Find all functions $f,g,h$ $\mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x) - g(y) = (x-y)h(x+y)$ $(\forall x,y \in \mathbb{R})$
Setting $y = x$ gives $f(x) - g(x) = 0$ for all $x$. Therefore, $f(x) = g(x)$. $f(x) - f(y) = (x - y)h(x+y)$. From here I tried a bunch of arbitrary values for $x$ and $y$ such as $0, kx, -x,$ etc. with little progress.
How do I continue and in general how do you approach a multi-variable functional equation like this? Thanks.
We have that $$f(x)-f(0)=xh(x)$$ from where $$f(x)-xh(x)=f(0).$$ Thus
$$xh(x)-yh(y)=(x-y)h(x+y).$$ Now, if $h$ is a solution then $h+c$ is also a solution. So we can assume $h(0)=0.$ Writing $y=-x$ we get $$x(h(x)+h(-x))=0$$ from where $$h(-x)=-h(x).$$ Now writing $-y$ instead of $y$ we get $$xh(x)+yh(-y)=(x+y)h(x-y).$$ That is $$xh(x)-yh(y)=(x+y)h(x-y).$$ Thus
$$(x-y)h(x+y)=(x+y)h(x-y)$$ from where $$\dfrac{h(x+y)}{x+y}=\dfrac{h(x-y)}{x-y}.$$ So there exists a constant such that $h(x)=ax.$ And the general solution is $$h(x)=ax+h(0).$$ Finally $$f(x)=f(0)+xh(x)=ax^2+h(0)x+f(0).$$
Hint: $$h(x+y)=\frac{f(x)-f(y)}{x-y}$$ Now take the limit $x\rightarrow y$. What you get is $$h(2x)=f'(x)$$ or $$h(x)=f'(x/2)$$
