Consider the initial value problem
\begin{align*}
y' + y& = u_3(t)\\
y(0) & = 1.
\end{align*}
If we take the Laplace transform of both sides of \(y' + y = u_3(t)\text{,}\) we obtain
\begin{equation*}
sY(s) - y(0) + Y(s) = \frac{e^{-3s}}{s}.
\end{equation*}
Using the fact that \(y(0) = 1\) and solving for \(Y(s)\text{,}\) we get
\begin{equation*}
Y(s) = \frac{1}{s + 1} + \frac{e^{-3s}}{s(s+1)}.
\end{equation*}
Therefore,
\begin{equation*}
y(t) = {\mathcal L}^{-1} \left( \frac{1}{s + 1} \right) + {\mathcal L}^{-1} \left( \frac{e^{-3s}}{s(s+1)} \right).
\end{equation*}
The inverse Laplace transform of the first term is
\begin{equation*}
{\mathcal L}^{-1} \left( \frac{1}{s + 1} \right) = e^{-t}.
\end{equation*}
To compute the inverse Laplace transform of the second term, recall from
Example 6.1.7 that if
\({\mathcal L}(f) = F(s)\text{,}\) then
\begin{equation*}
{\mathcal L}(u_a(t) f( t- a)) = e^{-as} F(s).
\end{equation*}
Using partial fractions to obtain
\begin{equation*}
\frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s - 1}.
\end{equation*}
Hence,
\begin{equation*}
{\mathcal L}^{-1} \left( \frac{e^{-3s}}{s(s+1)} \right) = {\mathcal L}^{-1} \left( \frac{e^{-3s}}{s} \right) - {\mathcal L}^{-1} \left(\frac{e^{-3s}}{s+1} \right) = u_3(t) - {\mathcal L}^{-1} \left( \frac{e^{-3s}}{s+1} \right).
\end{equation*}
If \(g(t) = u_3(t) e^{-(t - 3)}\text{,}\) then the Laplace transform of \(g(t)\) is
\begin{equation*}
{\mathcal L}(g) = e^{-3s} {\mathcal L}(e^{-t}) = \frac{e^{-3s}}{s + 1}.
\end{equation*}
Thus,
\begin{equation*}
y(t) = e^{-t} + u_3(t)\left( 1 - e^{-(t - 3)} \right).
\end{equation*}