Simple differential-difference equation

Question

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.

Then $f$ can be differentiated again: $$f’’(x)=(f(x)-f(x-1))-(f(x-1)-f(x-2))\=f(x)-2f(x-1)+f(x-2)$$ In general: $$f^{(n)}(x)=\sum_{k=0}^n (-1)^k \binom{n}k f(x-k)$$ — Thomas Andrews, May 29 '21 at 21:33
Does this answer your question? When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$? — Jean Marie, May 29 '21 at 21:39
Read the first two answers because the other question, because the first answer makes a conjecture the second answer proves incorrect. — Thomas Andrews, May 29 '21 at 21:51
If $,g(x):=f(-x),,$ then $,g(x+1)-g(x)=g'(x),$ is equivalent to $,f'(x) = f(x) - f(x-1).,$ — Somos, May 29 '21 at 22:13

Jean Marie · Accepted Answer · 2021-05-30T07:05:36.563

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.

Simple differential-difference equation

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