1.3.1 Example 1 \(y^{\prime }-ty=0,y\left ( 0\right ) =0\)

\begin{align*} y^{\prime }-ty & =0\\ y\left ( 0\right ) & =0 \end{align*}

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*}

Replacing initial conditions \(y\left ( 0\right ) =0\) the above becomes

\[ sY\left ( s\right ) +\frac {d}{ds}Y\left ( s\right ) =0 \]

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}\nonumber \\ Y & =c_{1}e^{\frac {-s^{2}}{2}} \tag {1}\end{align}

Taking the inverse Laplace gives

\begin{equation} y\left ( t\right ) =c_{1}\mathcal {L}^{-1}\left ( e^{\frac {-s^{2}}{2}}\right ) \tag {2}\end{equation}

And now apply IC which gives

\[ 0=c_{1}\mathcal {L}^{-1}\left ( e^{\frac {-s^{2}}{2}}\right ) \]

Hence \(c_{1}=0\). Therefore (2) becomes

\[ y\left ( t\right ) =0 \]