5.3.2.10 Example 10

Let

\begin{align} y^{\prime \prime }-x^{2}y^{\prime }-3xy & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

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

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

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

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {1}{4}x^{4}+2x\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 are no poles. Hence set \(\Gamma \) of poles is empty. Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =0-4=-4\). We are in case \(2v\leq 0\). Hence \(-2v=-4\) or

\[ v=2 \]

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 {x^{2}}{2}+\frac {2}{x}-\frac {4}{x^{4}}+\frac {16}{x^{7}}-\cdots \tag {7}\end{equation}

We want only terms for \(0\leq i\leq v\) but \(v=2\). Therefore need to sum terms \(x^{0},x^{1},x^{2}\). From the above we see that

\begin{align*} \left [ \sqrt {r}\right ] _{\infty } & =\frac {x^{2}}{2}+0x^{1}+0x^{0}\\ & =\frac {x^{2}}{2}\end{align*}

Which means

\[ a=\frac {1}{2}\]

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

\[ \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {x^{4}}{4}\]

Hence the coefficient is zero here. Now we find coefficient of \(x\) in \(r\). But \(r=\) \(\frac {1}{4}x^{4}+2x\) hence the coefficient of \(x\) is \(2\). Therefore

\begin{align*} b & =2-0\\ & =2 \end{align*}

Hence

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

Using \(d=1\) 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, since there are no poles, only last term above survives giving

\[ \omega =s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }=\left ( +\right ) \frac {x^{2}}{2}=\frac {x^{2}}{2}\]

Will finish the solution next.