1.3.11 Example 11 \(\left ( 1+at\right ) y^{\prime }+y=t,y(1)=0\)

\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

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