3.2.14 Example 14

Let

\begin{align} 3y^{\prime \prime }+xy^{\prime }-4y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

\begin{align*} a & =\frac {x}{3}\\ b & =-\frac {4}{3}\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}{3}\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{3}\right ) -\left ( -\frac {4}{3}\right ) \nonumber \\ & =\frac {x^{2}}{36}+\frac {3}{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 {x^{2}+54}{36}z \tag {5}\end{equation}

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}+54}{36}\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 \(\Gamma \) is empty.

Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =0-2=-2\). 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 {x}{6}+\frac {9}{2}\frac {1}{x}+\cdots \tag {7}\end{equation}

But we want only terms for \(0\leq i\leq v\) but \(v=1\). Therefore need to sum terms \(x^{0},x^{1}\). Therefore

\begin{equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {x}{6} \tag {8}\end{equation}

Which means

\[ a=\frac {1}{6}\]

Now we need to find \(b\). Which is given by the coefficient of \(x^{v-1}=x^{0}\) or the constant term in \(r\) minus coefficient of \(x^{0}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). But \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {x^{2}}{36}\). Hence the coefficient of \(x^{0}\) is zero here. Now we find coefficient of \(x^{0}\) in \(r\). Since \(r=\frac {x^{2}}{36}+\frac {54}{36}\) then the coefficient of \(x^{0}\) is \(\frac {54}{36}=\frac {3}{2}\). Hence \(b=\frac {3}{2}-0=\frac {3}{2}\). Therefore

\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {\frac {3}{2}}{\frac {1}{6}}-1\right ) =4\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {\frac {3}{2}}{\frac {1}{6}}-1\right ) =-5 \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 2 possible \(d\) values (since no poles). This gives

\begin{align*} d_{1} & =\alpha _{\infty }^{+}=4\\ d_{2} & =\alpha _{\infty }^{-}=-5 \end{align*}

Using \(d=4\) entry from above 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

\[ \omega =\left ( 0\right ) +\left ( +\right ) \left ( \frac {x}{6}\right ) =\frac {x}{6}\]