3.3.26.3 Examples with time varying coefficients

3.3.26.3.1 Example 1 IC \(y(0)=0\)

\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

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

3.3.26.3.2 Example 2 IC \(y(0)=0\)

\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

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

3.3.26.3.3 Example 3 IC \(y(0)=y_{0}\)

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

The following property is used

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

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

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

TThe ode becomes in Laplace domain becomes

\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 inverse Laplace gives

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

Applying initial conditions gives

\begin{align*} y_{0} & =\delta \left ( 0\right ) c_{1}\\ c_{1} & =\frac {y_{0}}{\delta \left ( 0\right ) }\end{align*}

The solution (2) becomes

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

3.3.26.3.4 Example 4 IC \(y(x_{0})=y_{0}\)

\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

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

3.3.26.3.5 Example 5 (no IC)

\[ ty^{\prime }+y=0 \]

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 inverse Laplace gives

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

Since no initial conditions are given, then the above is the final solution. Notice that \(y\left ( 0\right ) \) do not have to be known, since it cancels out in the above. What is left is the \(c_{1}\) which is generated from solve the ode in \(Y\left ( s\right ) \).

3.3.26.3.6 Example 6 IC \(y(1)=5\)

\begin{align*} ty^{\prime }+y & =0\\ y\left ( 1\right ) & =5 \end{align*}

method 1

Since IC given is not at zero, change of variables must be made so that the IC at zero. Let \(\tau =t-1\) then the ode becomes

\begin{align*} \left ( 1+\tau \right ) y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =0\\ y^{\prime }\left ( \tau \right ) +\tau y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =0\\ y\left ( 0\right ) & =5 \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 \(y\left ( \tau \right ) \))

\begin{align*} sY-y\left ( 0\right ) +\left ( -1\right ) \frac {d}{ds}\left ( sY-y\left ( 0\right ) \right ) +Y & =0\\ sY-y\left ( 0\right ) -\left ( Y+s\frac {d}{ds}Y\right ) +Y & =0\\ sY-5-Y-s\frac {d}{ds}Y+Y & =0\\ sY-s\frac {d}{ds}Y & =5\\ \frac {d}{ds}Y-Y & =-\frac {5}{s}\end{align*}

The solution is

\[ Y=c_{1}e^{s}+\left ( 5\operatorname {Ei}\left ( s\right ) \right ) e^{s}\]

Taking inverse Laplace transform gives

\begin{align} y\left ( \tau \right ) & =c_{1}\mathcal {L}^{-1}\left ( e^{s},s,\tau \right ) +\mathcal {L}^{-1}\left ( \left ( 5\operatorname {Ei}\left ( s\right ) \right ) e^{s}\right ) \tag {1}\\ & =c_{1}\mathcal {L}^{-1}\left ( e^{s},s,\tau \right ) +\frac {5}{1+\tau }\nonumber \end{align}

Applying IC \(y\left ( 0\right ) =5\) the above becomes

\begin{align*} 5 & =c_{1}\mathcal {L}^{-1}\left ( e^{s},s,0\right ) +5\\ 0 & =c_{1}\mathcal {L}^{-1}\left ( e^{s},s,0\right ) \end{align*}

Hence

\[ c_{1}=0 \]

Therefore the solution (1) becomes

\begin{equation} y\left ( \tau \right ) =\frac {5}{1+\tau } \tag {2}\end{equation}

Converting back to \(t\) the above becomes

\[ y\left ( t\right ) =\frac {5}{t}\]

Note that this ode can be solved much more easily but not using Laplace transform. Let see how. The given ode is

\[ y^{\prime }+\frac {y}{t}=0\hspace {0.5in}t\neq 0 \]

This is linear ode, its solution can be easily found as

\[ y=\frac {1}{t}c_{1}\]

Applying IC

\begin{align*} 5 & =\frac {1}{1}c_{1}\\ c_{1} & =5 \end{align*}

Hence the solution is

\[ y=\frac {5}{t}\]

method 2

This method shows what happens in the case of time varying ode whose IC is not at zero, and if we do not do change of variables as was done above.

Taking Laplace transform of original ode \(ty^{\prime }+y=0\) gives

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

Hence

\[ Y=c_{1}\]

Taking inverse Laplace transform gives

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

Applying IC \(y\left ( 1\right ) =5\) to the above

\begin{align*} 5 & =c_{1}\delta \left ( 1\right ) \\ c_{1} & =\frac {5}{\delta \left ( 1\right ) }\end{align*}

Which is off course is not valid, since \(\delta \left ( 1\right ) =0\). This shows that time varying ode, using Laplace transform, we must apply change of variables (as done in method 1) first. Notice that for constant coefficients, both methods work OK. See example above under constant coefficient for problem where IC was not at zero.

So to be consistent, it seems better to stick to one method which works for both time varying and constant coefficients, which is to do change of variables if the IC is given and it is not at zero.

3.3.26.3.7 Example 7 IC \(y(1)=0\)

\begin{align*} ty^{\prime }+y & =\sin \left ( t\right ) \\ y\left ( 1\right ) & =0 \end{align*}

Change of variables is made to make the IC at zero. Let \(\tau =t-1\). The ode becomes

\begin{align*} \left ( 1+\tau \right ) y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =\sin \left ( 1+\tau \right ) \\ y^{\prime }\left ( \tau \right ) +\tau y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =\sin \left ( 1+\tau \right ) \\ y\left ( 0\right ) & =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 \(y\left ( \tau \right ) \))

\begin{align*} \left ( sY-y\left ( 0\right ) \right ) +\left ( -1\right ) \frac {d}{ds}\left ( sY-y\left ( 0\right ) \right ) +Y & =\frac {\sin \left ( 1\right ) s+\cos \left ( 1\right ) }{1+s^{2}}\\ sY-y\left ( 0\right ) -\left ( Y+s\frac {dY}{ds}\right ) +Y & =\frac {\sin \left ( 1\right ) s+\cos \left ( 1\right ) }{1+s^{2}}\\ sY-0-Y-s\frac {dY}{ds}+Y & =\frac {\sin \left ( 1\right ) s+\cos \left ( 1\right ) }{1+s^{2}}\\ sY-s\frac {d}{ds}Y & =\frac {\sin \left ( 1\right ) s+\cos \left ( 1\right ) }{1+s^{2}}\\ \frac {d}{ds}Y-Y & =-\frac {\sin \left ( 1\right ) s+\cos \left ( 1\right ) }{s\left ( 1+s^{2}\right ) }\end{align*}

The above is linear ode. Solving it gives

\begin{align} Y & =\frac {e^{s}}{2}\left ( 2\operatorname {Ei}\left ( 1,s\right ) \cos \left ( 1\right ) -\operatorname {Ei}\left ( 1,s+i\right ) -\operatorname {Ei}\left ( 1,s-i\right ) +2c_{1}\right ) \nonumber \\ & =e^{s}\operatorname {Ei}\left ( 1,s\right ) \cos \left ( 1\right ) -\frac {e^{s}}{2}\operatorname {Ei}\left ( 1,s+i\right ) -\frac {e^{s}}{2}\operatorname {Ei}\left ( 1,s-i\right ) +c_{1}e^{s} \tag {1}\end{align}

Taking inverse Laplace transform gives

\begin{equation} y\left ( \tau \right ) =\frac {\cos 1}{\tau +1}-\frac {\cos \left ( \tau +1\right ) }{\tau +1}+c_{1}\mathcal {L}^{-1}\left ( e^{s}\right ) \tag {4}\end{equation}

Applying IC \(y\left ( 0\right ) =0\)

\begin{align*} 0 & =\cos \left ( 1\right ) -\cos \left ( 1\right ) +c_{1}\mathcal {L}^{-1}\left ( e^{s}\right ) \\ 0 & =c_{1}\mathcal {L}^{-1}\left ( e^{s}\right ) \end{align*}

Hence \(c_{1}=0\). Therefore (3) becomes

\[ y\left ( \tau \right ) =\frac {\cos 1}{\tau +1}-\frac {\cos \left ( \tau +1\right ) }{\tau +1}\]

Going back to \(t\) using \(\tau =t-1\) the above becomes

\[ y\left ( t\right ) =\frac {\cos 1}{t}-\frac {\cos \left ( t\right ) }{t}\]

3.3.26.3.8 Example 8 IC \(y(1)=0\)

\begin{align*} ty^{\prime }+y & =t\\ y\left ( 1\right ) & =0 \end{align*}

Applying change of variables to make the IC at zero. Let \(\tau =t-1\) the ode becomes

\begin{align*} \left ( \tau +1\right ) y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =\tau +1\\ y^{\prime }\left ( \tau \right ) +\tau y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =\tau +1\\ y\left ( 0\right ) & =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 \(y\left ( \tau \right ) \))

\begin{align*} \left ( sY-y\left ( 0\right ) \right ) +\left ( -1\right ) \frac {d}{ds}\left ( sY-y\left ( 0\right ) \right ) +Y & =\frac {s+1}{s^{2}}\\ sY-y\left ( 0\right ) -\left ( Y+s\frac {dY}{ds}\right ) +Y & =\frac {s+1}{s^{2}}\\ sY-0-Y-s\frac {dY}{ds}+Y & =\frac {s+1}{s^{2}}\\ sY-s\frac {d}{ds}Y & =\frac {s+1}{s^{2}}\\ \frac {d}{ds}Y-Y & =-\frac {s+1}{s^{3}}\end{align*}

The above is linear ode. Solving it gives

\begin{equation} Y=\frac {1}{2s^{2}}+\frac {1}{2s}-\frac {e^{s}\operatorname {Ei}\left ( 1,s\right ) }{2}+c_{1}e^{s}\nonumber \end{equation}

Taking the inverse Laplace transform gives

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

Applying \(y\left ( 0\right ) =0\)

\begin{align} 0 & =\frac {1}{2}-\frac {1}{2}+c_{1}\mathcal {L}^{-1}\left ( e^{s}\right ) \nonumber \\ 0 & =c_{1}\mathcal {L}^{-1}\left ( e^{s}\right ) \tag {2}\end{align}

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

\[ y\left ( \tau \right ) =\frac {\tau }{2}+\frac {1}{2}-\frac {1}{2\left ( 1+\tau \right ) }\]

Going back to \(t\) using \(\tau =t-1\) the above becomes

\begin{align*} y\left ( t\right ) & =\frac {t-1}{2}+\frac {1}{2}-\frac {1}{2t}\\ & =\frac {t}{2}-\frac {1}{2t}\end{align*}

We see in the above, we did not have to use initial value theorem to find \(c_{1}\). This is because the IC was \(y\left ( 0\right ) =0\). But if the IC was \(y\left ( 0\right ) =y_{0}\), where \(y_{0}\neq 0\) then (2) would becomes

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

And then we can not solve for \(c_{1}\). So the above method works for homogeneous IC. The following example solve this same problem but with IC \(y\left ( 1\right ) =1\) to show how to handle these cases.

3.3.26.3.9 Example 9 IC \(y(1)=1\) This is the same example as above, but with \(y(1)=1\) instead of homogeneous IC \(y\left ( 1\right ) =0\).

\begin{align*} ty^{\prime }+y & =t\\ y\left ( 1\right ) & =1 \end{align*}

Applying change of variables to make the IC at zero. Let \(\tau =t-1\) the ode becomes

\begin{align*} \left ( \tau +1\right ) y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =\tau +1\\ y^{\prime }\left ( \tau \right ) +\tau y^{\prime }\left ( \tau \right ) +y\left ( \tau \right ) & =\tau +1\\ y\left ( 0\right ) & =1 \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 \(y\left ( \tau \right ) \))

\begin{align*} \left ( sY-y\left ( 0\right ) \right ) +\left ( -1\right ) \frac {d}{ds}\left ( sY-y\left ( 0\right ) \right ) +Y & =\frac {s+1}{s^{2}}\\ sY-y\left ( 0\right ) -\left ( Y+s\frac {dY}{ds}\right ) +Y & =\frac {s+1}{s^{2}}\\ sY-1-Y-s\frac {dY}{ds}+Y & =\frac {s+1}{s^{2}}\\ sY-s\frac {d}{ds}Y & =\frac {s+1}{s^{2}}+1\\ \frac {d}{ds}Y-Y & =-\frac {s+1}{s^{3}}-\frac {1}{s}\end{align*}

The above is linear ode. Solving it gives

\begin{equation} Y=\frac {1}{2s^{2}}+\frac {1}{2s}+\frac {e^{s}\operatorname {Ei}\left ( 1,s\right ) }{2}+c_{1}e^{s} \tag {1}\end{equation}

Taking the inverse Laplace gives

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

Applying IC \(y\left ( 0\right ) =1\) gives

\begin{align*} 1 & =\frac {1}{2}+\frac {1}{2}+c_{1}\mathcal {L}^{-1}\left ( e^{s}\right ) \\ 0 & =c_{1}\mathcal {L}^{-1}\left ( e^{s}\right ) \\ c_{1} & =0 \end{align*}

Hence (2) becomes

\[ y\left ( \tau \right ) =\frac {\tau }{2}+\frac {1}{2}+\frac {1}{2\left ( 1+\tau \right ) }\]

Going back to \(t\) using \(\tau =t-1\) the above becomes

\begin{align*} y\left ( t\right ) & =\frac {t-1}{2}+\frac {1}{2}+\frac {1}{2t}\\ & =\frac {t}{2}+\frac {1}{2t}\end{align*}

3.3.26.3.10 Example 10 (time varying with \(t^{2}\)) IC \(y(0)=0\)

\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.

3.3.26.3.11 Example 11 IC \(y(1)=0\)

\begin{align*} \left ( 1+at\right ) y^{\prime }+y & =t\\ y\left ( 1\right ) & =0 \end{align*}

Applying change of variables to make the IC at zero. Let \(\tau =t-1\) the ode becomes

\begin{align*} \left ( 1+a\left ( \tau +1\right ) \right ) y^{\prime }+y & =\tau +1\\ y^{\prime }+a\left ( \tau +1\right ) y^{\prime }+y & =\tau +1\\ y^{\prime }+a\tau y^{\prime }+ay^{\prime }+y & =\tau +1\\ \left ( 1+a\right ) y^{\prime }+a\tau y^{\prime }+y & =\tau +1\\ y\left ( 0\right ) & =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 \(y\left ( \tau \right ) \))

\begin{align*} \left ( 1+a\right ) \left ( sY-y\left ( 0\right ) \right ) +a\left ( -1\right ) \frac {d}{ds}\left ( sY-y\left ( 0\right ) \right ) +Y & =\frac {s+1}{s^{2}}\\ \left ( 1+a\right ) sY-a\frac {d}{ds}\left ( sY\right ) +Y & =\frac {s+1}{s^{2}}\\ sY+asY-a\left ( Y+s\frac {dY}{ds}\right ) +Y & =\frac {s+1}{s^{2}}\\ sY+asY-aY-as\frac {dY}{ds}+Y & =\frac {s+1}{s^{2}}\\ -as\frac {dY}{ds}+Y\left ( 1+s+as-a\right ) & =\frac {s+1}{s^{2}}\\ \frac {dY}{ds}-Y\frac {\left ( 1+s+as-a\right ) }{as} & =-\frac {s+1}{as^{3}}\end{align*}

This is linear in \(Y\left ( s\right ) \). Solving gives

\[ Y\left ( s\right ) =\frac {1}{s^{2}\left ( a+1\right ) }+c_{1}\frac {s^{\frac {a+1}{a}}e^{s\frac {\left ( a+1\right ) }{a}}}{s^{2}}\]

Taking inverse Laplace gives

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

Applying IC \(y\left ( 0\right ) =0\) the above becomes

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

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

\[ y\left ( \tau \right ) =\frac {\tau }{a+1}\]

Going back to \(t\) using \(\tau =t-1\) gives

\[ y\left ( t\right ) =\frac {t-1}{a+1}\]