Let
Hence
It is first transformed to the following ode by eliminating the first derivative
Using what is known as the Liouville transformation given by
Where it can be found that \(r\) in (2) is given by
Hence the DE we will solve using Kovacic algorithm is Eq (2) which is
Step 0 We need to find which case it is. \(r=\frac {s}{t}\) where
The square free factorization of \(t\) is \(t=\left [ 1,\left ( x-1\right ) \right ] \). Hence
Since \(m\) is number of elements in the free square factorization. in this case we set
Now
There is one pole at \(x=1\) of order 2. Looking at the cases table
| 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 1,2 are possible. Hence \(L=\left [ 1,2\right ] \).
Step 1
This step has 4 parts (a,b,c,d).
part (a) Here the fixed parts \(e_{fixed},\theta _{fixed}\) are calculated using
Using \(O\left ( \infty \right ) =0,t=4\left ( x-1\right ) ^{2},t_{1}=1\) the above gives
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 \(t_{2}=\left ( x-1\right ) \). In other words, the number of poles of \(r\) that are of order \(2\). There is one pole of order 2. Hence \(k_{2}=1\). For each pole \(c_{i}\) then \(e_{i}=\sqrt {1+4b}\) where \(b\) is the coefficient of \(\frac {1}{\left ( x-c_{i}\right ) ^{2}}\) in the partial fraction expansion of \(r\) and \(\theta _{i}=\frac {e_{i}}{x-c_{i}}\). The partial fraction expansion of \(r\) is
The coefficient of \(\frac {1}{\left ( x-1\right ) ^{2}}\) is \(b_{1}=\frac {3}{4}\) from looking at the above. Hence \(e_{1}=\sqrt {1+4b}=\sqrt {1+4\left ( \frac {3}{4}\right ) }=\allowbreak 2\) and \(\theta _{1}=\frac {e_{1}}{x-c_{1}}=\frac {\allowbreak 2}{x-1}\).Therefore the lists \(e,\theta \) are
Part (c)
This part applied only to case 1. It is used to generate \(e_{i},\theta _{i}\) for poles of \(r\) order \(4,6,8,\cdots ,k\) if any exist. There are none. This step is skipped.
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 \(\infty \). Since \(O\left ( \infty \right ) =0\) here then none of these cases applies. For case 1 \(\left ( n=1\right ) \) we first find \(\left [ r\right ] _{\infty }\) the sum of terms \(x^{i}\) for \(i=-\frac {v}{2},\cdots 0\) where \(v\) is \(O\left ( \infty \right ) \) which is zero here. Hence \(v=0\). This sum of terms is from the Laurent series expansion of \(\sqrt {r}\) at \(x=\infty \) which is
We want only terms for \(0\leq i\leq v\) but \(v=0\). Therefore only the constant term. Hence
Then \(a\) is the coefficient of \(x^{-\frac {v}{2}}=x^{0}\) or constant term. Hence
And \(b\) is the coefficient of \(x^{\frac {-v}{2}+1}=x\) in \(r-\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). This comes out to be
Therefore
Hence now we have
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 \(n=1\). And since we have \(k_{2}=1\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=2^{2}=4\) sets \(s\) to try. The first set \(s\) is
Now we generate trial \(d\) using
Since \(k_{2}=1\) then the above becomes
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
Hence the first trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {1}{2x-2}\) gives using set \(\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \)
This will work. The corresponding \(\Theta \) is from (8)
Let us find all of the \(d\) and \(\Theta \) so to compare with the solution to same ode using original kovacic algorithm given earlier to see if we get same \(d^{\prime }s\). We try next set \(s=\left \{ \frac {-1}{2},\frac {+1}{2}\right \} \)
We skip this \(d\) since negative. Next is \(s=\left \{ \frac {+1}{2},\frac {-1}{2}\right \} \)
The corresponding \(\Theta \) is from (8)
The next set is \(\left \{ \frac {+1}{2},\frac {+1}{2}\right \} \)
OK, we have all \(d\) values. We now try the ones which are \(d\geq 0\) and these are \(d=0,d=1\). Let us use \(d=1\) case. 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 \(d=1\) and \(\Theta =-\frac {1}{2}\frac {x}{x-1}\) found from step 2 and also \(r=\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}\) which comes from \(z^{\prime \prime }=rz\). Since degree \(d=1\), then let \(p\left ( x\right ) =x+a\). Solving for \(p\left ( x\right ) \) from
gives \(p\left ( x\right ) =x\) as solution. Hence the solution is
Hence first solution to given ODE is
The second solution can be found by reduction of order.