1.3.2 Example 2 \(ty^{\prime }+y=0,y\left ( 0\right ) =0\)

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

We will use the property

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

Hence taking Laplace transform of each term of the ode gives

\begin{align*}\mathcal {L}\left ( ty^{\prime }\right ) & =-\frac {d}{ds}\left ( \mathcal {L}\left ( y^{\prime }\right ) \right ) \\ & =-\frac {d}{ds}\left ( sY-y\left ( 0\right ) \right ) \\ & =-\left ( Y+s\frac {dY}{ds}\right ) \\ & =-s\frac {dY}{ds}-Y \end{align*}

And

\[\mathcal {L}\left ( y\right ) =Y \]

Hence the ode becomes in Laplace domain as

\begin{align*} -s\frac {dY}{ds}-Y+Y & =0\\ -s\frac {dY}{ds} & =0\\ \frac {dY}{ds} & =0 \end{align*}

Solving this ode for \(Y\left ( s\right ) \) gives

\begin{equation} Y=c_{1} \tag {1}\end{equation}

Taking the inverse Laplace transform gives

\begin{equation} y\left ( t\right ) =c_{1}\delta \left ( t\right ) \tag {2}\end{equation}

Applying initial conditions

\[ 0=c_{1}\delta \left ( 0\right ) \]

Hence \(c_{1}=0\) and the solution (2) becomes

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