I would like to solve the functional differential equation $$f'(x)=f(x-e)$$ I've solved functional equations before, and I've solved differential equations, but I've never solved a functional differential equation before, so please be patient with me.
I've found a class of solutions through guesswork: $$f(x)=Ce^{x/e}$$ If this were just a differential equation, I'd be done, because the highest derivative of $f$ in the DE is $f'$, and I have one arbitrary constant in my class of solutions.
Are there solutions not in this form? When I'm dealing with an FDE, how many arbitrary constants I need to have an exhaustive solution?
If the equation is as in the main body of your question (not as in the title!) then it is a delay differetial equation.
The initial value problem has the form $$ \tag{$*$} \begin{cases} f'(x) = f(x - e) & \text{ for } x \in [0, \infty) \\ f(x) = u_0(x) & \text{ for } x \in [-e, 0] \end{cases} $$ where $u_0 \colon [-e, 0] \to \mathbb{R}$ is a continuous (say) function (initial condition).
How to find a solution to $(*)$? Notice that the derivative of the solution for $x \in [0, e]$ equals just the value of $u_0$ at $x - e$, therefore $$ f(x) = u_0(0) + \int\limits_{0}^{x} f(u_0(\xi-e)) \, d\xi, \quad x \in [0, e]. $$ We repeat this procedure for $x \in [e, 2e]$, taking the restriction of the solution to $[0, e]$ instead of $u_0$. And so on (Myshkis' method of steps).
As you can see, no finite number of arbitrary constants can give an exhaustive solution of the equation.
But if you are interested in the equation $$ f'(x) = f(x + e) $$ (as in the title), the situation changes dramatically.