5.3.2.12 Example 12

Let

\begin{align} y^{\prime \prime }+\frac {x^{2}-1}{x}y^{\prime }+x^{2}y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

\begin{align*} a & =\frac {x^{2}-1}{x}\\ b & =x^{2}\end{align*}

It is first transformed to the following ode by eliminating the first derivative

\begin{equation} z^{\prime \prime }=rz \tag {2}\end{equation}

Using what is known as the Liouville transformation given by

\begin{equation} y=ze^{\frac {-1}{2}\int adx} \tag {3}\end{equation}

Where it can be found that \(r\) in (2) is given by

\begin{align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {x^{2}-1}{x}\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x^{2}-1}{x}\right ) -x^{2}\nonumber \\ & =-\frac {3\left ( x^{4}-1\right ) }{4x^{2}} \tag {4}\end{align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is

\begin{equation} z^{\prime \prime }=-\frac {3\left ( x^{4}-1\right ) }{4x^{2}}z \tag {5}\end{equation}

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {-3\left ( x^{4}-1\right ) }{4x^{2}}\nonumber \end{align}

The necessary conditions for case 1 are met.

Step 1 In this we find all \(\left [ \sqrt {r}\right ] _{c}\) and associated \(\alpha _{c}^{\pm }\) for each pole. There is one pole at \(x=0\) of order 2. Hence

\begin{align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}\end{align*}

Where \(b\) is the coefficient of \(\frac {1}{\left ( x-0\right ) ^{2}}\) in the partial fraction decomposition of \(r\) which is (in Maple this can be found using fullparfrac).

\begin{equation} \frac {-3\left ( x^{4}-1\right ) }{4x^{2}}=-\frac {3}{4}x^{2}+\frac {3}{4}\frac {1}{x^{2}} \tag {6}\end{equation}

Hence \(b=\frac {3}{4}\). Therefore

\begin{align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=\frac {3}{2}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=-\frac {1}{2}\end{align*}

Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =2-4=-2\). We are in case \(2v\leq 0\). Hence \(-2v=-2\) or

\[ v=1 \]

Then now \(\left [ \sqrt {r}\right ] _{\infty }\) is the sum of all terms \(x^{i}\) for \(0\leq i\leq v\) in the Laurent series expansion of \(\sqrt {r}\) at \(\infty \) which is

\begin{equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {i\sqrt {3}}{2}x-\frac {i\sqrt {3}}{4}\frac {1}{x^{3}}+\cdots \tag {7}\end{equation}

We want only terms for \(0\leq i\leq v\) but \(v=1\). Therefore need to sum terms \(x^{0},x^{1}\). From the above we see that

\[ \left [ \sqrt {r}\right ] _{\infty }=\frac {i\sqrt {3}}{2}x \]

Which means

\[ a=\frac {i\sqrt {3}}{2}\]

As it is the the term that matches \(\left [ \sqrt {r}\right ] _{\infty }=ax\). \(\ \) Now we need to find \(b\). This will be the coefficient of \(x^{v-1}=x^{0}\) in \(r\) minus the coefficient of \(x^{0}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). But

\begin{align*} \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2} & =\left ( \frac {i\sqrt {3}}{2}x\right ) ^{2}\\ & =-\frac {3}{4}x^{2}\end{align*}

Hence the coefficient is zero here. Now we find coefficient of \(x^{0}\) in \(r\). Since \(r=-\frac {3}{4}x^{2}+\frac {3}{4}\frac {1}{x^{2}}\) then coefficient of \(x^{0}\) is zero also. Hence \(b=0-0=0\). Therefore

\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( 0-1\right ) =-\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( 0-1\right ) =-\frac {1}{2}\end{align*}

This completes step 1 of the solution. We have found \(\left [ \sqrt {r}\right ] _{c}\) and its associated \(\alpha _{c}^{\pm }\) and found \(\left [ \sqrt {r}\right ] _{\infty }\) and its associated \(\alpha _{\infty }^{\pm }\). Now we go to step 2 which is to find the \(d^{\prime }s\).

step 2 Now \(d\) is found using

\[ d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }\]

By trying all possible combinations. There are 4 possible \(d\) values. This gives

\begin{align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}\right ) =-\frac {1}{2}-\frac {3}{2}=-1\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}\right ) =-\frac {1}{2}-\left ( -\frac {1}{2}\right ) =0\\ d_{3} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}\right ) =-\frac {1}{2}-\frac {3}{2}=-2\\ d_{4} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}\right ) =-\frac {1}{2}-\left ( -\frac {1}{2}\right ) =0 \end{align*}

Using first \(d=0\) entry above now we find \(\omega \) using

\[ \omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{s\left ( c\right ) }}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\]

Hence

\[ \omega =\left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-0}+\left ( +\right ) \left ( \frac {i\sqrt {3}}{2}x\right ) =\frac {-1}{2x}+\frac {i\sqrt {3}}{2}x \]

Notice that if have taken the last \(d=0\) entry, we will get

\[ \omega =\left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-0}+\left ( -\right ) \left ( \frac {i\sqrt {3}}{2}x\right ) =\frac {-1}{2x}-\frac {i\sqrt {3}}{2}x \]

In practice, we will try the second one if the first fails. Will finish the solution next.