How to solve this Functional Equation: $f(x+1) - f(x) = e^{ax}$?

Call $T_a$ the set of functions such that $f(x+1)-f(x)=e^{ax}$ for all $x\in\Bbb R$, and consider the map $\Phi_a:T_a\to \Bbb R^{[0,1)}$ defined by $\Phi_a(f)=\left.f\right\rvert_{[0,1)}$. $\Phi_a$ is a bijection, and specifically $$[\Phi_a^{-1}(g)](x)=\begin{cases}g(\{x\})+\frac{e^{a\{ x\}}-e^{ax}}{1-e^a}&\text{if }a\ne 0\\ g(\{x\})+\lfloor x\rfloor&\text{if }a=0\end{cases}$$

Where $\{x\}$ is the real number in $[0,1)$ such that $x-\{x\}\in\Bbb Z$, and $\lfloor x\rfloor=x-\{x\}$ is the largest integer smaller or equal to $x$.

It is clear that $\Phi_a$ is injective and that $\left.\Phi_a^{-1}(g)\right\rvert_{[0,1)}=g$, therefore it is only a matter of verifying that $\Phi_a^{-1}(g)(x+1)-\Phi_a^{-1}(g)(x)=e^{ax}$ for all $x$, which is just algebra.

Added: Another approach is to notice that if you set $g(x)=f(x)+\frac{e^{ax}}{1-e^a}$ if $a\ne 0$ and $g(x)=f(x)-1$ if $a=0$, then you obtain that $g(x+1)=g(x)$, i.e. that the solutions are the functions in the form $\text{(1-periodic function)} +\frac{e^{ax}}{e^a-1}$.

Since $x+1$ is not cyclic, there's no way (that I know of) to arrive at a set of equations you could solve. Unfortunately my solution is on the intuitive side, not sure how easy this method is to generalize.

Let's write $f(0)=c$. Now, substituting $x=0,1,2,...$ into the equation we get $f(1)=c+1$, $f(2)=c+1+e^a$, $f(3)=c+1+e^a+e^{2a}$ etc. In general, for $k\in\mathbb{N}$ we have:

$$f(k)=c+\sum_{i=0}^{k-1}e^{ia}=c+\frac{e^{ak}-1}{e^a-1}$$

Plugging $f(x)=c+\frac{e^{ax}-1}{e^a-1}$ into the functional equation, it is easy to work out that the identity holds.

Again, sorry for the hand-waving technique. Though summing up a series in this way could work for other functional equations of this type.