3.2.2.1.8 Example 8 (non-constant coefficient) \(tx^{\prime \prime }\left ( t\right ) +x^{\prime }+tx=0\)

\begin{equation} tx^{\prime \prime }\left ( t\right ) +x^{\prime }+tx=0 \tag {1}\end{equation}

Assuming \(x\left ( 0\right ) =1,x^{\prime }\left ( 0\right ) =0\). In solving ode using Laplace where the coefficient in time varying, we will be using the relation

\begin{equation} L\left ( t^{n}f\left ( t\right ) \right ) =\left ( -1\right ) ^{n}F^{\left ( n\right ) }\left ( s\right ) \tag {2}\end{equation}

Where \(F\left ( s\right ) \) is the laplace transform of \(f\left ( t\right ) \). For example, if the input is \(tx\left ( t\right ) \) the Laplace tranaform is \(-X^{\prime }\left ( s\right ) =-\frac {dX\left ( s\right ) }{ds}\) and if the input is \(t^{2}x\left ( t\right ) \) then the Laplace transform is \(\frac {d^{2}X\left ( s\right ) }{s^{2}}\) and so on. This will generate an ODE in \(X\left ( s\right ) \) which we have to solve for \(X\left ( s\right ) \). Applying this to (1) gives

\begin{align*} L\left ( x^{\prime \prime }\right ) & =s^{2}X\left ( s\right ) -sx\left ( 0\right ) -x^{\prime }\left ( 0\right ) \\ L\left ( x^{\prime }\right ) & =sX\left ( s\right ) -x\left ( 0\right ) \\ L\left ( x\right ) & =X\left ( s\right ) \end{align*}

Hence using (2) on the above, then Laplace transform of (1) becomes

\[ -\frac {d}{ds}\left ( s^{2}X\left ( s\right ) -sx\left ( 0\right ) -x^{\prime }\left ( 0\right ) \right ) +\left ( sX\left ( s\right ) -x\left ( 0\right ) \right ) -\frac {d}{ds}X\left ( s\right ) =0 \]

Substituting initial conditions gives

\begin{align*} -\frac {d}{ds}\left ( s^{2}X\left ( s\right ) -s\right ) +\left ( sX\left ( s\right ) -1\right ) -\frac {d}{ds}X\left ( s\right ) & =0\\ -\left ( 2sX+s^{2}X^{\prime }-1\right ) +sX-1-X^{\prime }\left ( s\right ) & =0\\ X^{\prime }\left ( -s^{2}-1\right ) +X\left ( -2s+s\right ) & =0\\ \left ( s^{2}+1\right ) X^{\prime }+sX & =0 \end{align*}

This differential equation is now solved for \(X\left ( s\right ) \) which gives

\[ X\left ( s\right ) =\frac {c_{1}}{\sqrt {s^{2}+1}}\]

The inverse Laplace transform is

\[ x\left ( t\right ) =c_{1}\operatorname {BesselJ}_{0}\left ( t\right ) \]

Since \(x\left ( 0\right ) =1\) then

\[ 1=c_{1}\operatorname {BesselJ}_{0}\left ( 0\right ) \]

But \(\operatorname {BesselJ}_{0}\left ( 0\right ) =1\), hence \(c_{1}=1\) and the solution is

\[ x\left ( t\right ) =\operatorname {BesselJ}_{0}\left ( t\right ) \]