If $f(x)= \int_0^x e^t f(t)dt$ for all real $x$, then $f(x)=ce^{(e^x-1)}$ for some real $c$

Question

Let $f$ be a continuous function such that, for all $x \in \mathbb{R}$, it is true that $$f(x)= \int_0^x e^t f(t)dt$$ Prove that there exists $c \in \mathbb{R}$ such that $f(x)=ce^{(e^x-1)}$.

I am not quite sure how to proceed with this one. Maybe using logarithm properties, but just do not know.

Answer 1

Using Newton-lebinitz,
$$f'(x) = e^x f(x)$$
$$\cfrac{d(f(x))}{f(x)} = e^x dx$$
$$ln(f(x)) = e^x + C \implies f(x) = e^{e^x + C}$$

using $f(0) = 0$, you can find C

Answer 2

It is an ODE problem that $f'(x)=e^{x}f(x)$, now use separating method that $dy/y=e^{x}dx$ to get that $\log y=e^{x}+C$, so $f(x)=ce^{e^{x}}$.