Tintarn wrote: Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that \[ xf(x+y)=(x+\alpha y)f(x)+xf(y) \]for all $x,y\in\mathbb{R}$. Let $P(x,y)$ be the assertion $xf(x+y)=(x+\alpha y)f(x)+xf(y)$ $P(1,0)$ $\implies$ $f(0)=0$ Let $x,y\ne 0$ : $P(x,y)$ $\implies$ $f(x+y)=(1+\alpha \frac yx)f(x)+f(y)$ $P(y,x)$ $\implies$ $f(x+y)=(1+\alpha \frac xy)f(y)+f(x)$ Subtracting : $\frac yxf(x)=\frac xyf(y)$, which is $\frac{f(x)}{x^2}=\frac{f(y)}{y^2}$ $\forall x,y\ne 0$ And so $f(x)=cx^2$ $\forall x\ne 0$, still true when $x=0$ Plugging this back in original equation, we get any $c$ if $\alpha=2$ and $c=0$ if $\alpha\ne 2$ And so solutions : $\forall \alpha$: $\boxed{f(x)=0\quad\forall x}$, which indeed fits If $\alpha =2$ : $\boxed{f(x)=cx^2\quad\forall x}$, which indeed fits, whatever is $c$

Denote $P(x,y)$ the assertion of the given F.E. $P(0,x)$ gives $\alpha xf(0)=0$ therefore $f(0)=0$, now $P(x,-x)$ gives $xf(-x)=(\alpha-1)xf(x)$ so $f(-x)=(\alpha-1)f(x)$ for all $x \ne 0$, now if $f(c)=0$ for $c \ne 0$ then $P(c,x)$ gives $f(c+x)=f(x)$ and then $P(x,c)$ gives $\alpha c f(x)=0$ which means $f(x)=0$ for all reals $d$, so now if $f$ was injective at $0$ then swap $x$ with $-x$ to get $f(x)=(\alpha-1)f(-x)$ which gives $\alpha=2$ which means $f$ is even. Now by $P(x,y)-P(x,-y)$ we get $x(f(x+y)-f(x-y))=4yf(x)$ so swapping $x,y$ gives that $\frac{4yf(x)}{x}=\frac{4xf(y)}{y}$ therefore $f(x)=c \cdot x^2$ for some constant $c$ and we can verify that any such $c$ work, thus we are done .

a short Sketch

Interesting! We claim the only working function for $\alpha \ne 2$ is $f \equiv 0$; whereas for $\alpha = 2$, the previously mentioned function works as does $f(x) \equiv cx^2$ for some $c \in \mathbb{R}$. It can be seen that these solutions work, now we prove they're the only ones. Let $P(x, y)$ denote the assertion. Since the only constant function is the zero function, assume $f$ is nonconstant. $P(0, y) \implies f(0) = 0$. $P(x, -x)$ for $x \ne 0 \implies xf(-x) = (\alpha -1)xf(x) \implies f(-x) = (\alpha-1)f(x)$ $$\implies f(x) = (\alpha-1)^2f(x) \implies \alpha = 2.$$(as $\alpha \ne 0$) Now we work on $x, y \ne 0$. Note that $$(1+\frac{2x}{y})f(y)+f(x) = f(x+y) = (1+\frac{2y}{x})f(x) + f(y)$$$$\implies \frac{f(x)}{x^2} = \frac{f(y)}{y^2},$$which gives the second curve of solutions described for $x \ne 0$. But $f(0) = c*0^2$, so we're done.