functions - Can you solve the following functional equation?

I found the following functional equation:
Find all functions $f : \Bbb R \rightarrow \Bbb R $ such that:

$xf(x) - yf(y) = (x - y)f(x + y) $ for all $x, y \in \mathbb R $

Could you please help me? I think I proved that if $f(0) = 0$ then for each $x \in \Bbb Q$ $f(kx) = kf(x)$, but I don't know how to continue. Thanks in advance.

Answer

It's obvious that every function of the form $f(x)=ax+b$ satisfies the equation:
$$xf(x)-yf(y)=(x-y)f(x+y)\qquad(1)$$

It can be shown that those are the only solutions indeed. To show that, let $a=f(1)-f(0)$ and $b=f(0)$, and define $g(x)=f(x)-ax-b$. It's easy to see that by (1), $g$ satisfies
$$xg(x)-yg(y)=(x-y)g(x+y)\qquad(2)$$
and we have $g(0)=0$ and $g(1)=0$. Letting $x=1$ and $y=-1$ in (2) we get $g(-1)=0$. Now, letting $y=1$ and $y=-1$ in (2), we respectively get:
$$xg(x)=(x-1)g(x+1)\qquad(3)$$
$$xg(x)=(x+1)g(x-1)\qquad(4)$$
Substituting $x+1$ for $x$ in (4) we have:
$$(x+2)g(x)=(x+1)g(x+1)\qquad(5)$$
Subtracting (5) and (3) we get $2g(x)=2g(x+1)$ and thus $g(x)=g(x+1)$. Hence by (5) we have $(x+2)g(x)=(x+1)g(x)$ and therefore $g$ is the constant zero function. So $f(x)=ax+b$.