Solution: Recall that if $\mathcal{L} (y(t)) = Y(s)$, then \begin{equation}\label{eq:24Mar2023-1} \mathcal{L} \left( y^\prime (t) \right) = sY(s) -y(0). \end{equation} Some properties of the Laplace's transformation are as follows: \begin{align*} \mathcal{L} (f+g) & = \mathcal{L} (f) + \mathcal{L} (g)\\ \mathcal{L} (\alpha f) & =\alpha \mathcal{L} (f)\\ \mathcal{L} \left( e^{-nt} \right) & = \frac{1}{s+n}. \end{align*} So, \begin{align*} & y^\prime (t) + 5y(t) = e^{-5t} \\ \implies &\mathcal{L} \left( y^\prime (t) + 5y(t) \right) = \mathcal{L} \left( e^{-5t} \right) \\ \implies &\mathcal{L} \left( y^\prime (t) \right) + \mathcal{L} (5y(t)) = \frac{1}{(s+5)} \\ \overset{\text{from }\eqref{eq:24Mar2023-1}}{\implies} & sY(s) - 0 + 5Y(s) = \frac{1}{(s+5)} \\ \implies & Y(s) = \frac{1}{(s+5)^2} \\ \implies & y(t) = \mathcal{L}^{-1} \left( Y(s) \right) = \mathcal{L} ^{-1} \left( \frac{1}{(s+5)^2} \right) \\ \implies & y(t) = te^{-5t}. \end{align*} Therefore, the solution of the given differential equation will be \[ \textcolor{blue}{\boxed{y(t) = te^{-5t}}} \]