Let $y : \mathbb{R}\to\mathbb{R}$ be a differentiable function satisfying $y'(t)=y(t-1)$ for all $t$.
Is it possible to give an asymptotics for $y(t)$ as $t\to \infty$?
It is clear to me that there should be an asymptotics of the form $y(t) \sim C\frac{t^{\lfloor t \rfloor +1}}{(\lfloor t\rfloor +1)! }$, inducting on intervals of lenghts 1. How is it possible to prove it?
On
With the help of the Laplace transform
$$ Y(s) = \frac{\int_{-1}^0 e^{-s t} y(t) \, dt+e^s y(0)}{e^s s-1} $$
now considering $y(t)=0,\ \ -1\le t\le 0$ and $y(0)=1$
$$ Y(s) = \frac{1}{s-e^{-s}} $$
now taking the denominator $d(s) = s-e^{-s}$ and making $s=x + iy$ we have
$$ d(s) =x - e^{-x} \cos (y)+i \left(e^{-x} \sin (y)+y\right) $$
so the denominator has infinite zeros but we pick one of them which is located at $x = W(1), y = 0$ where $W(\cdot)$ indicates the Lambert function. Here $W(1) = 0.567143$ so $y(t)$ is unstable because $Y(s)$ has at least one pole in the right complex plane hence it grows exponentially.
This is a delayed DE. Let $y(t)=ce^{\lambda t}$ and then the characteristic equation is $$ \lambda=e^{-\lambda} $$ which has a unique solution $\lambda>0$. Therefore $y(t)\to\infty$ as $t\to\infty$.