When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$? The question is:

When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?

The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?
Edit
I should add that $f:\mathbb R\to\mathbb R$ to the question.

This is an example of a delay differential equation, and it has a very large number of solutions.  I will begin by discussing solutions $f\colon[0,\infty)\to\mathbb{R}$, and then extend to solutions on the real line.
Suppose first that we have defined $f$ on the interval $[0,1]$, say $f(x) = f_0(x)$ for some differentiable function $f_0\colon[0,1]\to\mathbb{R}$.  For extending solutions forwards, the best way to write the equation is
$$
f(x+1) \;=\; f(x) + f'(x).
$$
Using this equation, it is not hard to define $f$ on the interval $[1,2]$. Specifically, if $x\in[1,2]$, then
$$
f(x) \;=\; f_0(x-1) + f_0'(x-1).
$$
Note that this function will only be differentiable if $f_0$ was twice differentiable on $[0,1]$.  Moreover, the derivatives at $1$ will only match up if $f_0'(1) = f_0'(0) + f_0''(0)$.  Assuming this is the case, it is easy to extend again to $[2,3]$.  Specifically, if $x\in[2,3]$, then
$$
f(x) \;=\; f(x-1) + f'(x-1) \;=\; f_0(x-2) + 2 f_0'(x-2) + f_0''(x-2).
$$
In general, assuming that:


*

*The function $f_0$ is $C^\infty$ on $[0,1]$ and,

*$f_0^{(k)}(1) = f_0^{(k)}(0) + f_0^{(k+1)}(0)$ for all $k\geq 0$,


we can extend $f$ to all of $[0,\infty)$ by the formula
$$
f(x) \;=\; \sum_{k=0}^{\lfloor x\rfloor} \,\binom{\lfloor x\rfloor}{k}f_0^{(k)}(x-\lfloor x\rfloor)
$$
Now let's discuss how to extend backwards.  Assuming we have defined $f$ on $[0,\infty)$, the formula for $f$ on $[-1,0]$ is determined by the initial-value problem
$$
f'(x) + f(x) \;=\; f_0(x+1),\qquad f(0) = f_0(0).
$$
This is a first-order differential equation, so it must have a solution.  Moreover, assuming $f_0$ is $C^\infty$ on $[0,1]$, it follows from the Picard-Lindelöf Theorem that there is a unique $C^\infty$ solution for $f$ on $[-1,0]$.  Note that the derivatives at $0$ will automatically match, since $f_0(1) - f(0) = f_0'(0)$.  Continuing in this fashion, we can extend any solution backwards all the way to $-\infty$.
We conclude that any function $f_0\colon[0,1]\to\mathbb{R}$ satisfying the two conditions listed above extends to a solution on the entire real line.
Edit: Incidentally, there is a simple integral formula for $f$ on $(-\infty,0]$, which can be used to compute values for $f$ on a computer:
$$
f(x) \;=\; f(0) + e^{-x-1}\int_{x+1}^1 e^{t}\,f(t)\,dt
$$
This can also be used to provide an alternative proof that the function extends backwards, without appealing to existence and uniqueness of ODE's.  I found this formula by solving the initial value problem $f(x) + f'(x) = f(x+1)$ using integrating factors.
