XbenX wrote: Given a real number $\alpha$, find all pairs $(f,g)$ of functions $f,g :\mathbb{R} \to \mathbb{R}$ such that $$xf(x+y)+\alpha \cdot yf(x-y)=g(x)+g(y) \;\;\;\;\;\;\;\;\;\;\; ,\forall x,y \in \mathbb{R}.$$ Let $P(x,y)$ be the assertion $xf(x+y)+\alpha yf(x-y)=g(x)+g(y)$ Let $c=f(1)$ $\boxed{\text{S1 : }f(x)=g(x)=0\quad\forall x}$ is a solution. And so let us consider from now only non allzero solutions $P(0,0)$ $\implies$ $g(0)=0$ $P(x,0)$ $\implies$ $xf(x)=g(x)$ $P(0,x)$ $\implies$ $\alpha xf(-x)=g(x)$ and so $f(x)=\alpha f(-x)$ $\forall x\ne 0$ This implies, since non allzero, $\alpha^2=1$ Let $x\ne y$ : $f(x-y)=\alpha f(y-x)$ and so : $P(x,y)$ may be written $xf(x+y)+yf(y-x)=g(x)+g(y)$ $\forall x\ne y$ Then $P(y,x)$ implies $yf(x+y)+\alpha xf(y-x)=g(y)+g(x)$ Subtracting, we get $(x-y)f(x+y)+(y-\alpha x)f(y-x)=0$ Setting there $x=\frac{X-1}2$ and $y=\frac{X+1}2$, (note that $x\ne y$), this becomes $f(X)=cX\frac{1-\alpha}2+c\frac{1+\alpha}2$ And so : $\boxed{\text{If }\alpha\ne\pm 1\text{, No new solution}}$ $\boxed{\text{If }\alpha=1\text{, S2 : }(f,g)=(c,cx)\quad\forall x}$ which indeed is a new solution, whatever is $c\in\mathbb R\setminus 0$ $\boxed{\text{If }\alpha=-1\text{, S3 : }(f,g)=(cx,cx^2)\quad\forall x}$ which indeed is a new solution, whatever is $c\in\mathbb R\setminus 0$

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Let $P(x,y)$ denote assertion of given functional equation: 1) $P(0,0)$: $g(0)=0$ 2) $P(x,0)$: $xf(x)=g(x)$ . As a result original equation becomes $xf(x+y)+\alpha yf(x-y)=xf(x)+yf(y)$ 3) $P(0,x)$: $\alpha f(-y)=f(y)$ 4) $P(x,x)$: $f(2x)= 2f(x) - \alpha f(0)$ 5) $P(x,-x)$: $xf(0)- \alpha xf(2x) = xf(x) - xf(-x)= xf(x) - \alpha f(x) $. Using identity from point 4) we obtain that: $f(0)(\alpha^2+1)=f(x)(\alpha+1)$. Case 1 $\alpha = -1$ Original equation becomes $xf(x+y)- yf(x-y)=xf(x)+yf(y)$. Comparing $P(x,y)$ and $P(x,-y)$ yields: $(x+y)f(x-y)=(x-y)f(x+y)$. Replacing $x$ by $\frac{x+1}{2}$ and $y$ by $\frac{x-1}{2}$ gives us $f(x)=xf(1)$ Case 2 $f(x)=f(0)(\alpha-1)$ This implies that $f$ is constant Answers: $f(x)=Cx$, $g(x)=Cx^2$, when $\alpha =-1$ and $f(x)=C, g(x)=Cx$, where $C$ is real number