Let us consider a differential delay equation (DDE) with $a,b\in\mathbb{R}$:
$$ \frac{d}{dt}y(t)=ay(t)+bH(t-1)y(t-1),~0\le t<\infty, $$
where $H(t)=\int_{-\infty}^t\delta(t')dt'$ is the Heaviside step function. Let $\hat{y}(s)=\int_0^{\infty}e^{-st}y(t)dt$ be the Laplace transform of $y(t)$.
- i). Find $\hat{y}(s)$ and, using the inverse Laplace transform, solve the given DDE.
Here is what I have tried so far:
$$ \begin{aligned} \mathcal{L}\left(\frac{d}{dt}y(t)\right)(s)&=\mathcal{L}\left(ay(t)+bH(t-1)y(t-1)\right)(s)\\ s\hat{y}(s)-y(0^{-})&=a\hat{y}(s)+be^{-s}\hat{y}(s), \end{aligned} $$
therefore
$$ \begin{aligned} s\hat{y}(s)-a\hat{y}(s)-be^{-s}\hat{y}(s)&=y(0^{-})\\ \hat{y}(s)\left(s-a-be^{-s}\right)&=y(0^{-})\\ \hat{y}(s)&=\frac{y(0^{-})}{s-a-be^{-s}}\\ y(t)&=\mathcal{L}^{-1}\left(\hat{y}(s)\right)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{st}\hat{y}(s)ds\\ &=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{y(0^{-})e^{st}}{s-a-be^{-s}}ds\\ &=\frac{y(0^{-})}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{st}}{s-a-be^{-s}}ds \end{aligned} $$
I need help: I don't know how to get the exact value of the inverse Laplace transform of $\hat{y}(s)$