I'm trying to find functions $f(x)$ and $g(y)$ such that $$f(x)\cdot g(y) = x + y$$
I can't seem to find a single solution to this problem. Anything I try becomes of the form $f(x,y) \cdot g(y) or f(x) \cdot g(x,y)$
Here is my work so far:
$$f(x)g(y) = x + y$$
$$f'(x)g(y) + f(x)g'(y)\frac{dy}{dx} = 1 +\frac{dy}{dx} $$
$$f'(x)g(y) - 1 = \frac{dy}{dx} - f(x)g'(y)\frac{dy}{dx} $$
$$f'(x)g(y) - 1 = \frac{dy}{dx} (1 - f(x)g'(y))$$
$$\frac{f'(x)g(y) - 1}{1 - f(x)g'(y)} = \frac{dy}{dx} $$
But since $g(y)$ doesn't have a known or restricted order, I have no idea what $f(x)$ would be:
I am considering taking higher derivatives and then using substitutions (ex: $g(y) = (x+y)/f(x))$ but I'm not sure if that will work.
You have that $$\tag 1 g(0)f(0)=0$$ while $$f(1)g(0)=1$$ $$f(0)g(1)=1$$ The first equations says that either $g(0)=0$ or $f(0)=0$, or both. But this contradicts the last two equations.
ADD Note that what I wrote holds also when $x=-y$, and when $x=0$, $y=\alpha$ a constant, and vice-versa. In fact, the last observation means your function cannot be defined for any value of $x$ or $y$. Indeed, we have that for any $\alpha,\beta\in\Bbb R$
$$f(0)g(\alpha)=\alpha$$ $$f(\beta)g(0)=\beta$$
But beacuse of $(1)$, the last relations are impossible, so we cannot define either $f$ or $g$ for any $x\in\Bbb R$.