Functional equation in one variable: $(x-2)f(2x)+xf(x-2)=0$ Let $f$  be a function such that $$(x-2)f(2x)+xf(x-2)=0$$ where the domain of $f$ is set of real numbers. I want to find $f(x)$
 A: First of all, by letting $x=-2$, we get $-4\cdot f(-4)-2f(-4)=0$, so $f(-4)=0$.
Now let's define $y(x)=f(x-4)/(x-4)$, so $f(x)=x\cdot y(x+4)$. Then
$$(x-2)\cdot2x\cdot y(2x+4)=-x(x-2)y(x+2)\\
y(2x+4)=-{1\over2}y(x+2)\\
y(2t)=-{1\over2}y(t)$$
That's it. We're left with a great deal of freedom (that is, unless you insist on continuity). You define $y(t)$ arbitrarily at $(1,2]$, then apply the equation above to extend it to $(2,4]$, then to $(4,8]$ and further, then apply it in reverse to extend to $({1\over2},1]$ and so on, then repeat the whole thing with negative $t$, then work your way back to $f(x)$.
If you're willing to accept an answer which is continuous everywhere but one point, then choose an arbitrary number $a$, define $y(t)$ on $[a,2a]$ as a freehand curve with $y(a)=y(2a)=0$, and then continue as above.
If you need the answer to be continuous everywhere, then I suppose $f(x)=0$ is the only solution.
A: Hint:
$(x-2)f(2x)+xf(x-2)=0$
$\dfrac{f(2x)}{x}+\dfrac{f(x-2)}{x-2}=0$
Let $g(x)=\dfrac{f(x)}{x}$ ,
Then $2g(2x)+g(x-2)=0$
