Help needed showing that $f(x,y)=U(x+y)+V(x-y)$ Let $h(u,v)=f(u+v,u-v)$ and $f_{xx}=f_{yy}$ for every $(x,y)\in\mathbb{R}^2$. In addition, $f\in{C^2}.$ Show that $f(x,y)=U(x+y)+V(x-y)$.
I think applying the Taylor theorem could be useful.
$$f(x,y)=f(x+h_1,y+h_2)-\left(\frac{\partial{f(x,y)}}{\partial{x}}h_1+\frac{\partial{f(x,y)}}{\partial{y}}h_2\right)-\frac 1 2\left(\frac{\partial^2{f(x,y)}}{\partial^2{x}}h_1^2+\frac{\partial^2{f(x,y)}}{\partial{x}\partial{y}}h_1h_2+\frac{\partial^2{f(x,y)}}{\partial^2{y}}h_2^2\right)-R(h_1,h_2)$$

You define
$$
\xi=x+y\\
\eta=x-y
$$
then the partial differential equation becomes
$$
\frac{\partial^2f}{\partial \xi\partial \eta}=0
$$
The only way you can satisfy it is that it is a sum of 2 functions 
$$
U(\xi)+V(\eta)
$$
Such combination is always identically zero by the application of a second mixed derivative.
