It's easy to solve $\displaystyle f'(x) = xf(x) $, but is there a way to solve $\displaystyle f'(x) = xf(x-1)? $
Setting $\displaystyle \displaystyle f(s)= \mathcal{L} \{y(t)}\}$ the first order DE equation $\displaystyle y^{'}(t)= t\ y(t-1)$ became the following linear first order DE in the variable s...
$\displaystyle f^{'} (s) = (1 -s\ e^{s})\ f(s) + e^{s}\ y(0)$ (1)
... that can be solved with a 'standard' approach...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$