuck on this part see section 2 of this link