How to solve differential equations of the form $f'(x) = f(x + a)$

What could one do to find analytic solutions for $f'(x) = f(x + a)$ for various values of $a$?

I know that $c_1\sin(x + c_2)$ is solution when $a = \frac{1}{2}\pi$, and of course $c_1e^x$ when $a = 0$.

For instance, is there a function satisfying $f'(x) = f(x + 1)$? What about negative or imaginary $a$?

Is there possibly a generalization?

Note that $f(x) = e^{cx}$ is a solution if $c = e^{ca}$. This can be solved for $c$ in terms of the Lambert W function: $c = - W(-a)/a$. The Lambert W function has infinitely many branches, each of which gives a solution. There are two real solutions if $0 < a < 1/e$, one if $a \le 0$ or $a = 1/e$, none if $a > 1/e$. Of course, linear combinations of solutions of your delay-differential equation are solutions. In particular, given a pair of complex conjugate $c = r \pm i s$ (for real $a$), we get real solutions of the delay-differential equation by taking the real and imaginary parts of $e^{cx}$, namely $e^{rx} \cos(sx)$ and $e^{rx} \sin(sx)$.

For example, for $a = 1$ the smallest values of $c$ are approximately $ 0.3181315052 \pm 1.337235701\,i$, $2.062277730 \pm 7.588631178\,i$, $ 2.653191974 \pm 13.94920833\,i$, $ 3.020239708 \pm 20.27245764\,i$, $3.287768612 \pm 26.58047150\,i$, $3.498515212 \pm 32.88072148\,i$, $3.672450069 \pm 39.17644002\,i$.

To make things a little more natural, I will consider the equation $f'(x)=f(x-a)$, $a>0$. As observed in the comments, this is a (linear) delay differential equation. Given a continuous function $\phi\colon[-a,0]\to\mathbb{R}$, there is a unique continuous function $f\colon[-a,\infty)\to\mathbb{R}$, with continuous derivative on $(0,\infty)$, such that $f(x)=\phi(x)$ for $x\in[-a,0]$ and $f'(x)=f(x-a)$ for $x>0$.

The solution $f$ is constructed recursively on the intervals $[na,(n+1)a]$, $n=0,1,2\dots$

If $x\in[0,a]$, then $$ f(x)=f(0)+\int_0^xf'(t)dt=f(0)+\int_0^xf(t-a)dt=\phi(0)+\int_{-a}^{x-a}\phi(t)dt. $$ If $x\in[a,2a]$, then $$ f(x)=f(a)+\int_a^xf'(t)dt=f(a)+\int_a^xf(t-a)dt=f(a)+\int_0^{x-a}f(t)dt, $$ which is well defined since in the previous step we calculated $f$ on $[0,a]$. Iterating this procedure, we can find $f$ on any interval $[0,na]$. Only on rare ocasions it is possible to obtain a closed formula for $f$.

Example: $a=1$, $\phi(x)=x^2$.

For $x\in[0,1]$ $$ f(x)=\phi(0)+\int_{-1}^{x-1}t^2dt=x-x^2+\frac{x^3}{3}. $$ For $x\in[1,2]$ $$ f(x)=f(1)+\int_{0}^{x-1}\Bigl(t-t^2+\frac{t^3}{3}. \Bigr)dt=\frac{5}{4} - \frac{7 x}{3} + 2 x^2 - \frac{2 x^3}{3} + \frac{x^4}{12}. $$ For $x\in[2,3]$ $$ f(x)=f(2)+\int_{1}^{x-1}f(t)dt=\frac{x^5}{60}-\frac{x^4}{4}+\frac{3 x^3}{2}-\frac{13 x^2}{3}+\frac{19 x}{3}-\frac{197}{60}. $$ The graph shows the smoothness of $f$.