3.2.8 Example 8

Let

\[ \left ( x^{2}+1\right ) y^{\prime \prime }+2xy^{\prime }-2y=0 \]

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

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

Hence

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

Therefore

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

\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 \(\frac {1}{\left ( x+i\right ) ^{2}}\) in the partial fraction decomposition of \(r\) which is (in Maple this can be found using fullparfrac.

\begin{equation} \frac {2x^{2}+3}{\left ( x^{2}+1\right ) ^{2}}=-\frac {1}{4}\frac {1}{\left ( x-i\right ) ^{2}}-\frac {1}{4}\frac {1}{\left ( x+i\right ) ^{2}}-\frac {5i}{4}\frac {1}{\left ( x-i\right ) }+\frac {5i}{4}\frac {1}{x+i} \tag {6}\end{equation}

Hence \(b=-\frac {1}{4}\). Therefore

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

And for the pole at \(x=+i\) 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 \(\frac {1}{\left ( x-i\right ) ^{2}}\) in the partial fraction decomposition of \(r\) given in Eq (6). Hence \(b=-\frac {1}{4}\). 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 {1}{4}\right ) }=\frac {1}{2}\\ \alpha _{c=2}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( -\frac {1}{4}\right ) }=\frac {1}{2}\end{align*}

We are done with all the poles. Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2\). Since \(O\left ( \infty \right ) =2\), then \(\left [ \sqrt {r}\right ] _{\infty }=0\). Now \(b\) is the coefficient of \(\frac {1}{x^{2}}\) in \(r\) minus coefficient of \(\frac {1}{x^{2}}\) in \(\left [ \sqrt {r}\right ] _{\infty }^{2}\) which is zero. the coefficient of \(\frac {1}{x^{2}}\) in \(r\) is found from \(\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }\) which from Eq (5A) above is \(\frac {2}{1}=2\). Hence \(b=2-0=2.\)

\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=2\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=-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 \(x=-i\) and pole at \(x=+i\) each of order 2, 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} & =2-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{2} & =2-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{3} & =2-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{4} & =2-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{5} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2\\ d_{6} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2\\ d_{7} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2\\ d_{8} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2 \end{align*}

Need to complete the solution next.