2.2.21.1.1 Example 1 \[ y^{\prime }-2y=6e^{5t}\] With initial conditions \(y\left ( 0\right ) =3\). 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\[ y\left ( t\right ) =2e^{5t}+e^{2t}\]
2.2.21.1.2 Example 2 \[ y^{\prime }-ty=0 \] With initial conditions \(y\left ( 0\right ) =0\). For this we will use relation \(\mathcal {L}\left ( tf\left ( t\right ) \right ) =-\frac {d}{ds}F\left ( s\right ) \). Hence taking the Laplace transform gives\begin {align*} \mathcal {L}\left ( ty\right ) & =-\frac {d}{ds}\mathcal {L}\left ( y\right ) \\ & =-\frac {d}{ds}Y\left ( s\right ) \\\mathcal {L}\left ( y^{\prime }\right ) & =sY\left ( s\right ) -y\left ( 0\right ) \end {align*}
The ode becomes\begin {align*} sY\left ( s\right ) -y\left ( 0\right ) +\frac {d}{ds}Y\left ( s\right ) & =0\\ sY\left ( s\right ) +\frac {d}{ds}Y\left ( s\right ) & =0 \end {align*}
This is linear ode in \(Y\left ( s\right ) \). The integrating factor is \(e^{\int sds}=e^{\frac {s^{2}}{2}}\). Hence the above becomes\[ \frac {d}{ds}\left ( Ye^{\frac {s^{2}}{2}}\right ) =0 \] Integrating gives\begin {align*} Ye^{\frac {s^{2}}{2}} & =c_{1}\\ Y & =c_{1}e^{\frac {-s^{2}}{2}} \end {align*}
Applying inverse Laplace transform and using \[ y\left ( t\right ) =c_{1}\mathcal {L}^{-1}\left ( e^{\frac {-s^{2}}{2}}\right ) \] Using Laplace transform on time varying coefficient ode is not good idea. I need to look more into this. There is no \(\mathcal {L}^{-1}\left ( e^{\frac {-s^{2}}{2}}\right ) \). Solving this in time domain is much easier of course.
2.2.21.1.3 Example 3 \begin {align*} y^{\prime }-6y & =0\\ y\left ( -1\right ) & =4 \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 ) \end {align*}
The ode becomes\[ sY\left ( s\right ) -y\left ( 0\right ) -6Y=0 \] Since IC is not at zero, we let \(y\left ( 0\right ) =c_{1}\) and solving for \(Y\) gives\begin {align*} Y\left ( s-6\right ) -c_{1} & =0\\ Y & =\frac {c_{1}}{s-6} \end {align*}
Taking inverse Laplace transform gives\[ y\left ( t\right ) =c_{1}e^{6t}\] At \(t=-1\), from IC, we obtain\begin {align*} 4 & =c_{1}e^{-6}\\ c_{1} & =4e^{6} \end {align*}
Hence solution is\begin {align*} y\left ( t\right ) & =4e^{6}e^{6t}\\ & =4e^{6t+6} \end {align*}