5.3.2.7 Example 7

Let

\[ \left ( x^{2}-2x\right ) y^{\prime \prime }+\left ( 2-x^{2}\right ) y^{\prime }+\left ( 2x-2\right ) y=0 \]

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

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

Hence

\begin{align*} a & =\frac {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }\\ b & =\frac {\left ( 2x-2\right ) }{\left ( x^{2}-2x\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 {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }\right ) -\frac {\left ( 2x-2\right ) }{\left ( x^{2}-2x\right ) }\nonumber \\ & =\frac {\left ( x^{4}-8x^{3}+24x^{2}-24x+12\right ) }{4x^{2}\left ( x-2\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 {\left ( x^{4}-8x^{3}+24x^{2}-24x+12\right ) }{4x^{2}\left ( x-2\right ) ^{2}}z \tag {5}\end{equation}

Therefore

\begin{align*} r & =\frac {s}{t}\\ & =\frac {\left ( x^{4}-8x^{3}+24x^{2}-24x+12\right ) }{4x^{2}\left ( x-2\right ) ^{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 \(x=2\) 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 \(r\) which is

\begin{equation} \frac {x^{4}-8x^{3}+24x^{2}-24x+12}{4x^{2}\left ( x-2\right ) ^{2}}=\frac {1}{4}-\frac {3}{4}\frac {1}{x}-\frac {1}{4}\frac {1}{\left ( x-2\right ) }+\frac {3}{4}\frac {1}{\left ( x-2\right ) ^{2}}+\frac {3}{4}\frac {1}{x^{2}} \tag {6}\end{equation}

Hence \(b=\frac {3}{4}\). 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}{4}\right ) }=\frac {3}{2}\\ \alpha _{c=0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=-\frac {1}{2}\end{align*}

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

\begin{align*} \left [ \sqrt {r}\right ] _{c=2} & =0\\ \alpha _{c=2}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c=2}^{-} & =\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 \(r\) given in Eq (6). Hence \(b=\frac {3}{4}\). Therefore the above becomes

\begin{align*} \left [ \sqrt {r}\right ] _{c=2} & =0\\ \alpha _{c=2}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=\frac {3}{2}\\ \alpha _{c=2}^{-} & =\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 \(\deg \left ( t\right ) -\deg \left ( s\right ) =4-4=0\). 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}{x}+\frac {2}{x^{3}}+\frac {11}{x^{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 the term that matches \(\left [ \sqrt {r}\right ] _{\infty }=ax^{v}+\cdots \). Hence \(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {1}{4}\) and the coefficient of \(\frac {1}{x}\) is zero. To find the coefficient of \(\frac {1}{x}\) in \(r\) long division is done

\begin{align*} r & =\frac {s}{t}\\ & =\frac {x^{4}-8x^{3}+24x^{2}-24x+12}{4x^{4}-16x^{3}+16x^{2}}\\ & =Q+\frac {R}{4x^{4}-16x^{3}+16x^{2}}\end{align*}

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

\[ r=\frac {1}{4}+\frac {-4x^{3}+20x^{2}-24x+12}{4x^{4}-16x^{3}+16x^{2}}\]

Since \(v=0\) then the coefficient of \(x^{-1}\) in \(r\) is found using \(\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }\). This gives \(-1\) as seen from above. Hence \(b=-1-0=-1\). Therefore

\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {-1}{\frac {1}{2}}-0\right ) =-1\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {-1}{\frac {1}{2}}-0\right ) =1 \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} & =1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =1-\left ( \frac {3}{2}+\frac {3}{2}\right ) =-2\\ d_{2} & =1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =1-\left ( \frac {3}{2}-\frac {1}{2}\right ) =0\\ d_{3} & =1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =1-\left ( -\frac {1}{2}+\frac {3}{2}\right ) =0\\ d_{4} & =1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =1-\left ( -\frac {1}{2}-\frac {1}{2}\right ) =2\\ d_{5} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( \frac {3}{2}+\frac {3}{2}\right ) =-4\\ d_{6} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( \frac {3}{2}-\frac {1}{2}\right ) =-2\\ d_{7} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( -\frac {1}{2}+\frac {3}{2}\right ) =-2\\ d_{8} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( -\frac {1}{2}-\frac {1}{2}\right ) =0 \end{align*}

Need to complete the solution next.