I am not sure whether this question is even well-posed. But today I learnt that $e^Df(x) = f(x+1)$ where $D$ is differential operator and
$e^D \triangleq \sum_{i=0}^{\infty} \frac{D^i}{i!}.$
(ref. Dan Piponi's answer)
So I was curious as to whether the differential equation
$\frac{df(t)}{dt} = e^Df(t) = f(t+1)$
has any solutions apart from $f = 0$?