5.3.2.13 Example 13

Let

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

Hence

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

\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*}

Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =2-2=0\). We are in case \(2v\leq 0\). Hence \(-2v=0\) or

\[ v=0 \]

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}}+\cdots \tag {7}\end{equation}

But we want only terms for \(0\leq i\leq v\) but \(v=0\). Therefore need to sum terms \(x^{0}\). Which is the constant term

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

Which means

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

Now we need to find \(b\). Which is given by the coefficient of \(\frac {1}{x}\) in \(r\) minus coefficient of \(\frac {1}{x}\) in \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). But \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {1}{4}\) Hence the coefficient is zero here.  To find the coefficient of \(\frac {1}{x}\) in \(r\) long division is done (here paper is not clear at all what it means by coefficient of \(x^{v-1}\) in \(r\) as that depends on the form of \(r\) and how it is represented). This method of using long division to find the coefficient works to obtain the correct result. But it is still not clear what the paper actually means by this.

\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

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

For the case of \(v=0\) then the coefficient of \(x^{-1}\) is  \(\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }=\frac {-2}{4}=-\frac {1}{2}\). Notice that if we just expanded \(r\) it will give \(\frac {x^{2}}{4\left ( x-1\right ) ^{2}}-\frac {x}{\left ( x-1\right ) ^{2}}+\frac {3}{2\left ( x-1\right ) ^{2}}\) and we see there is no coefficient of \(\frac {1}{x}\) in this representation. So we would have obtain wrong value of \(b\) if we just used what the paper said. Now \(b=-\frac {1}{2}-0=-\frac {1}{2}\). Therefore

\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}-\left ( \frac {3}{2}\right ) =-1\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}\right ) =\frac {1}{2}-\left ( -\frac {1}{2}\right ) =1\\ d_{3} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}\right ) =-\frac {1}{2}-\left ( \frac {3}{2}\right ) =-2\\ d_{4} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}\right ) =-\frac {1}{2}-\left ( -\frac {1}{2}\right ) =0 \end{align*}

Using first \(d=1\) 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

\[ \omega =\left ( \left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-1}\right ) +\left ( +\right ) \left ( \frac {1}{2}\right ) =\frac {-1}{2(x-1)}+\frac {1}{2}\]

And if use the last entry \(d=0\) then

\[ \omega \left ( =\left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-1}\right ) +\left ( -\right ) \left ( \frac {1}{2}x\right ) =\frac {-1}{2(x-1)}-\frac {1}{2}\]

In practice, we will try the second one if the first fails. Will finish the solution next.