Transformada de Laplace para resolver ecuación diferencial [Ejercicio resuelto de Ecuaciones Diferenciales]

Sea la siguiente ecuación diferencial $$\left\{
\begin{array}{l}
y''-3y'=e^{2t}\\
y(0)=1,\ y'(0)=2
\end{array}
\right.$$
Aplicando transformada de Laplace
$$\begin{align}
\mathscr{L}\left\{\dfrac{d^2y}{dt^2}\right\}-3\mathscr{L}\left\{\dfrac{dy}{dt}\right\}&=\mathscr{L}\left\{e^{2t}\right\}\\
s^2Y(s)-sy(0)-y'(0)-3[sY(s)-y(0)]&=\dfrac{1}{s-2}\\
s^2Y(s)-s(1)-2-3[sY(s)-1]&=\dfrac{1}{s-2}\\
[s^2-3s]Y(s)&=s-1+\dfrac{1}{s-2}\\
Y(s)&=\dfrac{s-1}{s(s-3)}+\dfrac{1}{s(s-3)(s-2)}\\
Y(s)&=\dfrac{1}{3}\cdot\dfrac{1}{s}+\dfrac{2}{3}\cdot\dfrac{1}{s-3}+\dfrac{1}{6}\cdot\dfrac{1}{s}\\
& \qquad\qquad +\dfrac{1}{3}\cdot\dfrac{1}{s-3}-\dfrac{1}{2}\cdot\dfrac{1}{s-2}\\
Y(s)&=\dfrac{1}{2}\cdot\dfrac{1}{s}+\dfrac{1}{s-3}-\dfrac{1}{2}\cdot\dfrac{1}{s-2}
\end{align}$$
Aplicamos transformada inversa de Laplace
$$\begin{align}
\mathscr{L}^{-1}\{Y(s)\}&=\dfrac{1}{2}\mathscr{L}^{-1}\left\{\dfrac{1}{s}\right\}+\mathscr{L}^{-1}\left\{\dfrac{1}{s-3}\right\}-\dfrac{1}{2}\mathscr{L}^{-1}\left\{\dfrac{1}{s-2}\right\}\\
y(t)&=\dfrac{1}{2}+e^{3t}-\dfrac{1}{2}e^{2t}
\end{align}$$