2. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ so that
$$
(x+y)(f(x)-f(y))=(x-y) f(x+y)
$$
for all $x, y \in \mathbb{R}$.
2. Suppose that $f$ is a solution. Let
$$
a=\frac{1}{2}(f(1)-f(-1)), \quad b=\frac{1}{2}(f(1)+f(-1))
$$
and $g(x)=f(x)-a x-b x^{2}$. Then
$$
(x+y)(g(x)-g(y))=(x-y) g(x+y)
$$
and $g(1)=g(-1)=0$. Letting $y=1$ and $y=-1$ above give
$$
\begin{aligned}
(x+1) g(x) & =(x-1) g(x+1) \\
x g(x+1) & =(x+2) g(x) .
\end{aligned}
$$

Thus
$$
x(x+1) g(x)=x(x-1) g(x+1)=(x-1)(x+2) g(x)
$$
for all $x$. So $g(x)=0$ for all $x$. Hence $f(x)=a x+b x^{2}$. We can check directly that any function of this form (for some $a, b \in \mathbb{R}$ ) satisfies the given equation.