1.2.1 Example 1 \(y^{\prime }-2y=6e^{5t},y\left ( 0\right ) =3\)

\begin{align*} y^{\prime }-2y & =6e^{5t}\\ y\left ( 0\right ) & =3 \end{align*}

Taking the Laplace transform gives

\begin{align*}\mathcal {L}\left ( y\right ) & =Y\left ( s\right ) \\\mathcal {L}\left ( y^{\prime }\right ) & =sY\left ( s\right ) -y\left ( 0\right ) \\\mathcal {L}\left ( 6e^{5t}\right ) & =\frac {6}{s-5}\end{align*}

The ode becomes

\begin{align*} sY\left ( s\right ) -y\left ( 0\right ) -2Y\left ( s\right ) & =\frac {6}{s-5}\\ Y\left ( s\right ) \left ( s-2\right ) -y\left ( 0\right ) & =\frac {6}{s-5}\\ Y\left ( s\right ) \left ( s-2\right ) & =\frac {6}{s-5}+y\left ( 0\right ) \\ Y\left ( s\right ) \left ( s-2\right ) & =\frac {6}{s-5}+3\\ Y\left ( s\right ) \left ( s-2\right ) & =\frac {6+3\left ( s-5\right ) }{s-5}\\ Y\left ( s\right ) \left ( s-2\right ) & =\frac {3s-9}{s-5}\\ Y\left ( s\right ) & =\frac {3s-9}{\left ( s-5\right ) \left ( s-2\right ) }\\ & =\frac {2}{s-5}+\frac {1}{s-2}\end{align*}

Applying inverse Laplace transform and using \(\mathcal {L}^{-1}\left ( \frac {2}{s-5}\right ) =2e^{5t},\mathcal {L}^{-1}\left ( \frac {1}{s-2}\right ) =e^{2t}\) then the above gives

\[ y\left ( t\right ) =2e^{5t}+e^{2t}\]