Determining all $f : \mathbb R \to \mathbb R$ that satisfy $xf(x) - yf(y) = (x-y)f(x+y)$

Solution 1:

Write the given functional equation for the three pairs $(x,y)$, $(y,z)$, $(z,x)$, and add up. You then obtain $$0=(x-y)f(x+y)+(y-z)f(y+z)+(z-x)f(z+x)\ .$$ We now have a functional equation with three free variables. Put $$x:={t+1\over2},\quad y:={t-1\over2},\quad z:={1-t\over2}\ ,$$ and you get $$0=1\cdot f(t)+(t-1)\cdot f(0)-t\cdot f(1)\ ,$$ or $$f(t)=f(0)+t\bigl(f(1)-f(0)\bigr)\qquad\forall t\in{\mathbb R}\ .$$ This shows that $f$ has to be of the form $f(t)=at+b$ with arbitrary constants $a$, $b$.

Solution 2:

Here are a few hints : use $\frac{xf(x)-(y+z)\bigg(\frac{yf(y)-zf(z)}{y-z}\bigg)}{x-(y+z)}=\frac{(x+y)\bigg(\frac{xf(x)-yf(y)}{x-y}\bigg)-zf(z)}{(x+y)-z}$ to show that $f(z)=\frac{xf(y)-yf(x)+z(f(x)-f(y))}{x-y}$. Deduce that there are two constants $a,b$ such $f(z)=az+b$, then plug into the original equation.

Solution 3:

Here's another interesting solution: $$xf(x)−yf(y)=(x−y)f(x+y)\iff x(f(x+y)-f(x))=y(f(x+y)-f(y))$$ Let, $$g(x)=\frac{f(x)-f(0)}{f(1)-f(0)} \iff g(x)(f(1)-f(0))+f(0)=f(x)$$ So, now, $g(1)=1$ and $g(0)=0$ and $$x(g(x+y)-g(x))=y(g(x+y)-g(y))$$ $$y=1 \implies x(g(x+1)-g(x))=g(x+1)-1 \tag{1}$$ $$y=-x \implies-g(x)=g(-x) \tag{since $g(0)=0$}$$ $$y=1-x \implies x(1-g(x))=(1-x)(1-g(1-x))$$ Replacing $x$ with $-x$ in the previous equation yields (knowing the fact that $g$ is odd) $$ -x(1+g(x))=(x+1)(1-g(x+1))\iff 2x+1=x(g(x+1)-g(x))+g(x+1)$$ $$\implies2g(x+1)=2(x+1) \tag{by (1)}$$ $$\implies g(y)=y \ \ \ \ \forall \ y \in \mathbb{R}$$ $$\implies f(x)=ax+b \ \ \ \ \forall \ a,b,x \in \mathbb{R}$$