5.3.2.5 Example 5

Let

\[ y^{\prime \prime }=\frac {32x^{2}-27x+27}{144x^{4}-288x^{3}+144x^{2}}y \]

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {32x^{2}-27x+27}{144x^{4}-288x^{3}+144x^{2}}\tag {1}\\ & =\frac {3}{16x}-\frac {3}{16}\frac {1}{\left ( x-1\right ) }+\frac {3}{16}\frac {1}{x^{2}}+\frac {2}{9}\frac {1}{\left ( x-1\right ) ^{2}} \tag {2}\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 and one pole at \(x=1\) of order 2. For the pole at \(x=0\) since order is \(2\) then

\begin{align*} \left [ \sqrt {r}\right ] _{c=0} & =0\\ \alpha _{c=0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c=0}^{-} & =\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 \(r\) given in Eq (2) which is \(\frac {3}{16}\). Hence \(b=\frac {3}{16}\). Therefore

\begin{align*} \left [ \sqrt {r}\right ] _{c=0} & =0\\ \alpha _{c=0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{16}\right ) }=\frac {1}{4}\sqrt {7}+\frac {1}{2}\\ \alpha _{c=0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{16}\right ) }=\frac {1}{2}-\frac {1}{4}\sqrt {7}\end{align*}

And for the pole at \(x=1\) which is order 2,

\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 \(\frac {1}{\left ( x-1\right ) ^{2}}\) in the partial fraction decomposition of \(r\) given in Eq (2) which is \(\frac {2}{9}\). Hence \(b=\frac {2}{9}\). Therefore the above becomes

\begin{align*} \left [ \sqrt {r}\right ] _{c=1} & =0\\ \alpha _{c=1}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{6}\sqrt {17}+\frac {1}{2}\\ \alpha _{c=1}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{2}-\frac {1}{6}\sqrt {17}\end{align*}

We are done with all the poles. Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2\). Since \(O\left ( \infty \right ) =2\) then from the algorithm above

\[ \left [ \sqrt {r}\right ] _{\infty }=0 \]

Now we calculate \(b\) for this case. This is given by the leading coefficient of \(s\) divided by the leading coefficient of \(t\) when \(\gcd \left ( s,t\right ) =1\). In this case \(r=\frac {32x^{2}-27x+27}{144x^{4}-288x^{3}+144x^{2}}\) from Eq (1) , hence \(b=\frac {32}{144}=\frac {2}{9}\). Therefore

\begin{align*} \left [ \sqrt {r}\right ] _{\infty } & =0\\ \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{6}\sqrt {17}+\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{2}-\frac {1}{6}\sqrt {17}\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 Since we have a pole at \(x=c_{1}=0\) and pole at \(x=c_{2}=1\), and we have one \(O\left ( \infty \right ) \), each with \(\pm \) signs. The following now implements

\[ d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{2}\alpha _{c_{i}}^{\pm }\]

By trying all possible combinations. There are 8 possible \(d\) values. This gives

\begin{align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =-\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =\frac {1}{3}\sqrt {17}-\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{3} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{4} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =\frac {1}{4}\sqrt {7}+\frac {1}{3}\sqrt {17}-\frac {1}{2}\\ d_{5} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =-\frac {1}{4}\sqrt {7}-\frac {1}{3}\sqrt {17}-\frac {1}{2}\\ d_{6} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =-\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{7} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =\frac {1}{4}\sqrt {7}-\frac {1}{3}\sqrt {17}-\frac {1}{2}\\ d_{8} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =\frac {1}{4}\sqrt {7}-\frac {1}{2}\end{align*}

None of the \(d\) found are integer. Hence case 1 did not work we need to try case 2 and if that also fail, try case 3. We will find all three cases fail on this ode..