1.3.4 Example 4 \(ty^{\prime }+y=0,y\left ( x_{0}\right ) =y_{0}\)

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

Since IC given is not at zero, change of variables must be made so that the IC at zero. Let \(\tau =t-x_{0}\) then the ode becomes

\begin{align*} \left ( x_{0}+\tau \right ) y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =0\\ x_{0}y^{\prime }\left ( \tau \right ) +\tau y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =0\\ y\left ( 0\right ) & =y_{0}\end{align*}

Converting the above new ode to Laplace domain using

\[\mathcal {L}\left ( tf\left ( \tau \right ) \right ) =-\frac {d}{ds}F\left ( s\right ) \]

Gives (using \(Y\left ( s\right ) \) as the Laplace of \(y\left ( \tau \right ) \)) and simplifying using \(y\left ( 0\right ) =y_{0}\)

\begin{align*} x_{0}\left ( sY-y\left ( 0\right ) \right ) +\left ( -1\right ) \frac {d}{ds}\left ( sY-y\left ( 0\right ) \right ) +Y & =0\\ x_{0}\left ( sY-y_{0}\right ) -\left ( Y+s\frac {dY}{ds}\right ) +Y & =0\\ x_{0}sY-x_{0}y_{0}-Y-s\frac {dY}{ds}+Y & =0\\ x_{0}sY-s\frac {dY}{ds} & =x_{0}y_{0}\\ \frac {dY}{ds}-x_{0}Y & =-\frac {x_{0}y_{0}}{s}\end{align*}

The solution is

\[ Y=c_{1}e^{sx_{0}}+\left ( x_{0}y_{0}\operatorname {Ei}\left ( sx_{0}\right ) \right ) e^{sx_{0}}\]

Taking inverse Laplace gives

\begin{equation} y\left ( \tau \right ) =\frac {x_{0}y_{0}}{\tau +x_{0}}+c_{1}\mathcal {L}^{-1}\left ( e^{sx_{0}}\right ) \tag {1}\end{equation}

Applying initial conditions gives \(y\left ( 0\right ) =y_{0}\) gives

\begin{align*} y_{0} & =\frac {x_{0}y_{0}}{x_{0}}+c_{1}\mathcal {L}^{-1}\left ( e^{sx_{0}}\right ) \\ y_{0} & =y_{0}+c_{1}\mathcal {L}^{-1}\left ( e^{sx_{0}}\right ) \\ 0 & =c_{1}\mathcal {L}^{-1}\left ( e^{sx_{0}}\right ) \\ c_{1} & =0 \end{align*}

Hence the solution (1) becomes

\[ y\left ( \tau \right ) =\frac {x_{0}y_{0}}{\tau +x_{0}}\]

Converting back to \(t\) using \(\tau =t-x_{0}\) the above becomes

\[ y\left ( \tau \right ) =\frac {x_{0}y_{0}}{t}\]