Solving the inverse laplace transform of $\frac{1}{s - e^{-s}}$

Question

I'm trying to solve de differential equation $$ y'(t) = y(t-1) H(t-1) $$ where $H(t)$ is the Heaviside step function, but I got stuck solving an Inverse Laplace Transform: $$ \mathcal{L}\{y'(t)\} = \mathcal{L}\{y(t-1)H(t-1)\} \\ sY(s)-y(0) = e^{-s}Y(s) \\ s - \frac{y(0)}{Y(s)} = e^{-s} \\ \frac{y(0)}{Y(s)} = s - e^{-s} \\ \frac{Y(s)}{y(0)} = \frac{1}{s - e^{-s}} \\ \frac{\mathcal{L}^{-1}\{Y(s)\}}{y(0)} = \mathcal{L}^{-1}\left\{ \frac{1}{s - e^{-s}} \right\} \\ \frac{y(s)}{y(0)} = \mathcal{L}^{-1}\left\{ \frac{1}{s - e^{-s}} \right\} $$ Can anyone help? Thanks

Answer 1

For the backward transform, you need all the zeros of $s-e^{-s}$. These are given by the $k$-th branch of the Lambert W function $W_k(1)$. The function is thus given by $$ \mathcal{L}^{-1} \left\{ \frac{1}{s-e^{-s}} \right\} = \sum_k \mathop{\rm Res}_{s= W_k(1)} \frac{e^{s t}}{s-e^{-s}} .$$ As the poles are simple, we have that $$ \mathcal{L}^{-1} \left\{ \frac{1}{s-e^{-s}} \right\} = \sum_k \frac{\exp\left[ W_k(1) t \right]}{1+ W_k(1)} $$