5.4.2.7 Example 7 case two

Solve

\begin{align} y^{\prime \prime } & =\frac {16x-3}{16x^{2}}y\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

\begin{align*} a & =0\\ b & =-\frac {16x-3}{16x^{2}}\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 {16x-3}{16x^{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 {16x-3}{16x^{2}}z \tag {5}\end{equation}

Step 0 We need to find which case it is. \(r=\frac {s}{t}\) where

\begin{align*} s & =16x-3\\ t & =16x^{2}\end{align*}

The free square factorization of \(t\) is \(t=\left [ 1,x\right ] \). Hence

\begin{equation} m=2 \tag {6}\end{equation}

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

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

Now

\begin{align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =2-1\\ & =1 \end{align*}

There is pole \(x=0\) of order 2. Looking at the cases table, reproduced here

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

Shows that only case 2 is possible (due to \(O\left ( \infty \right ) =1\) which is not allowed other than for case 2). Hence \(L=\left [ 2\right ] \).

Step 1

This step has 4 parts (a,b,c,d).

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*}

Using \(O\left ( \infty \right ) =1,t=16x^{2},t_{1}=1\) the above gives

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 1,2\right ) -2-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 1-2\right ) \\ & =-\frac {1}{4}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 16x^{2}\right ) }{16x^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{2x}\end{align*}

part (b)

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 \(2\).

\[ r=\frac {16x-3}{16x^{2}}\]

These will be the zeros of \(t_{2}\) in the above square free factorization of \(t\). From above we found that

\[ t_{2}=x \]

Label these zeros of \(t_{2}\) as \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ 0\right \} \). Therefore \(k_{2}=1\). Hence

\[ M=1 \]

Now we iterate over each zero \(c_{i}\) times finding \(e_{i}\) and \(\theta _{i}\) from each. These are found to be (following formula in paper) to be

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

And

\[ \theta _{1}=\frac {1}{2x}\]

Part (c)

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. Since only case 2 exist in this example. This is skipped. Hence \(M\) stays \(1\).

Part(d)

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 \(\infty \). None of these apply, and this is not case 1. Hence

\begin{align*} e_{0} & =0\\ \theta _{0} & =0 \end{align*}

Hence now we have

\begin{align*} e & =\left \{ 0,\frac {1}{2}\right \} \\ \theta & =\left \{ 0,\frac {1}{2x}\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.

Step 2

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

Starting with \(n=2\).  Since case 2 only applies here. And since we have \(M=1\) then there are \(\left ( n+1\right ) ^{M+1}=3^{2}=9\) sets \(s\) to try. The first set \(s\) is

\[ s=\left \{ \frac {-n}{2},\frac {-n}{2}\right \} =\left \{ -1,-1\right \} \]

Now we generate trial \(d\) using

\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}

Since \(M=1\) then the above becomes

\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 \(e_{fix}=-\frac {1}{4},\theta _{fixed}=\frac {1}{2x}\) gives

\begin{align*} d & =\left ( 2\right ) \left ( -\frac {1}{4}\right ) +\left ( -1\right ) \left ( 0\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) \\ & =0 \end{align*}

Since \(d\geq 0\), then we can use it. Using Eq (8) gives (using \(M=1\))

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

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

Step 3

The input to this step is the integer \(d=0\) and \(\Theta =\frac {1}{2x}\) found from step 2 and also \(r=\frac {16x-3}{16x^{2}}\) which comes from \(z^{\prime \prime }=rz\). We need now to find \(P\left ( x\right ) \) of degree \(d=0\) which is a constant such that

\[ P^{\prime \prime \prime }+3\Theta P^{\prime \prime }+\left ( 3\Theta ^{2}+3\Theta ^{\prime }-4r\right ) P^{\prime }+\left ( \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }\right ) P=0 \]

Since \(P=1\) then above simplifies to

\[ \left ( \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }\right ) =0 \]

We know \(\Theta \) and \(r\). If this verifies, then we can use \(P=1\). Substituting the above becomes

\begin{align*} \left ( \left ( \frac {1}{2x}\right ) ^{\prime \prime }+3\frac {1}{2x}\left ( \frac {1}{2x}\right ) ^{\prime }+\left ( \frac {1}{2x}\right ) ^{3}-4\left ( \frac {16x-3}{16x^{2}}\right ) \left ( \frac {1}{2x}\right ) -2\left ( \frac {16x-3}{16x^{2}}\right ) ^{\prime }\right ) & =0\\ \left ( \frac {d^{2}}{dx^{2}}\left ( \frac {1}{2x}\right ) +3\frac {1}{2x}\frac {d}{dx}\left ( \frac {1}{2x}\right ) +\left ( \frac {1}{2x}\right ) ^{3}-4\left ( \frac {16x-3}{16x^{2}}\right ) \left ( \frac {1}{2x}\right ) -2\frac {d}{dx}\left ( \frac {16x-3}{16x^{2}}\right ) \right ) & =0\\ 0 & =0 \end{align*}

Verified. Hence

\[ P\left ( x\right ) =1 \]

Let

\begin{align*} \phi & =\Theta +\frac {P^{\prime }}{P}\\ & =\frac {1}{2x}\end{align*}

Now we solve for \(\omega \) from

\begin{align*} \omega ^{2}-\phi \omega +\left ( \frac {1}{2}\phi ^{\prime }+\frac {1}{2}\phi -r\right ) & =0\\ \omega ^{2}-\frac {1}{2x}\omega +\left ( \frac {1}{2}\left ( \frac {1}{2x}\right ) ^{\prime }+\frac {1}{2}\left ( \frac {1}{2x}\right ) -\frac {16x-3}{16x^{2}}\right ) & =0\\ \omega ^{2}-\frac {1}{2x}\omega +\frac {1}{16x^{2}}-\frac {1}{x} & =0 \end{align*}

The solution \(\ \omega =\frac {1}{4x}\pm \frac {1}{\sqrt {x}}\). We pick either solution. Hence the solution is

\begin{align*} z & =e^{\int \omega dx}\\ & =e^{\int \frac {1}{4x}+\frac {1}{\sqrt {x}}dx}\\ & =x^{\frac {1}{4}}e^{2\sqrt {x}}\end{align*}

Hence first solution to given ODE is

\begin{align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =x^{\frac {1}{4}}e^{2\sqrt {x}}e^{-\frac {1}{2}\int 0dx}\\ & =x^{\frac {1}{4}}e^{2\sqrt {x}}\end{align*}