1.3.10 Example 10 \(y^{\prime }+t^{2}y=0,y(0)=0\)

\begin{align*} y^{\prime }+t^{2}y & =0\\ y\left ( 0\right ) & =0 \end{align*}

Using the property

\[\mathcal {L}\left ( t^{n}f\left ( t\right ) \right ) =\left ( -1\right ) ^{n}\frac {d^{n}}{ds^{n}}F\left ( s\right ) \]

Taking Laplace transform of each term of the ode gives

\[\mathcal {L}\left ( y^{\prime }\right ) =sY-y\left ( 0\right ) \]

And

\begin{align*}\mathcal {L}\left ( t^{2}y\right ) & =\left ( -1\right ) ^{2}\frac {d^{2}}{ds^{2}}L\left ( y\right ) \\ & =\frac {d^{2}}{ds^{2}}Y \end{align*}

Hence the ode becomes in Laplace domain as

\begin{align*} sY-y\left ( 0\right ) +\frac {d^{2}}{ds^{2}}Y & =0\\ \frac {d^{2}}{ds^{2}}Y+sY & =y\left ( 0\right ) \end{align*}

Replacing \(y\left ( 0\right ) \) from initial conditions

\[ \frac {d^{2}}{ds^{2}}Y+sY=0 \]

This is Airy ode. The solution is

\begin{equation} Y=c_{1}\operatorname {AiryAi}\left ( -s\right ) +c_{2}\operatorname {AiryBi}\left ( -s\right ) \tag {1}\end{equation}

Taking inverse Laplace transform gives

\begin{equation} y=c_{1}\mathcal {L}^{-1}\operatorname {AiryAi}\left ( -s\right ) +c_{2}\mathcal {L}^{-1}\operatorname {AiryBi}\left ( -s\right ) \tag {2}\end{equation}

Since \(y_{0}=0\) at \(t=0\), the above becomes

\[ 0=c_{1}\mathcal {L}^{-1}\operatorname {AiryAi}\left ( -s\right ) +c_{2}\mathcal {L}^{-1}\operatorname {AiryBi}\left ( -s\right ) \]

if we take \(c_{1}=0,c_{2}=0\), this will make the LHS equal to RHS. Hence (2) becomes

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

I need to double check I could do the above or not. If not, then this is not possible to solve using Laplace, since there is no inverse Laplace transform for Airy functions.