Solving an ODE where the function is evaluated on different points How does one solve a differential equation of the form
$$f'(x) = f(x) - f(x-1)? $$
Is there any analytic way to obtain a solution?
 A: I haven't found any good general ways to analytically construct solutions to DDEs (I have asked about it before here, not to much avail). In this case, we can do kind-of a guess-and-check type approach by looking for solutions of the form $f(x)=e^{rx}$. Linear combinations of these solutions will give more general solutions to the DDE. We obtain the equation $$
re^{rx} = e^{rx} e^{-r} - e^{rx}
$$
hence $$
r + 1 = e^{-r}
$$
This has a unique real solution $r = 0$, corresponding to the solution $f(x) = 1$. Not very exciting, but fear not, the non-real solutions to $r+1 = e^{-r}$ can give us additional real-analytic solutions. The equation solved gives us $r = W_n(e) - 1$, where $W_n$ is the $n$th branch of the Lambert-W function. Let $r_n = W_n(e) - 1$. The conjugate-symmetry of the branches of the W-function implies that $r_n = \overline{r_{-n}}$, hence $e^{r_n x} + e^{r_{-n} x}$ is a real-valued function, which also solves the equation, as is $i\cdot\left(e^{r_n x} - e^{r_{-n} x}\right)$. We will have the general solutions to the DDE should have the Fourier-series-like form $$
f(x) = \sum_{n=-\infty}^{\infty} a_n e^{r_n x}
$$
where $a_n$ is some sequence of constants such that the series converges. If $a_{-n} = \overline{a_n}$, then the series will be something real-valued. If $a_{n}$ decays fast enough, then it will be analytic.
