The exercise is the following:
Using Laplace Transforms, find the solution to $y'(t) + y(t-1) = t^2$, with $y(t) = 0$ for $t \leq 0$.
Being the first exercise I saw that showed something like $y(t-1)$ and knowing that $\mathcal{L}\{u_1(t)y(t-1)\}(s)=e^{-s}\mathcal{L}\{y(t)\}(s)$ where $u_1(t)$ is the Heaviside step function, I tried multiplying the whole expression by $u_1(t)$, but then didn't know how to proceed with $\mathcal{L}\{u_1(t)y(t)\}(s)$.
Performing the Laplace transform of the equation you obtain $$sY(s)-\underbrace{y(0^{-})}_{=0}+e^{-s}Y(s)=\frac{2}{s^{3}}$$ It follows $$Y(s)=\frac{2}{s^{3}}\frac{1}{s+e^{-s}}=\frac{2}{s^{4}}\frac{1}{1+\frac{1}{s}e^{-s}}=2\sum_{n=0}^{\infty}\frac{(-1)^{n}}{s^{n+4}}e^{-ns}$$ It follows $$y(t)=\int_{\eta-i\infty}^{\eta+i\infty}\frac{ds}{2\pi i}e^{st}Y(s)=2\sum_{n=0}^{\infty}(-1)^{n}\int_{\eta-i\infty}^{\eta+i\infty}\frac{ds}{2\pi i}e^{st}\frac{1}{s^{n+4}}e^{-ns}$$ $$=\sum_{n=0}^{\infty}\frac{2(-1)^{n}}{(n+3)!}(t-n)^{n+3}\Theta(t-n).$$ Where we've used Cauchy theorem in the last line.