Example D.19. Example 1: Solving a First-Order Linear Differential Equation.
Consider the differential equation:
\begin{equation*}
y' + 3y = 6e^{2t}, \quad y(0) = 1
\end{equation*}
Applying the Laplace transform to both sides, we get:
\begin{equation*}
sY(s) - y(0) + 3Y(s) = \frac{6}{s-2}
\end{equation*}
Substituting the initial condition \(y(0) = 1\text{,}\) the equation becomes:
\begin{equation*}
sY(s) - 1 + 3Y(s) = \frac{6}{s-2}
\end{equation*}
Rearranging and solving for \(Y(s)\text{:}\)
\begin{equation*}
Y(s)(s + 3) = 1 + \frac{6}{s-2}
\end{equation*}
\begin{equation*}
Y(s) = \frac{1}{s+3} + \frac{6}{(s-2)(s+3)}
\end{equation*}
We can now decompose the second term using partial fractions:
\begin{equation*}
\frac{6}{(s-2)(s+3)} = \frac{A}{s-2} + \frac{B}{s+3}
\end{equation*}
Solving for \(A\) and \(B\text{,}\) we get \(A = 2\) and \(B = 4\text{.}\) Therefore:
\begin{equation*}
Y(s) = \frac{1}{s+3} + \frac{2}{s-2} + \frac{4}{s+3}
\end{equation*}
\begin{equation*}
Y(s) = \frac{5}{s+3} + \frac{2}{s-2}
\end{equation*}
Taking the inverse Laplace transform, we obtain the solution:
\begin{equation*}
y(t) = 5e^{-3t} + 2e^{2t}
\end{equation*}
