5.3.2.15 Example 15

Let

\begin{align} \left ( 4-x^{2}\right ) y^{\prime \prime }+xy^{\prime }+2y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

\begin{align*} a & =\frac {x}{\left ( 4-x^{2}\right ) }\\ b & =\frac {2}{\left ( 4-x^{2}\right ) }\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}{\left ( 4-x^{2}\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{\left ( 4-x^{2}\right ) }\right ) -\left ( \frac {2}{\left ( 4-x^{2}\right ) }\right ) \nonumber \\ & =\frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{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 {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}z \tag {5}\end{equation}

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}=\frac {11x^{2}-24}{4x^{4}-32x^{2}+64} \tag {5A}\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 two poles at \(\pm 2\) each of order 2. For pole at \(x=2=c_{1}\)

\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 \(\frac {1}{\left ( x-2\right ) ^{2}}\) in the partial fraction decomposition of \(r\) which is

\begin{equation} \frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}=\frac {5}{16}\frac {1}{\left ( x+2\right ) ^{2}}+\frac {5}{16}\frac {1}{\left ( x-2\right ) ^{2}}-\frac {17}{32}\frac {1}{\left ( x+2\right ) }+\frac {17}{32}\frac {1}{\left ( x-2\right ) } \tag {6}\end{equation}

Hence \(b=\frac {5}{16}\). Therefore

\begin{align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=\frac {5}{4}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=-\frac {1}{4}\end{align*}

And for pole at \(x=-2=c_{2}\)

\begin{align*} \left [ \sqrt {r}\right ] _{c_{2}} & =0\\ \alpha _{c_{2}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=\frac {5}{4}\\ \alpha _{c_{2}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=-\frac {1}{4}\end{align*}

Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2\). Therefore \(v=1\). In this case

\begin{equation} \left [ \sqrt {r}\right ] _{\infty }=0 \tag {7}\end{equation}

And

\[ \alpha _{\infty }^{\pm }=\frac {1}{2}\pm \frac {1}{2}\sqrt {1+4b}\]

The coefficient of \(x^{v-1}=x^{0}\) is zero in \(\left [ \sqrt {r}\right ] _{\infty }^{2}\). To find coefficient of \(x^{0}\) in \(r=\frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}\) and since \(v=0\) then using \(b=\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }\) gives \(\frac {11}{4}\). Hence \(b=\frac {11}{4}-0=\frac {11}{4}\). Therefore

\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {11}{4}\right ) }=\frac {1}{2}+\sqrt {3}\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {11}{4}\right ) }=\frac {1}{2}-\sqrt {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 8 possible \(d\) values. These are

\begin{align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}+\sqrt {3}-\left ( \frac {5}{4}+\frac {5}{4}\right ) =\sqrt {3}-2\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}+\sqrt {3}-\left ( \frac {5}{4}-\frac {1}{4}\right ) =\sqrt {3}-\frac {1}{2}\\ d_{3} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}+\sqrt {3}-\left ( -\frac {1}{4}+\frac {5}{4}\right ) =\sqrt {3}-\frac {1}{2}\\ d_{4} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}+\sqrt {3}-\left ( -\frac {1}{4}-\frac {1}{4}\right ) =\sqrt {3}+1\\ d_{5} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}-\sqrt {3}-\left ( \frac {5}{4}+\frac {5}{4}\right ) =-\sqrt {3}-2\\ d_{6} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}-\sqrt {3}-\left ( \frac {5}{4}-\frac {1}{4}\right ) =-\sqrt {3}-\frac {1}{2}\\ d_{7} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}-\sqrt {3}-\left ( -\frac {1}{4}+\frac {5}{4}\right ) =-\sqrt {3}-\frac {1}{2}\\ d_{8} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}-\sqrt {3}-\left ( -\frac {1}{4}-\frac {1}{4}\right ) =1-\sqrt {3}\end{align*}

There are no \(d\geq 0\) integers. This means case 1 does not apply. We need to try case 2 now.