Simple differential-difference equation I was doing some exam practice when I almost surely went the wrong way in my method and ended up with the following equation:
$$f'(x) = f(x) - f(x-1)$$
Although not what I was intended to find, I'm still curious to know if there are methods to solve this. We can spot the obvious solution $f(x) = Ax+B$ , but I could not find a way to prove this was the only solution, or find any other solutions. Any insight would be helpful. For context I am a first-year undergrad.
 A: Preliminary remark: We can multiply LHS and RHS by the same quantity; therefore we can assume WLOG that $f(0)=1$.
Laplace Transform (L.T.) you may have studied is especially adapted to delay differential equations like this one.
Using tables (I don't give details) its L.T. is
$$sF(s)-f(0)=F(s)-e^{-s}F(s)$$
We can extract $F(s)$ under the form:
$$F(s)=\frac{1}{(s-1)+e^{-s}}=\frac{1}{(s-1)}\frac{1}{1+\frac{e^{-s}}{s-1}}$$
The second fraction can be expanded into a (geometric) series:
$$F(s)=\frac{1}{(s-1)}\left(1-\frac{e^{-s}}{(s-1)}+\frac{e^{-2s}}{(s-1)^2}-\cdots \right)$$
$$F(s)=\sum_{k=0}^{\infty}(-1)^k\frac{e^{-ks}}{(s-1)^{k+1}}$$
whose inverse Laplace Transform is, using again tables:
$$f(x)=\sum_{k=0}^{\infty}H(x-k)(-1)^k \frac{(x-k)^k}{k!}e^{x-k}\tag{1}$$
where $H$ is the Heaviside step function ($H(x)=0$ if $x<0$; $H(x)=1$ if $x \ge 0$).
In fact, for a given $x$, (1) is a finite sum: intuitively, proceeding by increasing values of $x$, each time variable $x$ crosses an integer $k$, the corresponding factor $H(x-k)$ "activates" a new term.
