Example 6 case one

\begin{equation} y^{\prime \prime }=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}y \tag {1}\end{equation}

\begin{align*} a & =0\\ b & =-\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\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}

\begin{equation} y=ze^{\frac {-1}{2}\int adx} \tag {3}\end{equation}

\begin{align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}} \tag {4}\end{align}

\begin{equation} z^{\prime \prime }=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}z \tag {5}\end{equation}

\begin{align*} s & =4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4\\ t & =4x^{4}\end{align*}

\begin{equation} m=4 \tag {6}\end{equation}

Since \(m\) is number of elements in the free square factorization. in this case we set

\begin{align*} t_{1} & =1\\ t_{2} & =1 \end{align*}

\begin{align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =4-6\\ & =-2 \end{align*}


case	allowed pole order for \(r=\frac {s}{t}\)	allowed \(O\left ( \infty \right ) \) order	\(L\)

1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-8,-6,-4,-2,0,2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 1\right ] \)

2	\(\left \{ 2,3,5,7,9,\cdots \right \} \)	no condition	\(\left [ 2\right ] \)

3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

part (a) Here the ﬁxed parts \(e_{fixed},\theta _{fixed}\) are calculated using

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( O\left ( \infty \right ) ,2\right ) -\deg \left ( t\right ) -3\deg \left ( t_{1}\right ) \right ) \\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {t^{\prime }}{t}+3\frac {t_{1}^{\prime }}{t_{1}}\right ) \end{align*}

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( -2,2\right ) -4-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( -2-4\right ) \\ & =-\frac {3}{2}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4x^{4}\right ) }{4x^{4}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{x}\end{align*}

Here the values \(e_{i},\theta _{i}\) are found for \(i=1\cdots k_{2}\) where \(k_{2}\) is the number of roots of \(t_{2}\). In other words, the number of poles of \(r\) that are of order \(2\). Since \(t_{2}=1\) then there are poles. Hence \(k_{2}=0\) and

This part applied only to case 1. It is used to generate \(e_{i},\theta _{i}\) for poles of \(r\) order \(4,6,8,\cdots ,M\) if any exist. Here we have pole \(x=0\) of order 4. Following the paper (need to document), we ﬁnd \(e_{1}=-5,\theta _{1}=\frac {2}{x^{2}}-\frac {5}{x}\). We start from index \(1\) since \(k_{2}=0\) from part (b). Now \(k_{1}=1\). Note that for case 1, we use \(k_{1}\). Hence

And now not \(M=0\) (for case 1 only). For other cases, we use \(k_{2}\) for \(M\).

Now we need to ﬁnd \(e_{0},\theta _{0}\). If \(O\left ( \infty \right ) >2\) then \(e_{0}=1,\theta _{0}=0\). But if \(O\left ( \infty \right ) =2\) then \(\theta _{0}=0\) and \(e_{0}=\sqrt {1+4b}\) where \(b\) is the coeﬃcient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). Since \(O\left ( \infty \right ) =0\) here then none of these cases applies. For case 1 \(\left ( n=1\right ) \) following the method in the paper we ﬁnd (need to document)

\begin{align*} e_{0} & =2\\ \theta _{0} & =-2+2x \end{align*}

\begin{align*} e & =\left \{ 2,-5\right \} \\ \theta & =\left \{ -2+2x,\frac {2}{x^{2}}-\frac {5}{x}\right \} \end{align*}

The above are arranged such that \(e_{0}\) is the ﬁrst entry. Same for \(\theta \). This to keep the same notation as in the paper. The above complete step 1, which is to generates the candidate \(e^{\prime }s\) and \(\theta ^{\prime }s\). In step 2, these are used to generate trials \(d\) and \(\theta \) and ﬁnd from them \(P\left ( x\right ) \) polynomial if possible.

In this step, we now have all the \(e_{i},\theta _{i}\) values found above in addition to \(e_{fix},\theta _{fix}\).

Starting with \(n=1\). And since we have \(M=k_{1}=1\) then there are \(\left ( n+1\right ) ^{M+1}=2^{2}=4\) sets \(s\) to try. The ﬁrst set \(s\) is

\begin{equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\nonumber \end{equation}

\begin{equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1} \tag {7}\end{equation}

If \(d\geq 0\) then we go and ﬁnd a trial \(\Theta \). We need to have both \(d,\Theta \) to go to the next step. \(\Theta \) is found using

\begin{equation} \Theta =\left ( n\right ) \left ( \theta _{fix}\right ) +\sum _{i=0}^{M}s_{i}\theta _{i} \tag {8}\end{equation}

Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-\frac {3}{2},\theta _{fixed}=\frac {1}{x}\) gives

\begin{align*} d & =\left ( 1\right ) \left ( -\frac {3}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 2\right ) -\left ( \frac {-1}{2}\right ) \left ( -5\right ) \\ & =-5 \end{align*}

Since this is not \(\geq 0\), we go to next set \(s\) \(\left \{ \frac {+1}{2},\frac {+1}{2}\right \} \) and try again

\begin{align*} d & =\left ( 1\right ) \left ( -\frac {3}{2}\right ) +\left ( \frac {+1}{2}\right ) \left ( 2\right ) -\left ( \frac {+1}{2}\right ) \left ( -5\right ) \\ & =2 \end{align*}

\begin{align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}\\ & =\left ( 1\right ) \left ( \frac {1}{x}\right ) +\left ( \frac {+1}{2}\right ) \left ( -2+2x\right ) +\left ( \frac {+1}{2}\right ) \left ( \frac {2}{x^{2}}-\frac {5}{x}\right ) \\ & =-\frac {\left ( -2x^{3}+2x^{2}+3x-2\right ) }{2x^{2}}\\ & =x-1-\frac {3}{2x}+\frac {1}{x^{2}}\end{align*}

Now that we have good trial \(d\) and \(\Theta \), then step 3 is called to generate \(P\left ( x\right ) \) if possible.

The input to this step is the integer \(d=2\) and \(\Theta =x-1-\frac {3}{2x}+\frac {1}{x^{2}}\) found from step 2 and also \(r=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\) which comes from \(z^{\prime \prime }=rz\). Since degree \(d=2\), then let \(p\left ( x\right ) =x^{2}+ax+b\). Therefore we need to now ﬁnd \(P\left ( x\right ) \) that solves

\begin{align*} P^{\prime \prime }+2\Theta P^{\prime }+\left ( \Theta ^{\prime }+\Theta ^{2}-r\right ) P & =0\\ 2+2\Theta \left ( 2x+a\right ) +\left ( \Theta ^{\prime }+\Theta ^{2}-r\right ) \left ( x^{2}+ax+b\right ) & =0\\ 2+2\left ( x-1-\frac {3}{2x}+\frac {1}{x^{2}}\right ) \left ( 2x+a\right ) +\left ( \left ( x-1-\frac {3}{2x}+\frac {1}{x^{2}}\right ) ^{\prime }+\left ( x-1-\frac {3}{2x}+\frac {1}{x^{2}}\right ) ^{2}-\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\right ) \left ( x^{2}+ax+b\right ) & =0 \end{align*}

\begin{align*} -\frac {1}{x^{2}}\left ( 2ax^{3}-4x-2ax^{2}-2a+4bx^{2}+3ax-4bx+4x^{2}\right ) & =0\\ -2ax+\frac {4}{x}+2a+2\frac {a}{x^{2}}-4b-3a\frac {1}{x}+4\frac {b}{x}-4 & =0 \end{align*}

\[ x\left ( -2a\right ) +\frac {1}{x}\left ( 4-3a+4b\right ) +\frac {1}{x^{2}}\left ( 2a\right ) +\left ( 2a-4b-4\right ) =0 \]

Therefore \(a=0\). And \(4-3a+4b=0\) gives \(b=-1\). Same if we used \(2a-4b-4=0\), So consistent equations. Therefore

\begin{align*} z & =P\left ( x\right ) e^{\int \Theta dx}\\ & =\left ( x^{2}-1\right ) e^{\int x-1-\frac {3}{2x}+\frac {1}{x^{2}}dx}\\ & =\left ( x^{2}-1\right ) x^{-\frac {3}{2}}e^{-\frac {1}{x}+\frac {x^{2}}{2}-x}\end{align*}

\begin{align*} y & =\left ( x^{2}-1\right ) x^{-\frac {3}{2}}e^{-\frac {1}{x}+\frac {x^{2}}{2}-x}e^{\frac {-1}{2}\int adx}\\ & =\left ( x^{2}-1\right ) x^{-\frac {3}{2}}e^{-\frac {1}{x}+\frac {x^{2}}{2}-x}e^{-\frac {1}{2}\int 0dx}\\ & =\left ( x^{2}-1\right ) x^{-\frac {3}{2}}e^{-\frac {1}{x}+\frac {x^{2}}{2}-x}\end{align*}

4.1.6 Example 6 case one