3.2.9 Example 9

Let

\[ \left ( 1-x\right ) y^{\prime \prime }+xy^{\prime }-y=0 \]

Normalizing so that coefficient of \(y^{\prime \prime }\) is one gives

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

Hence

\begin{align*} a & =\frac {x}{\left ( 1-x\right ) }\\ b & =-\frac {1}{\left ( 1-x\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+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {x}{\left ( 1-x\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{\left ( 1-x\right ) }\right ) -\left ( -\frac {1}{\left ( 1-x\right ) }\right ) \nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\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 {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}z \tag {5}\end{equation}

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}\nonumber \\ & =\frac {x^{2}-4x+6}{4x^{2}-8x+4} \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 is one pole at \(x=1\) 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 \(\frac {1}{\left ( x-1\right ) ^{2}}\) in the partial fraction decomposition of \(r\) which is (in Maple this can be found using fullparfrac).

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

We are done with all the poles. Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =2-2=0\). Since \(O\left ( \infty \right ) =0\), we are in case \(2v\leq 0\). Hence

\[ v=0 \]

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 {1}{2}-\frac {1}{2x}+\frac {1}{x^{3}}+\frac {11}{4x^{4}}+\cdots \tag {7}\end{equation}

We want only terms for \(0\leq i\leq v\) but \(v=0\). Therefore only the constant term. Hence

\begin{equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2} \tag {8}\end{equation}

Which means

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

As it is the term that matches \(\left [ \sqrt {r}\right ] _{\infty }=ax^{v}+\cdots \). \(\ \) Now we need to find \(b\). This will be the coefficient of \(x^{v-1}=\frac {1}{x}\) in \(r-\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). But \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {1}{4}\). So coefficient of \(\frac {1}{x}\) is zero in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\).  To find the coefficient of \(\frac {1}{x}\) in \(r\) long division is done

\begin{align*} r & =\frac {s}{t}\\ & =\frac {x^{2}-4x+6}{4x^{2}-8x+4}\\ & =Q+\frac {R}{4x^{2}-8x+4}\end{align*}

Where \(Q\) is the quotient and \(R\) is the remainder. This gives

\[ r=\frac {1}{4}+\frac {-2x+5}{4x^{2}-8x+4}\]

Hence the coefficient of \(\frac {1}{x}\) is \(\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }=\frac {-2}{4}=-\frac {1}{2}\). Therefore \(b=-\frac {1}{2}-0=-\frac {1}{2}\). Hence

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

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

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

Using entry \(d=1\) 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

\begin{align*} \omega & =\left ( 0+\frac {\alpha _{c_{1}}^{-}}{x-c_{1}}\right ) +\left ( -\right ) \frac {1}{2}\\ & =\frac {-\frac {1}{2}}{x-1}-\frac {1}{2}\\ & =-\frac {1}{2}\frac {1}{x-1}-\frac {1}{2}\end{align*}

If this \(\omega \) fails to find \(p\left ( x\right ) \), then we will try the entry \(d=0\). Which will give

\[ \omega =-\frac {1}{2}\frac {1}{x-1}+\frac {1}{2}\]

Will finish the solution later.