What $f(x)$ functions satisfy the following: $$f(x+y)-f(x)=g(y)f'(x)$$ I know this is satisfied for $f(x)=c_1+c_2 e^{c_3 x}$. Note that $f(x+y)-f(x)=(e^{c3 y}-1) c_2 e^{c_3 x}$, and that $f'(x)=c_2 e^{c_3 x}$. Thus in this case $g(y)=(e^{c3 y}-1)$.
My question is, is this the only function for which the above holds?
On
Set $x=0$, \begin{align} f(y) - f(0) = g(y)f'(0) = g(y) c \end{align} where $c=f'(0)$. Further taking derivatives in $y$ gives \begin{align} f'(x) = g'(x) c \end{align} Let's work with $c\neq 0$. (If $c=0$, it is obvious from the above that $f(x) = f(0) = constant.$)
Substituting $f(y) = g(y) c + f(0)$ into the original equation, we can get an equation that contains the funcion $g$ only, \begin{align} g(x+y) - g(x) = g(y) \frac{f'(x)}{c} = g(y)g'(x) \end{align} Further taking derivative in $y$, we have \begin{align} g'(x+y) = g'(y) g'(x) \end{align} This is a standard problem of the form that $F(x+y)=F(x)F(y)$. With the conditions that $g'(0)=1$ from $f'(x) = g'(x) c$, \begin{align} g'(x) = e^{ax} \end{align} for some constant $a$. Therefore \begin{align} g(x) = \frac{1}{a}(e^{ax}-1) \end{align} since we require that $g(0)=0$ from the first equation. Then $f(x)$ can only take the form you suggested after taking into account the constant.
Differentiate the equality with respect to $y$, and you get that $$f'(x+y)=g'(y)f'(x),$$ so that $$\frac{f'(x+y)}{f'(x)}=g'(y).$$ The left hand side is thus independent of $x$. It's derivative with respect to $x$ is thus zero, so $$\frac{f''(x+y)f'(x)-f''(x)f'(x+y)}{f''(x)^2}=0$$ or just $$f''(x+y)f'(x)-f''(x)f'(x+y)=0.$$ This is true for all $x$ an $y$, so we can put $x=0$ to see that $$f''(y)f'(0)-f''(0)f'(y)=0.$$ We thus see that $f$ satisfies a second order linear differential equation with constant coefficients, which we can solve.
One has to take care of the possibility that $f'(x)$ is zero at some point, though.
But if $f'(x_0)=0$ for some $x_0$, then $f(x)-f(x_0)=g(x-x_0)f'(x_0)=0$ for all $x$, so $f$ is constant.