3.9.8.5.1 Example 7. \(xy^{\prime \prime }+\left ( x^{2}-1\right ) y^{\prime }+x^{3}y=0\)

Writing the ode in normal form gives

\begin{align*} y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y & =r\qquad \\ p & =\frac {x^{2}-1}{x}\qquad x\neq 0\\ q & =x^{2}\\ r & =0 \end{align*}

Trying change of variable on the independent variable as above.  Let \(\tau =g\left ( x\right ) \) where \(\tau \) will be the new independent variable. Applying this transformation results in

\begin{equation} y^{\prime \prime }+p_{1}y^{\prime }+q_{1}y=r_{1} \tag {1}\end{equation}

Where

\begin{align} p_{1}\left ( \tau \right ) & =\frac {\tau ^{\prime \prime }\left ( x\right ) +p\left ( x\right ) \tau ^{\prime }\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\tag {2}\\ q_{1}\left ( \tau \right ) & =\frac {q\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\tag {3}\\ r_{1}\left ( \tau \right ) & =\frac {r\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}} \tag {4}\end{align}

Approach 1. Let \(q_{1}=c^{2}\) where \(c^{2}\) is some constant. This implies

\begin{align} \frac {q}{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}} & =c^{2}\nonumber \\ \tau ^{\prime } & =\frac {1}{c}\sqrt {q} \tag {5}\end{align}

If \(p_{1}\) turns out to be constant with this \(\tau \) then it implies (1) is second order constant coefficient ode. Eq (5) becomes

\begin{align*} \tau ^{\prime } & =\frac {1}{c}\sqrt {x^{2}}\\ \tau ^{\prime \prime }\left ( x\right ) & =\frac {1}{2c}\frac {2x}{\sqrt {x^{2}}}\end{align*}

Hence from (2)

\begin{align*} p_{1}\left ( \tau \right ) & =\frac {\tau ^{\prime \prime }\left ( x\right ) +p\tau ^{\prime }\left ( x\right ) }{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\\ & =\frac {\frac {1}{2c}\frac {2x}{\sqrt {x^{2}}}+\left ( \frac {x^{2}-1}{x}\right ) \frac {1}{c}\sqrt {x^{2}}}{\left ( \frac {1}{c}\sqrt {x^{2}}\right ) ^{2}}\\ & =c \end{align*}

Which is a constant. Then (1) becomes second order of constant coefficient

\[ y^{\prime \prime }\left ( \tau \right ) +cy^{\prime }\left ( \tau \right ) +c^{2}y\left ( \tau \right ) =0 \]

Which has the solution

\[ y\left ( \tau \right ) =e^{-\frac {c\tau }{2}}\left ( A\sin \left ( \frac {c\sqrt {3}\tau }{2}\right ) +B\sin \left ( \frac {c\sqrt {3}\tau }{2}\right ) \right ) \]

But from earlier \(\tau =\frac {x^{2}}{2c}\). Hence the above becomes

\begin{align*} y\left ( x\right ) & =Ae^{-\frac {c\frac {x^{2}}{2c}}{2}}\sin \left ( \frac {c\sqrt {3}\frac {x^{2}}{2c}}{2}\right ) +Be^{-\frac {c\frac {x^{2}}{2c}}{2}}\sin \left ( \frac {c\sqrt {3}\frac {x^{2}}{2c}}{2}\right ) \\ & =e^{-\frac {x^{2}}{4}}\left ( A\sin \left ( \frac {\sqrt {3}x^{2}}{4}\right ) +B\sin \left ( \frac {\sqrt {3}x^{2}}{4}\right ) \right ) \end{align*}

Approach 2  

Let \(p_{1}=0\). If with this choice now \(q_{1}\) becomes constant or a constant divided by \(\tau ^{2}\) then (2) can be easily integrated.  \(p_{1}=0\) implies from (2) that

\begin{align*} \tau ^{\prime \prime }+p\tau ^{\prime } & =0\\ \tau & =\int e^{-\int pdx}dx\\ & =\int e^{-\int \frac {x^{2}-1}{x}dx}dx\\ & =\int xe^{-\frac {x^{2}}{2}}dx\\ & =-e^{-\frac {x^{2}}{2}}\end{align*}

Therefore

\begin{align*} q_{1} & =\frac {q}{\left ( \tau ^{\prime }\left ( x\right ) \right ) ^{2}}\\ & =\frac {x^{2}}{\left ( xe^{-\frac {x^{2}}{2}}\right ) ^{2}}\\ & =e^{x^{2}}\end{align*}

Which is not constant. Now it is checked to see if it is constant divided by \(\tau ^{2}\). Since \(\tau ^{2}=\left ( -e^{-\frac {x^{2}}{2}}\right ) ^{2}=\allowbreak e^{-x^{2}}\) then \(q_{1}=\frac {1}{\tau ^{2}}\). Therefore this approach also worked.

Eq (2) becomes

\begin{align} y^{\prime \prime }+p_{1}y^{\prime }+q_{1}y & =0\tag {1}\\ y^{\prime \prime }+\frac {1}{\tau ^{2}}y & =0\nonumber \\ \tau ^{2}y^{\prime \prime }+y & =0\nonumber \end{align}

Which is standard Euler ode which can be solved easily. Giving

\[ y\left ( \tau \right ) =A\sqrt {\tau }\cos \left ( \frac {\sqrt {3}}{2}\ln \left ( \tau \right ) \right ) +B\sqrt {\tau }\sin \left ( \frac {\sqrt {3}}{2}\ln \left ( \tau \right ) \right ) \]

But \(\tau =-e^{-\frac {x^{2}}{2}}\). Hence the above becomes

\[ y\left ( x\right ) =A\sqrt {-e^{-\frac {x^{2}}{2}}}\cos \left ( \frac {\sqrt {3}}{2}\ln \left ( -e^{-\frac {x^{2}}{2}}\right ) \right ) +B\sqrt {-e^{-\frac {x^{2}}{2}}}\sin \left ( \frac {\sqrt {3}}{2}\ln \left ( -e^{-\frac {x^{2}}{2}}\right ) \right ) \]

This looks different from the solution obtained in approach 1, but it verifies also as correct solution.  This is an example where change of independent variable using \(q_{1}=c^{2}\) works and also change of independent variable using \(p_{1}=0\) works as well.