Worked examples

Solve \[ \left ( 1-x^{2}\right ) y^{\prime \prime }-2xy^{\prime }+6y=0 \] Normalizing so that coeﬃcient of \(u^{\prime \prime }\) is one gives (assuming \(x\neq 1\))\begin {align} y^{\prime \prime }-2\frac {x}{\left ( 1-x^{2}\right ) }y^{\prime }+\frac {6}{\left ( 1-x^{2}\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 ( 1-x^{2}\right ) }\\ b & =\frac {6}{\left ( 1-x^{2}\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 ( 1-x^{2}\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {-2x}{\left ( 1-x^{2}\right ) }\right ) -\frac {6}{\left ( 1-x^{2}\right ) }\nonumber \\ & =\frac {6x^{2}-7}{\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 {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\). The free square factorization of \(t\) is \(t=t_{1}t_{2}^{2}\). Hence \begin {equation} m=2 \tag {6} \end {equation} And \(t_{1}=1,t_{2}=\left ( x^{2}-1\right ) \). Now \(O\left ( \infty \right ) =\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2\). The poles of \(r\) are \(x=1,-1\) each of order \(2\). Looking at the cases table giving up, 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 all cases are possible. Hence \(L=\left \{ 1,2,4,6,12\right \} \). So \(n=1,n=2,n=4,n=6,n=12\) will be tried until one is successful.

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 \(O\left ( \infty \right ) =2,t=\left ( x^{2}-1\right ) ^{2},t_{1}=1\) the above gives\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 {1}{2}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( \left ( x^{2}-1\right ) ^{2}\right ) }{\left ( x^{2}-1\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {x}{x^{2}-1} \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 \(2\). These will be the zeros of \(t_{2}\) in the above square free factorization of \(t\). From above we found that \(t_{2}=x^{2}-1\). Label these poles \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ 1,-1\right \} \) therefore \(c_{1}=1,c_{2}=-1\) and \(k_{2}=2\). For each \(c_{i}\) 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 \(r\) which is\begin {align*} \frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}} & =\left ( \sum _{i}\frac {\alpha _{i}}{\left ( x-c_{i}\right ) ^{2}}\right ) +\left ( \sum _{j}\frac {\beta _{j}}{x-d_{j}}\right ) \\ & =\left ( -\frac {1}{4}\frac {1}{\left ( x-1\right ) ^{2}}-\frac {1}{4}\frac {1}{\left ( x+1\right ) ^{2}}\right ) +\left ( \frac {13}{4}\frac {1}{\left ( x-1\right ) }-\frac {13}{4}\frac {1}{x+1}\right ) \end {align*}

Therefore for \(c_{1}=1\), looking at the above, we see that the coeﬃcient of \(\frac {1}{\left ( x-1\right ) ^{2}}\) is \(\frac {-1}{4}\). Hence \(b=\frac {-1}{4}\) and \(e_{1}=\sqrt {1+4b}=\sqrt {1-4\frac {1}{4}}=0\). For \(c_{2}=-1\ \)looking at the above, we see that the coeﬃcient of \(\frac {1}{\left ( x+1\right ) ^{2}}\) is \(\frac {-1}{4}\). Hence \(b=\frac {-1}{4}\) and \(e_{2}=\sqrt {1+4b}=\sqrt {1-4\frac {1}{4}}=0\). Therefore\begin {align*} e_{1} & =0\\ e_{2} & =0 \end {align*}

Now, \(\theta _{i}=\frac {e_{i}}{x-c_{i}}\). For \(c_{1}=1\) this gives \(\theta _{1}=\frac {e_{1}}{x-1}=0\) since \(e_{1}=0\). And \(\theta _{2}=\frac {e_{2}}{x-c_{2}}\). For \(c_{2}=-1\) this gives \(\theta _{2}=\frac {e_{2}}{x+1}=0\) since \(e_{2}=0\). Hence\begin {align*} \theta _{1} & =0\\ \theta _{2} & =0 \end {align*}

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 non exist here. This is skipped. This means \[ M=2 \] since \(2\) is the number of poles of order 2.

Now we need to ﬁnd \(b\). The Laurent series expansion of \(r\) at \(\infty \) is \(\frac {6}{x^{2}}+\frac {5}{x^{4}}+\frac {4}{x^{6}}+\cdots \). Hence \(b=6\). Therefore\begin {align*} e_{0} & =\sqrt {1+4\left ( 6\right ) }\\ & =\sqrt {25}\\ & =5 \end {align*}

Now we have found all \(e_{i},\theta _{i}\). They are\begin {align*} e & =\left \{ 5,0,0\right \} \\ \theta & =\left \{ 0,0,0\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 \(e_{fix},\theta _{fix}\). 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} \tag {7} \end {equation} Where \(n\) is the case number. For case 1, it will be \(n=1\). For case 2 it will be \(n=2\). For case 3 it will be \(4\) and \(6\) and \(12.\) 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} We need to ﬁrst generate \(s\) sets. For \(n=1\) and since \(M=2\) in this example (number of poles of order 2), then these these are given by\[ \Lambda =\begin {bmatrix} -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2}\\ -\frac {1}{2} & -\frac {1}{2} & +\frac {1}{2}\\ -\frac {1}{2} & +\frac {1}{2} & -\frac {1}{2}\\ -\frac {1}{2} & +\frac {1}{2} & +\frac {1}{2}\\ +\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2}\\ +\frac {1}{2} & -\frac {1}{2} & +\frac {1}{2}\\ +\frac {1}{2} & +\frac {1}{2} & -\frac {1}{2}\\ +\frac {1}{2} & +\frac {1}{2} & +\frac {1}{2}\end {bmatrix} \] We go over each row one at a time. Trying the ﬁrst row \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \) which means \(s_{0}=\frac {-1}{2},s_{1}=\frac {-1}{2},s_{2}=-\frac {1}{2}\). Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {x}{x^{2}-1}\) then\begin {align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {-1}{2}\left ( 0\right ) -\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end {align*}

Since this is negative, then we skip this set \(s\). Now we try the second row of \(\Lambda \) which is \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \). Then above now gives\begin {align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {-1}{2}\left ( 0\right ) +\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end {align*}

Since this is negative, then we skip this set \(s\). Now we try the third row of \(\Lambda \) which is \(s=\left \{ \frac {-1}{2},\frac {+1}{2},\frac {-1}{2}\right \} \). Then above now gives\begin {align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {+1}{2}\left ( 0\right ) -\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end {align*}

Since this is negative, then we skip this set \(s\). Now we try the row 4 of \(\Lambda \) which is \(s=\left \{ \frac {-1}{2},\frac {+1}{2},\frac {+1}{2}\right \} \). Then above now gives\begin {align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {+1}{2}\left ( 0\right ) +\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end {align*}

Since this is negative, then we skip this set \(s\). Now we try the row 5 of \(\Lambda \) which is \(s=\left \{ \frac {+1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \). Then above now gives\begin {align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {+1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {+1}{2}\right ) \left ( 5\right ) -\left ( \frac {+1}{2}\left ( 0\right ) +\frac {1}{2}\left ( 0\right ) \right ) \\ & =+2 \end {align*}

Since \(d\geq 0\), then we can use it. Now using Eq (8) gives\begin {align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +\sum _{i=0}^{M}s_{i}\theta _{i}\\ & =\left ( 1\right ) \left ( \frac {x}{x^{2}-1}\right ) +\left ( s_{0}\theta _{0}+s_{1}\theta _{1}+s_{2}\theta _{2}\right ) \\ & =\frac {x}{x^{2}-1}+\left ( \frac {+1}{2}\theta _{0}-\frac {1}{2}\theta _{1}-\frac {1}{2}\theta _{2}\right ) \end {align*}

But all \(\theta _{i}=0\). Therefore \[ \Theta =\frac {x}{x^{2}-1}\] Now that we have good trial \(d\) and \(\Theta \), then step 3 is called to generate \(\omega \) if possible.

The input to this step is the integer \(d=2\) and \(\Theta =\frac {x}{x^{2}-1}\) found from step 2 and also \(r=\frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}}\) which comes from \(z^{\prime \prime }=rz\). This step is broken into these parts. First we ﬁnd the \(p_{-1}\left ( x\right ) \) polynomial. If we are to solve for its coeﬃcients, then next we build the minimal polynomial from the \(p_{i}\left ( x\right ) \) polynomials constructed during ﬁnding \(p_{-1}\left ( x\right ) \). The minimal polynomial \(p_{\min }\left ( x\right ) \) will be a function of \(\omega \). Next we solve for \(\omega \) from \(p_{\min }\left ( x\right ) =0\). If this is successful, then we have found \(\omega \) and the ﬁrst solution to the ode \(z^{\prime \prime }=rz\) is \(z=e^{\int \omega dx}\,\). Below shows how this is done.

We start by forming a polynomial\begin {align*} p\left ( x\right ) & =x^{d}+a_{d-1}x^{d-1}+\cdots +a_{0}\\ & =x^{2}+a_{1}x+a_{0} \end {align*}

The goal is to solve for the \(a_{i}\) coeﬃcient. Now depending on case number \(n\), we do the following. Since we are in case \(n=1\) then\begin {align*} p_{1} & =-p\\ & =-x^{2}-a_{1}x-a_{0}\\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}\\ & =-\left ( -x^{2}-a_{1}x-a_{0}\right ) ^{\prime }-\left ( \frac {x}{x^{2}-1}\right ) \left ( -x^{2}-a_{1}x-a_{0}\right ) \\ & =\frac {\left ( xa_{0}-a_{1}-2x+2x^{2}a_{1}+3x^{3}\right ) }{x^{2}-1}\\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 1\right ) \left ( 1\right ) rp_{1}\\ & =-\left ( \frac {\left ( xa_{0}-a_{1}-2x+2x^{2}a_{1}+3x^{3}\right ) }{x^{2}-1}\right ) ^{\prime }-\left ( \frac {x}{x^{2}-1}\right ) \left ( \frac {\left ( xa_{0}-a_{1}-2x+2x^{2}a_{1}+3x^{3}\right ) }{x^{2}-1}\right ) -\frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}}\left ( -x^{2}-a_{1}x-a_{0}\right ) \\ & =2\frac {\left ( 2xa_{1}+3a_{0}+1\right ) }{x^{2}-1} \end {align*}

Now we try to solve for \(a_{i}\) using \(p_{-1}\left ( x\right ) =0\). This gives \(2xa_{1}+3a_{0}+1=0\) which gives \(a_{1}=0,a_{0}=-\frac {1}{3}\). Hence this implies\begin {align*} p\left ( x\right ) & =x^{2}+a_{1}x+a_{0}\\ & =x^{2}-\frac {1}{3} \end {align*}

Since this is case \(n=1\) then\begin {align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =\frac {2x}{x^{2}-\frac {1}{3}}+\frac {x}{x^{2}-1}\\ & =x\frac {9x^{2}-7}{3x^{4}-4x^{2}+1} \end {align*}

Before using this, we will verify it is correct. For case 1 the above should satisfy\[ \omega ^{\prime }+\omega ^{2}=r \] Let us see if this is the case or not.\begin {align*} \frac {d}{dx}\left ( x\frac {9x^{2}-7}{3x^{4}-4x^{2}+1}\right ) +\left ( x\frac {9x^{2}-7}{3x^{4}-4x^{2}+1}\right ) ^{2} & =\frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}}\\ \frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}} & =\frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}} \end {align*}

Veriﬁed. Since solution \(\omega \) is found and veriﬁed, then ﬁrst solution to the ode is \begin {align*} z & =e^{\int \omega dx}\\ & =e^{\int \frac {x\left ( 9x^{2}-7\right ) }{3x^{4}-4x^{2}+1}dx}\\ & =e^{\frac {1}{2}\ln \left ( x^{2}-1\right ) +\ln \left ( x^{2}-\frac {1}{3}\right ) }\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) \end {align*}

Hence ﬁrst solution to the original ode is\begin {align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) e^{\frac {-1}{2}\int \frac {-2x}{\left ( 1-x^{2}\right ) }dx}\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) e^{\int \frac {x}{\left ( 1-x^{2}\right ) }dx}\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) e^{\frac {-1}{2}\ln \left ( x^{2}-1\right ) }\\ & =\frac {\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) }{\sqrt {x^{2}-1}}\\ & =\left ( x^{2}-\frac {1}{3}\right ) \end {align*}

4.1.2 Example 2 case one

Solve \[ x\left ( x-1\right ) ^{2}y^{\prime \prime }-2y=0 \] Normalizing so that coeﬃcient of \(u^{\prime \prime }\) is one gives (assuming \(x\neq 1\) and \(x\neq 0\))\begin {align} y^{\prime \prime }-\frac {2}{x\left ( x-1\right ) ^{2}}y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0 \tag {1} \end {align}

Hence \begin {align*} a & =0\\ b & =-\frac {2}{x\left ( x-1\right ) ^{2}} \end {align*}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {2}{x\left ( x-1\right ) ^{2}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\). The free square factorization of \(t\) is \(t=t_{1}t_{2}^{2}\). Hence \begin {equation} m=2 \tag {6} \end {equation} And\begin {align*} t_{1} & =x\\ t_{2} & =x-1 \end {align*}

Now \(O\left ( \infty \right ) =\deg \left ( t\right ) -\deg \left ( s\right ) =3-0=3\). The poles of \(r\) are \(x=0\) of order 1 and \(x=1\) of order 2. Looking at the cases table giving up, 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 all cases are possible. Hence \(L=\left \{ 1,2,4,6,12\right \} \). So \(n=1,n=2,n=4,n=6,n=12\) will be tried until one is successful. Starting with \(n=1\).

Using \(O\left ( \infty \right ) =3,t=x\left ( x-1\right ) ^{2},t_{1}=x\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 3,2\right ) -3-3\left ( 1\right ) \right ) \\ & =\frac {1}{4}\left ( 2-3-3\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( x\left ( x-1\right ) ^{2}\right ) }{x\left ( x-1\right ) ^{2}}+3\left ( \frac {x^{\prime }}{x}\right ) \right ) \\ & =\frac {1}{4}\left ( \frac {3x^{2}-4x+1}{x\left ( x-1\right ) ^{2}}+\frac {3}{x}\right ) \\ & =\frac {1}{2x}\frac {3x-2}{x-1} \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 \(2\). These will be the zeros of \(t_{2}\) in the above square free factorization of \(t\). From above we found that \(t_{2}=x-1\). Label these poles \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ 1\right \} \) therefore \(c_{1}=1\) and \(k_{2}=1\) since one zero. Hence \[ M=1 \] For each \(c_{i}\) 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 \(r\) which is\begin {align*} r & =\frac {2}{x\left ( x-1\right ) ^{2}}\\ & =\left ( \sum _{i}\frac {\alpha _{i}}{\left ( x-c_{i}\right ) ^{2}}\right ) +\left ( \sum _{j}\frac {\beta _{j}}{x-d_{j}}\right ) \\ & =\left ( \frac {2}{\left ( x-1\right ) ^{2}}\right ) +\left ( -\frac {2}{x-1}+\frac {2}{x}\right ) \end {align*}

Therefore for \(c_{1}=1\), looking at the above, we see that the coeﬃcient of \(\frac {1}{\left ( x-1\right ) ^{2}}\) is \(2\). Hence \(b=2\) and \(e_{1}=\sqrt {1+4b}=\sqrt {1+8}=3\). Hence\[ e_{1}=3 \] Now, \(\theta _{i}=\frac {e_{i}}{x-c_{i}}\). For \(c_{1}=1\) this gives \(\theta _{1}=\frac {e_{1}}{x-1}=\frac {3}{x-1}\). Hence\[ \theta _{1}=\frac {3}{x-1}\] 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 non exist here. This is skipped. Hence \(M\) stays \(1\).

Now we have found all \(e_{i},\theta _{i}\). They are\begin {align*} e & =\left \{ 1,3\right \} \\ \theta & =\left \{ 0,\frac {3}{x-1}\right \} \end {align*}

In this step, we now have all the \(e_{i},\theta _{i}\) values found above in addition to \(e_{fix},\theta _{fix}\). 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} \tag {7} \end {equation} Where \(n\) is the case number. For case 1, it will be \(n=1\). For case 2 it will be \(n=2\). For case 3 it will be \(4\) and \(6\) and \(12.\) 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} We need to ﬁrst generate \(s\) sets. For \(n=1\) and since \(M=1\) in this example, then these these are given by\[ \Lambda =\begin {bmatrix} -\frac {1}{2} & -\frac {1}{2}\\ -\frac {1}{2} & +\frac {1}{2}\\ +\frac {1}{2} & -\frac {1}{2}\\ +\frac {1}{2} & +\frac {1}{2}\end {bmatrix} \] We go over each row one at a time. Trying the ﬁrst row \(s=\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \) which means \(s_{0}=\frac {-1}{2},s_{1}=\frac {-1}{2}\). Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-1,\theta _{fixed}=\frac {1}{2x}\frac {3x-2}{x-1}\) then\begin {align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\\ & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( 1\right ) -\left ( s_{1}e_{1}\right ) \\ & =-1-\frac {1}{2}-\left ( \frac {-1}{2}\left ( 3\right ) \right ) \\ & =0 \end {align*}

Since \(d\geq 0\), then we can use it. Now using Eq (8) gives\begin {align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +\sum _{i=0}^{M}s_{i}\theta _{i}\\ & =\left ( 1\right ) \left ( \frac {1}{2x}\frac {3x-2}{x-1}\right ) +\left ( s_{0}\theta _{0}+s_{1}\theta _{1}\right ) \\ & =\frac {1}{2x}\frac {3x-2}{x-1}+\left ( \frac {-1}{2}\theta _{0}-\frac {1}{2}\theta _{1}\right ) \\ & =\frac {1}{2x}\frac {3x-2}{x-1}-\frac {1}{2}\left ( 0\right ) -\frac {1}{2}\frac {3}{x-1} \end {align*}

Therefore\[ \Theta =-\frac {1}{x\left ( x-1\right ) }\] Now that we have good trial \(d\) and \(\Theta \), then step 3 is called to generate \(\omega \) if possible.

The input to this step is the integer \(d=0\) and \(\Theta =-\frac {1}{x\left ( x-1\right ) }\) found from step 2 and also \(r=\frac {2}{x\left ( x-1\right ) ^{2}}\) which comes from \(z^{\prime \prime }=rz\). This step is broken into these parts. First we ﬁnd the \(p_{-1}\left ( x\right ) \) polynomial. If we are to solve for its coeﬃcients, then next we build the minimal polynomial from the \(p_{i}\left ( x\right ) \) polynomials constructed during ﬁnding \(p_{-1}\left ( x\right ) \). The minimal polynomial \(p_{\min }\left ( x\right ) \) will be a function of \(\omega \). Next we solve for \(\omega \) from \(p_{\min }\left ( x\right ) =0\). If this is successful, then we have found \(\omega \) and the ﬁrst solution to the ode \(z^{\prime \prime }=rz\) is \(z=e^{\int \omega dx}\,\). Below shows how this is done.

We start by forming a polynomial\begin {align*} p\left ( x\right ) & =x^{d}+a_{d-1}x^{d-1}+\cdots +a_{0}\\ & =1 \end {align*}

Since this is case \(n=1\) then\begin {align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =0-\frac {1}{x\left ( x-1\right ) }\\ & =\frac {-1}{x\left ( x-1\right ) } \end {align*}

Before using this, we will verify it is correct. For case 1 the above should satisfy\[ \omega ^{\prime }+\omega ^{2}=r \] Let us see if this is the case or not.\begin {align*} \frac {d}{dx}\left ( -\frac {1}{x\left ( x-1\right ) }\right ) +\left ( -\frac {1}{x\left ( x-1\right ) }\right ) ^{2} & =\frac {2}{x\left ( x-1\right ) ^{2}}\\ \frac {1}{x^{2}}\frac {2x-1}{\left ( x-1\right ) ^{2}}+\left ( -\frac {1}{x\left ( x-1\right ) }\right ) ^{2} & =\frac {2}{x\left ( x-1\right ) ^{2}}\\ \frac {2}{x\left ( x-1\right ) ^{2}} & =\frac {2}{x\left ( x-1\right ) ^{2}} \end {align*}

Veriﬁed. Since solution \(\omega \) is found and veriﬁed, then ﬁrst solution to the ode is \begin {align*} z & =e^{\int -\frac {1}{x\left ( x-1\right ) }dx}\\ & =e^{\ln x-\ln \left ( x-1\right ) }\\ & =\frac {x}{x-1} \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =\frac {x}{x-1}e^{-\frac {1}{2}\int 0dx}\\ & =\frac {x}{x-1} \end {align*}

4.1.3 Example 3 case one

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

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\left ( \frac {x^{4}+4x^{2}-4x}{4}\right ) z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =x^{4}+4x^{2}-4x\\ t & =4 \end {align*}

The free square factorization of \(t\) is \(t=\left [ \left [ {}\right ] \right ] \). Hence \begin {equation} m=0 \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} & =1 \end {align*}

Now \begin {align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =0-4\\ & =-4 \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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Shows that only case 1 meets the necessary conditions. Hence \(L=\left [ 1\right ] \). So \(n=1.\)

Using \(O\left ( \infty \right ) =-4,t=4,t_{1}=4\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( -4,2\right ) -0-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( -4\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4\right ) }{\left ( 4\right ) }+3\left ( \frac {\left ( 4\right ) ^{\prime }}{4}\right ) \right ) \\ & =0 \end {align*}

In this step, we now have all the \(e_{i},\theta _{i}\) values found above in addition to \(e_{fix},\theta _{fix}\). Since \(n=1\) and \(M=0\) then we have \(\left ( n+1\right ) ^{M+1}=2^{1}=2\) sets \(s\) to try. These are given by\[ \Lambda =\begin {bmatrix} -\frac {1}{2}\\ +\frac {1}{2}\end {bmatrix} \] 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 \(M=0\) then the above becomes\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0} \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} Since \(M=0\). Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-1,\theta _{fixed}=0\) then\begin {align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +\left ( \frac {-1}{2}\right ) \left ( -2\right ) \\ & =\left ( 1\right ) \left ( -1\right ) +1\\ & =0 \end {align*}

Since \(d\geq 0\), then we can use it. Using Eq (8) gives\begin {align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +s_{0}\theta _{0}\\ & =0-\frac {1}{2}\left ( 2+x^{2}\right ) \\ & =-1-\frac {1}{2}x^{2} \end {align*}

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

The input to this step is the integer \(d=0\) and \(\Theta =-1-\frac {1}{2}x^{2}\) found from step 2 and also \(r=\frac {x^{4}+4x^{2}-4x}{4}\) which comes from \(z^{\prime \prime }=rz\). This step is broken into these parts. First we ﬁnd the \(p_{-1}\left ( x\right ) \) polynomial. If we are to solve for its coeﬃcients, then next we build the minimal polynomial from the \(p_{i}\left ( x\right ) \) polynomials constructed during ﬁnding \(p_{-1}\left ( x\right ) \). The minimal polynomial \(p_{\min }\left ( x\right ) \) will be a function of \(\omega \). Next we solve for \(\omega \) from \(p_{\min }\left ( x\right ) =0\). If this is successful, then we have found \(\omega \) and the ﬁrst solution to the ode \(y^{\prime \prime }=ry\) is \(e^{\int \omega dx}\,\). Below shows how this is done.

We start by forming a polynomial\begin {align*} p\left ( x\right ) & =x^{d}+a_{d-1}x^{d-1}+\cdots +a_{0}\\ & =1 \end {align*}

Since this is case \(n=1\) then\begin {align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =-1-\frac {1}{2}x^{2} \end {align*}

Before using this, we will verify it is correct. For case 1 the above should satisfy\[ \omega ^{\prime }+\omega ^{2}=r \] Let us see if this is the case or not.\begin {align*} \frac {d}{dx}\left ( -1-\frac {1}{2}x^{2}\right ) +\left ( -1-\frac {1}{2}x^{2}\right ) ^{2} & =\left ( \frac {x^{4}+4x^{2}-4x}{4}\right ) \\ -x+\frac {1}{4}x^{4}+x^{2}+1 & =\frac {x^{4}+4x^{2}-4x}{4}\\ \frac {x^{4}+4x^{2}-4x}{4}+\frac {1}{4} & =\frac {x^{4}+4x^{2}-4x}{4} \end {align*}

It did not verify. This means this solution can not be used. If we try the next row in \(\Lambda \) we will ﬁnd it gives negative \(d\). This means there is no Liouvillian solution. This is an example where even if we ﬁnd \(d\geq 0\) we still can end up not ﬁnding a solution.

4.1.4 Example 4 case one

Solve \begin {align} \left ( x^{3}+1\right ) y^{\prime \prime }+7x^{2}y^{\prime }+9xy & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =\frac {7x^{2}}{\left ( x^{3}+1\right ) }\\ b & =\frac {9x}{\left ( x^{3}+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 {7x^{2}}{\left ( x^{3}+1\right ) }\right ) ^{2}+\frac {1}{2}\left ( \frac {d}{dx}\left ( \frac {7x^{2}}{\left ( x^{3}+1\right ) }\right ) \right ) -\frac {9x}{\left ( x^{3}+1\right ) }\nonumber \\ & =\frac {1}{4}\left ( \frac {7x^{2}}{\left ( x^{3}+1\right ) }\right ) ^{2}+\frac {1}{2}\left ( \frac {-7x\left ( x^{3}-2\right ) }{\left ( x^{3}+1\right ) ^{2}}\right ) -\frac {9x}{\left ( x^{3}+1\right ) }\nonumber \\ & =\frac {-x\left ( x^{3}+8\right ) }{4\left ( x^{3}+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 {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =-x\left ( x^{3}+8\right ) \\ t & =4\left ( x^{3}+1\right ) ^{2} \end {align*}

The free square factorization of \(t\) is \(t=\left [ 1,x^{3}+1\right ] \). 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^{3}+1 \end {align*}

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

There are three poles each of order 2. Looking at the cases table giving up, 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 all three cases are possible. Hence \(L=\left [ 1,2,4,6,12\right ] \).

Using \(O\left ( \infty \right ) =2,t=4\left ( x^{3}+1\right ) ^{2},t_{1}=1\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 2,2\right ) -6-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 2-6\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4\left ( x^{3}+1\right ) ^{2}\right ) }{4\left ( x^{3}+1\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {3}{2}\frac {x^{2}}{x^{3}+1} \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 \(2\).

\[ r=-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\] These will be the zeros of \(t_{2}\) in the above square free factorization of \(t\). From above we found that \[ t_{2}=x^{3}+1 \] Label these zeros of \(t_{2}\) as \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ -1,\left ( -1\right ) ^{\frac {1}{3}},-\left ( -1\right ) ^{\frac {1}{3}}\right \} =\left \{ -1,\frac {1}{2}-\frac {1}{2}i\sqrt {3},\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right \} \). Therefore \(k_{2}=3\). Hence \[ M=3 \] 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 \begin {align*} b_{1} & =\frac {7}{36}\\ e_{1} & =\frac {4}{3}\\ b_{2} & =\frac {7}{36}\\ e_{2} & =\frac {4}{3}\\ b_{3} & =\frac {7}{36}\\ e_{3} & =\frac {4}{3} \end {align*}

And\begin {align*} \theta _{1} & =\frac {4}{3\left ( x+1\right ) }\\ \theta _{2} & =\frac {8}{3\left ( i\sqrt {3}+2x-1\right ) }\\ \theta _{3} & =\frac {-8}{3\left ( i\sqrt {3}-2x+1\right ) } \end {align*}

Hence now we have\begin {align*} e & =\left \{ 0,\frac {4}{3},\frac {4}{3},\frac {4}{3}\right \} \\ \theta & =\left \{ 0,\frac {4}{3\left ( x+1\right ) },\frac {8}{3\left ( i\sqrt {3}+2x-1\right ) },\frac {-8}{3\left ( i\sqrt {3}-2x+1\right ) }\right \} \end {align*}

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 \(M=3\) then there are \(\left ( n+1\right ) ^{M+1}=2^{4}=16\) sets \(s\) to try. The ﬁrst set \(s\) is\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \] 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 \(M=3\) then the above becomes\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1}-s_{2}e_{2}-s_{3}e_{3} \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 \(e_{fix}=-1,\theta _{fixed}=\frac {3}{2}\frac {x^{2}}{x^{3}+1}\) gives\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {4}{3}\right ) -\left ( -\frac {1}{2}\right ) \left ( \frac {4}{3}\right ) -\left ( -\frac {1}{2}\right ) \left ( \frac {4}{3}\right ) \\ & =1 \end {align*}

Since \(d\geq 0\), then we can use it. Using Eq (8) gives (using \(M=3\))\begin {align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}+s_{2}\theta _{2}+s_{3}\theta _{3}\\ & =\left ( 1\right ) \left ( \frac {3}{2}\frac {x^{2}}{x^{3}+1}\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {4}{3\left ( x+1\right ) }\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {8}{3\left ( i\sqrt {3}+2x-1\right ) }\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {-8}{3\left ( i\sqrt {3}-2x+1\right ) }\right ) \\ & =\left ( \frac {3}{2}\frac {x^{2}}{x^{3}+1}\right ) -\frac {2}{3\left ( x+1\right ) }-\frac {4}{3\left ( i\sqrt {3}+2x-1\right ) }+\frac {4}{3\left ( i\sqrt {3}-2x+1\right ) }\\ & =\frac {2x^{2}}{\left ( x+1\right ) \left ( i\sqrt {3}+2x-1\right ) \left ( i\sqrt {3}-2x+1\right ) }\\ & =\frac {-x^{2}}{2x^{3}+2} \end {align*}

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

The input to this step is the integer \(d=1\) and \(\Theta =\frac {-x^{2}}{2x^{3}+2}\) found from step 2 and also \(r=-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\) which comes from \(z^{\prime \prime }=rz\). This step is broken into these parts. First we ﬁnd the \(p_{-1}\left ( x\right ) \) polynomial. If we are to solve for its coeﬃcients, then next we build the minimal polynomial from the \(p_{i}\left ( x\right ) \) polynomials constructed during ﬁnding \(p_{-1}\left ( x\right ) \). The minimal polynomial \(p_{\min }\left ( x\right ) \) will be a function of \(\omega \). Next we solve for \(\omega \) from \(p_{\min }\left ( x\right ) =0\). If this is successful, then we have found \(\omega \) and the ﬁrst solution to the ode \(y^{\prime \prime }=ry\) is \(e^{\int \omega dx}\,\). Below shows how this is done.

We start by forming a polynomial\begin {align*} p\left ( x\right ) & =x^{d}+a_{d-1}x^{d-1}+\cdots +a_{0}\\ & =x \end {align*}

Since this is case \(n=1\) then\begin {align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =\frac {1}{x}-\frac {x^{2}}{2x^{3}+2}\\ & =\frac {x^{3}+2}{2x\left ( x^{3}+1\right ) } \end {align*}

Before using this, we will verify it is correct. For case 1 the above should satisfy

\[ \omega ^{\prime }+\omega ^{2}=r \] Let us see if this is the case or not.\begin {align*} \frac {d}{dx}\left ( \frac {x^{3}+2}{2x\left ( x^{3}+1\right ) }\right ) +\left ( \frac {x^{3}+2}{2x\left ( x^{3}+1\right ) }\right ) ^{2} & =-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\\ -\frac {\left ( x^{6}+6x^{3}+2\right ) }{2x^{2}\left ( x^{3}+1\right ) ^{2}}+\frac {\left ( x^{3}+2\right ) ^{2}}{4x^{2}\left ( x^{3}+1\right ) ^{2}} & =-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\\ -\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}} & =-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}} \end {align*}

Veriﬁed. Since solution \(\omega \) is found and veriﬁed, then ﬁrst solution to the ode is \begin {align*} z & =e^{\int \omega dx}\\ & =e^{\int \frac {x^{3}+2}{2x\left ( x^{3}+1\right ) }dx}\\ & =\frac {x}{\sqrt [6]{x^{3}+1}} \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =\frac {x}{\sqrt [6]{x^{3}+1}}e^{-\frac {1}{2}\int \frac {7x^{2}}{\left ( x^{3}+1\right ) }dx}\\ & =\frac {x}{\sqrt [6]{x^{3}+1}}\frac {1}{\left ( x^{3}+1\right ) ^{\frac {7}{6}}}\\ & =\frac {x}{\left ( x^{3}+1\right ) ^{\frac {4}{3}}} \end {align*}

4.1.5 Example 5 case one

Solve Bessel ode (from Kovacic original paper) \begin {align} y^{\prime \prime }-\frac {4\left ( m^{2}-x^{2}\right ) -1}{4x^{2}}y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }+by & =0\nonumber \end {align}

Hence \begin {align*} a & =0\\ b & =\frac {4\left ( m^{2}-x^{2}\right ) -1}{4x^{2}} \end {align*}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {4\left ( m^{2}-x^{2}\right ) -1}{4x^{2}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =4\left ( m^{2}-x^{2}\right ) -1\\ t & =4x^{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 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 ) \\ & =2-2\\ & =0 \end {align*}

There is one pole at \(x=0\) of order 2. Looking at the cases table giving up, 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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Using \(O\left ( \infty \right ) =0,t=4x^{2},t_{1}=1\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 0,2\right ) -2-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 0-2\right ) \\ & =-\frac {1}{2}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4x^{2}\right ) }{4x^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{2x} \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 \(2\).

\[ r=\frac {4\left ( m^{2}-x^{2}\right ) -1}{4x^{2}}\] 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 \begin {align*} b_{1} & =m^{2}-\frac {1}{4}\\ e_{1} & =\sqrt {1+4b}=\sqrt {1+4\left ( m^{2}-\frac {1}{4}\right ) }=2m\qquad m>0 \end {align*}

Where \(b_{1}\) is the coeﬃcient of \(\frac {1}{\left ( x-c_{1}\right ) ^{2}}\) in the partial fractions decomposition of \(r\) which is \(r=-1+\left ( m^{2}-\frac {1}{4}\right ) \frac {1}{x^{2}}\). And\[ \theta _{1}=\frac {e_{1}}{x-c_{1}}=\frac {2m}{x}\] Part (c)

Hence now we have\begin {align*} e & =\left \{ 0,2m\right \} \\ \theta & =\left \{ 2i,\frac {2m}{x}\right \} \end {align*}

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 \(M=1\) then there are \(\left ( n+1\right ) ^{M+1}=2^{2}=4\) sets \(s\) to try. The ﬁrst set \(s\) is\[ s=\left \{ \frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \] 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 \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {1}{2x}\) gives\begin {align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) -\left ( \frac {-1}{2}\right ) \left ( 2m\right ) \\ & =m-\frac {1}{2} \end {align*}

We now have to assume something about \(m\) to be to continue otherwise we will not be able to decide if \(d\) is integer and \(d\geq 0\). If we assume that \(m\) is half of all positive odd integers (\(\frac {1}{2},\frac {3}{2},\frac {5}{2},\cdots \)) then \(d\geq 0\). We can also assume that \(m\) is half of all negative odd integers and the other \(s\) set will match. So under the assumption that \(m\) is half of all positive odd integers the above \(d\) can be used for the next step. To continue, we assume \(m\) takes some speciﬁc value to simplify the steps. Let \(m=\frac {3}{2}\) from now on. Hence \(d=1\). Therefore \(e,\theta \) become

\begin {align*} e & =\left \{ 0,3\right \} \\ \theta & =\left \{ 2i,\frac {3}{x}\right \} \end {align*}

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 ( 1\right ) \left ( \frac {1}{2x}\right ) +\left ( \frac {-1}{2}\right ) \left ( 2i\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {3}{x}\right ) \\ & =-\frac {1}{x}\left ( ix+1\right ) \end {align*}

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

The input to this step is the integer \(d=1\) and \(\Theta =-\frac {1}{x}\left ( ix+1\right ) \) found from step 2 and also \(r=\frac {4\left ( n^{2}-x^{2}\right ) -1}{4x^{2}}\) which comes from \(z^{\prime \prime }=rz\). This step is broken into these parts. First we ﬁnd the \(p_{-1}\left ( x\right ) \) polynomial. If we are to solve for its coeﬃcients, then next we build the minimal polynomial from the \(p_{i}\left ( x\right ) \) polynomials constructed during ﬁnding \(p_{-1}\left ( x\right ) \). The minimal polynomial \(p_{\min }\left ( x\right ) \) will be a function of \(\omega \). Next we solve for \(\omega \) from \(p_{\min }\left ( x\right ) =0\). If this is successful, then we have found \(\omega \) and the ﬁrst solution to the ode \(y^{\prime \prime }=ry\) is \(e^{\int \omega dx}\,\). Below shows how this is done.

We start by forming a polynomial \begin {align*} p\left ( x\right ) & =x^{d}+a_{d-1}x^{d-1}+\cdots +a_{0}\\ & =x+a_{0} \end {align*}

The goal is to solve for the \(a_{0}\) coeﬃcient. Now depending on case number \(n\), we do the following. Since we are in case \(n=1\) then\begin {align*} p_{1} & =-p\\ & =-x-a_{0}\\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}\\ & =-\left ( -x-a_{0}\right ) ^{\prime }-\left ( -\frac {1}{x}\left ( ix+1\right ) \right ) \left ( -x-a_{0}\right ) \\ & =1-\left ( -\frac {1}{x}\left ( ix+1\right ) \right ) \left ( -x-a_{0}\right ) \\ & =-\frac {1}{x}\left ( ix^{2}+ia_{0}x+a_{0}\right ) \\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 1\right ) \left ( 1\right ) rp_{1}\\ & =-\left ( -\frac {1}{x}\left ( ix^{2}+ia_{0}x+a_{0}\right ) \right ) ^{\prime }-\left ( -\frac {1}{x}\left ( ix+1\right ) \right ) \left ( -\frac {1}{x}\left ( ix^{2}+ia_{0}x+a_{0}\right ) \right ) -\frac {4\left ( n^{2}-x^{2}\right ) -1}{4x^{2}}\left ( -x-a_{0}\right ) \\ & =-\frac {2}{x}\left ( ia_{0}-1\right ) \end {align*}

Now we try to solve for \(a_{i}\) using \(p_{-1}\left ( x\right ) =0\). This gives \(a_{0}=-i\). Hence this implies\[ p\left ( x\right ) =x-i \] Since this is case \(n=1\) then\begin {align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =\frac {\left ( x-i\right ) ^{\prime }}{x-i}-\frac {1}{x}\left ( ix+1\right ) \\ & =\frac {-i\left ( ix-x^{2}+1\right ) }{\left ( -x+i\right ) x}\\ & =-\frac {\left ( ix^{3}+1\right ) }{x^{3}+x} \end {align*}

Before using this, we will verify it is correct. For case 1 the above should satisfy\[ \omega ^{\prime }+\omega ^{2}=r \] Let us see if this is the case or not. \begin {align*} \frac {d}{dx}\left ( \frac {-i\left ( ix-x^{2}+1\right ) }{\left ( -x+i\right ) x}\right ) +\left ( \frac {-i\left ( ix-x^{2}+1\right ) }{\left ( -x+i\right ) x}\right ) ^{2} & =\frac {4\left ( \left ( \frac {3}{2}\right ) ^{2}-x^{2}\right ) -1}{4x^{2}}\\ \frac {\left ( -2ix^{3}+3x^{2}+1\right ) }{x^{2}\left ( x^{2}+1\right ) ^{2}}+\frac {\left ( ix^{2}+x-i\right ) ^{2}}{x^{2}\left ( x-i\right ) ^{2}} & =-\frac {1}{x^{2}}\left ( x^{2}-2\right ) \\ -\frac {1}{x^{2}}\left ( x^{2}-2\right ) & =-\frac {1}{x^{2}}\left ( x^{2}-2\right ) \end {align*}

Veriﬁed. Hence \(\omega =\frac {-i\left ( ix-x^{2}+1\right ) }{\left ( -x+i\right ) x}=-\frac {\left ( ix^{3}+1\right ) }{x^{3}+x}\) will give the solution to the ode \(y^{\prime \prime }-\frac {4\left ( m^{2}-x^{2}\right ) -1}{4x^{2}}y=0\) when \(m=\frac {3}{2}\). Since solution \(\omega \) is found and veriﬁed, then ﬁrst solution to the ode is \begin {align*} z & =e^{\int \omega dx}\\ & =e^{\int -\frac {\left ( ix^{3}+1\right ) }{x^{3}+x}dx}\\ & =\frac {1}{x}e^{-ix}\left ( x-i\right ) \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =\frac {1}{x}e^{-ix}\left ( x-i\right ) e^{-\frac {1}{2}\int 0dx}\\ & =\frac {1}{x}e^{-ix}\left ( x-i\right ) \end {align*}

One diﬃculty in implementation of Kovacic algorithm using an ode with a parameter \(m\) like in this Bessel ode example, is that it makes it hard to decide if \(d\geq 0\) or not. So in practice, it is better to use this algorithm for speciﬁc values of any parameters that can be involved.

4.1.6 Example 6 case one

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

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \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} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4\\ t & =4x^{4} \end {align*}

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

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

Using \(O\left ( \infty \right ) =0,t=4x^{4},t_{1}=1\) the above gives\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*}

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

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

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\[ s=\left \{ \frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \] 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 \(e_{fix}=-\frac {3}{2},\theta _{fixed}=\frac {1}{x}\) 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*}

This works. 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 ( 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*}

Which simpliﬁes to\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*}

Hence by comparing coeﬃcients\[ x\left ( -2a\right ) +\frac {1}{x}\left ( 4-3a+4b\right ) +\frac {1}{x^{2}}\left ( 2a\right ) +\left ( 2a-4b-4\right ) =0 \] Therefore \(a=0\). And \(4-3a+4b=0\) gives \(b=-1\). Same if we used \(2a-4b-4=0\), So consistent equations. Therefore \[ P\left ( x\right ) =x^{2}-1 \] And the solution is\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*}

Hence ﬁrst solution to given ODE is\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*}

4.1.7 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 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*}


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 \(L=\left [ 2\right ] \).

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*}

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 \(1\).

Hence now we have\begin {align*} e & =\left \{ 0,\frac {1}{2}\right \} \\ \theta & =\left \{ 0,\frac {1}{2x}\right \} \end {align*}

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.

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*}

4.1.8 Example 8 case two

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

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {1}{x^{3}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =1\\ t & =x^{3} \end {align*}

The free square factorization of \(t\) is \(t=\left [ 1,1,x\right ] \). Hence\begin {equation} m=3 \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} & =1 \end {align*}

Now \begin {align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =3-0\\ & =3 \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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Shows that only case 2 is possible (since odd pole is only allowed in case 2). Hence \(L=\left [ 2\right ] \).

Using \(O\left ( \infty \right ) =1,t=x^{3},t_{1}=1\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 3,2\right ) -3-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 2-3\right ) \\ & =-\frac {1}{4}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( x^{3}\right ) }{x^{3}}+3\left ( 0\right ) \right ) \\ & =\frac {3}{4x} \end {align*}

This part applied only to case 1. It is used to generate \(e_{i},\theta _{i}\) for poles of \(r\) order \(4,6,8,\cdots \) if any exist. Since only case 2 exist in this example. This is skipped. Hence \(k_{2}\) stays \(0\).

Now we need to ﬁnd \(e_{0},\theta _{0}\). If \(O\left ( \infty \right ) >2\) then \(e_{0}=1,\theta _{0}=0\). Hence \begin {align*} e_{0} & =1\\ \theta _{0} & =0 \end {align*}

Hence now we have\begin {align*} e & =\left \{ 1\right \} \\ \theta & =\left \{ 0\right \} \end {align*}

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 \(k_{2}=0\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=3\) sets \(s\) to try. The ﬁrst set \(s\) is\[ s=\left \{ \frac {-n}{2}\right \} =\left \{ -1\right \} \] Now we generate trial \(d\) using\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{k_{2}}s_{i}e_{i}\nonumber \end {equation} Since \(k_{2}=0\) then the above becomes\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0} \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}^{k_{2}}s_{i}\theta _{i} \tag {8} \end {equation} Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-\frac {1}{4},\theta _{fixed}=\frac {3}{4x}\) gives\begin {align*} d & =\left ( 2\right ) \left ( -\frac {1}{4}\right ) +\left ( -1\right ) \left ( 1\right ) \\ & =-\frac {3}{2} \end {align*}

Since negative then we can not use it. Now we try the next set \(s=\left \{ 0\right \} \). Then Eq(7) gives\begin {align*} d & =\left ( 2\right ) \left ( -\frac {1}{4}\right ) +\left ( 0\right ) \left ( 1\right ) \\ & =-\frac {1}{2} \end {align*}

Since negative then we can not use it. Now we try the last set \(s=\left \{ +1\right \} \). Then Eq(7) gives\begin {align*} d & =\left ( 2\right ) \left ( -\frac {1}{4}\right ) +\left ( +1\right ) \left ( 1\right ) \\ & =\frac {1}{2} \end {align*}

Since not an integer, then we can not use it. We are run out of sets \(s\) to try. Therefore there is no Liouvillian solution.

4.1.9 Example 9 case two

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

Hence \begin {align*} a & =-\frac {1}{2x}\\ b & =\frac {\left ( 1-2x\right ) }{2x^{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*}


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 (\(O\left ( \infty \right ) =1\) is only possible for case 2). Hence \(L=\left [ 2\right ] \).

Here the values \(e_{i},\theta _{i}\) are found for \(i=1\cdots k_{2}\) where \(k_{2}\) is the number of roots of \(t_{2}=x\). In other words, the number of poles of \(r\) that are of order \(2\). There is one pole of order \(2.\) Hence \(k_{2}=1\). the coeﬃcient of \(\frac {1}{\left ( x-0\right ) ^{2}}\) in the partial fractions of \(r=\frac {16x-3}{16x^{2}}=\frac {1}{x}-\frac {3}{16}\frac {1}{\left ( x-0\right ) ^{2}}\). Therefore \(b=-\frac {3}{16}\). Hence \(e_{1}=\sqrt {1+4b}=\sqrt {1+4\left ( -\frac {3}{16}\right ) }=\allowbreak \frac {1}{2}\) and \(\theta _{1}=\frac {e_{1}}{x-0}=\frac {1}{2x}\). Hence\begin {align*} e & =\left \{ \frac {1}{2}\right \} \\ \theta & =\left \{ \frac {1}{2x}\right \} \end {align*}

Now we need to ﬁnd \(e_{0},\theta _{0}\). Since this is not case 1 and since it is not \(O\left ( \infty \right ) >2\) and not \(O\left ( \infty \right ) =2\), then\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*}

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 \(k_{2}=1\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=3^{2}=9\) 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}^{k_{2}}s_{i}e_{i}\nonumber \end {equation} Since \(k_{2}=1\) 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}^{k_{2}}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*}

We can use this \(d.\)From Eq (8)\begin {align*} \Theta & =\left ( 2\right ) \left ( \frac {1}{2x}\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 ) \frac {1}{2x}\\ & =\frac {1}{2x} \end {align*}

Since this is case 2 (\(n=2\)) then we need to ﬁrst ﬁnd \(P\left ( x\right ) \). The degree is \(d=0\). Hence constant. Say \(P=1\). But we need to verify this is valid. Setting up the equation\[ P^{\prime \prime \prime }+3\Theta P^{\prime \prime }+\left ( 3\Theta ^{2}+3\Theta ^{\prime }-4r\right ) P^{\prime }+\left ( \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }\right ) P=0 \] Which simpliﬁes to (since \(P=1\))\[ \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }=0 \] Using \(\Theta =\frac {1}{2x},r=\frac {16x-3}{16x^{2}}\) the above reduces to \[ 0=0 \] Hence \(P\left ( x\right ) =1\) can be used. Now let \begin {align*} \phi & =\Theta +\frac {P^{\prime }}{P}\\ & =\frac {1}{2x} \end {align*}

Solving (and picking ﬁrst root) gives \[ \omega =\frac {1}{4x}\left ( 1+4\sqrt {x}\right ) \] Before using this, we verify it satisﬁes \(\omega ^{\prime }+\omega ^{2}=r\)\begin {align*} \frac {d}{dx}\left ( \frac {1}{4x}\left ( 1+4\sqrt {x}\right ) \right ) +\left ( \frac {1}{4x}\left ( 1+4\sqrt {x}\right ) \right ) ^{2} & =\frac {16x-3}{16x^{2}}\\ \frac {1}{16x^{2}}\left ( 16x-3\right ) & =\frac {16x-3}{16x^{2}} \end {align*}

Veriﬁed OK. Hence solution is \begin {align*} z & =e^{\int \omega dx}\\ & =x^{\frac {1}{4}}e^{2\sqrt {x}} \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y_{1} & =ze^{\frac {-1}{2}\int adx}\\ & =x^{\frac {1}{4}}e^{2\sqrt {x}}e^{-\frac {1}{2}\int -\frac {1}{2x}dx}\\ & =\sqrt {x}e^{2\sqrt {x}} \end {align*}

4.1.10 Example 10 case two

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

Hence \begin {align*} a & =\frac {3x}{\left ( x^{2}+2\right ) }\\ b & =-\frac {1}{\left ( x^{2}+2\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 {3x}{\left ( x^{2}+2\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {3x}{\left ( x^{2}+2\right ) }\right ) +\frac {1}{\left ( x^{2}+2\right ) }\nonumber \\ & =\frac {7x^{2}+20}{4\left ( x^{2}+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 {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =7x^{2}+20\\ t & =4\left ( x^{2}+2\right ) ^{2}=16+16x^{2}+4x^{4} \end {align*}

The free square factorization of \(t\) is \(t=\left [ 1,\left ( x^{2}+2\right ) \right ] \). 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} & =\left ( x^{2}+2\right ) \end {align*}

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

There is pole \(x=\pm i\sqrt {2}\) 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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Using \(O\left ( \infty \right ) =2,t=4\left ( x^{2}+2\right ) ^{2},t_{1}=1\) the above gives\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 {1}{2}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4\left ( x^{2}+2\right ) ^{2}\right ) }{4\left ( x^{2}+2\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {x}{x^{2}+2} \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 \(t_{2}=\left ( x^{2}+2\right ) \). In other words, the number of poles of \(r\) that are of order \(2\). There are two poles of order \(2.\) Hence \(k_{2}=2\). These poles at \(x=\pm i\sqrt {2}\). The coeﬃcient of \(\frac {1}{\left ( x-c_{1}\right ) ^{2}}\) where \(c_{1}\) is ﬁrst pole is \(b_{1}=-\frac {3}{16}\). Hence \(e_{1}=\sqrt {1+4b}=\sqrt {1+4\left ( -\frac {3}{16}\right ) }=\frac {1}{2}\) and \(\theta _{1}=\frac {e_{1}}{x-c_{1}}=\frac {1}{2\left ( x-i\sqrt {2}\right ) }\). The coeﬃcient of \(\frac {1}{\left ( x-c_{2}\right ) ^{2}}\) where \(c_{2}\) is second pole is \(b_{2}=-\frac {3}{16}\). Hence \(e_{2}=\sqrt {1+4b}=\sqrt {1+4\left ( -\frac {3}{16}\right ) }=\frac {1}{2}\) and \(\theta _{2}=\frac {e_{1}}{x-c_{2}}=\frac {1}{2\left ( x+i\sqrt {2}\right ) }\) Hence\begin {align*} e & =\left \{ \frac {1}{2},\frac {1}{2}\right \} \\ \theta & =\left \{ \frac {1}{2\left ( x-i\sqrt {2}\right ) },\frac {1}{2\left ( x+i\sqrt {2}\right ) }\right \} \end {align*}

This part applied only to case 1. It is used to generate \(e_{i},\theta _{i}\) for poles of \(r\) order \(4,6,8\) if any exist. Since only order 2 pole exist, then this is skipped. Hence \(k_{2}\) stays \(2\).

Now we need to ﬁnd \(e_{0},\theta _{0}\). Since this is case \(O\left ( \infty \right ) =2\), then \(e_{0}=\sqrt {1+4b}\) where \(b=\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }\) where \(\,lcoeff\left ( s\right ) \) is leading coeﬃcient of \(s=7x^{2}+20\) which is \(7\) and \(lcoeff\left ( t\right ) \) is leading coeﬃcient of \(t=16+16x^{2}+4x^{4}\) which is \(4\). Hence \(b=\frac {7}{4}\). Therefore \begin {align*} e_{0} & =\sqrt {1+4b}=\sqrt {1+4\left ( \frac {7}{4}\right ) }=2\sqrt {2}\\ \theta _{0} & =0 \end {align*}

Hence now we have\begin {align*} e & =\left \{ 2\sqrt {2},\frac {1}{2},\frac {1}{2}\right \} \\ \theta & =\left \{ 0,\frac {1}{2\left ( x-i\sqrt {2}\right ) },\frac {1}{2\left ( x+i\sqrt {2}\right ) }\right \} \end {align*}

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 \(k_{2}=2\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=2^{3}=8\) sets \(s\) to try. The ﬁrst set \(s\) is\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \] Now we generate trial \(d\) using\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{k_{2}}s_{i}e_{i}\nonumber \end {equation} Since \(k_{2}=2\) then the above becomes\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1}-s_{2}e_{2} \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}^{k_{2}}s_{i}\theta _{i} \tag {8} \end {equation} Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {x}{x^{2}+2}\) gives\begin {align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 2\sqrt {2}\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {1}{2}\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {1}{2}\right ) \\ & =-\sqrt {2} \end {align*}

Since not an integer, we try next set \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \) and now Eq (7) gives\begin {align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 2\sqrt {2}\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {1}{2}\right ) -\left ( \frac {+1}{2}\right ) \left ( \frac {1}{2}\right ) \\ & =-\sqrt {2}-\frac {1}{2} \end {align*}

Since not an integer, we try next set \(s=\left \{ \frac {-1}{2},\frac {+1}{2},\frac {-1}{2}\right \} \). If we continue this way we will ﬁnd that all sets \(s\) will fail to generate a \(d\geq 0\). Hence case one did not work. Now we go to case 2 \(\left ( n=2\right ) \).

Starting with \(n=2\). And since we have \(k_{2}=2\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=3^{3}\) sets \(s\) to try. The ﬁrst set \(s\) is\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ -1,-1,-1\right \} \] Now we generate trial \(d\) using\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{k_{2}}s_{i}e_{i}\nonumber \end {equation} Since \(k_{2}=2\) then the above becomes\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1}-s_{2}e_{2} \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}^{k_{2}}s_{i}\theta _{i} \tag {8} \end {equation} Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {x}{x^{2}+2}\) gives\begin {align*} d & =\left ( 2\right ) \left ( -\frac {1}{2}\right ) +\left ( -1\right ) \left ( 2\sqrt {2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) \\ & =-2\sqrt {2} \end {align*}

Since not an integer, we try next set \(s=\left \{ -1,-1,+1\right \} \). If we continue this way we will ﬁnd that set \(s=\left \{ 0,-1,-1\right \} \) works. \begin {align*} d & =\left ( 2\right ) \left ( -\frac {1}{2}\right ) +\left ( 0\right ) \left ( 2\sqrt {2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) \\ & =0 \end {align*}

We can use this \(d.\ \)From Eq (8)\begin {align*} \Theta & =\left ( 2\right ) \left ( \frac {1}{2x}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}+s_{2}\theta _{2}\\ & =\left ( 2\right ) \left ( \frac {x}{x^{2}+2}\right ) \left ( 0\right ) \left ( 0\right ) -\left ( -1\right ) \left ( \frac {1}{2\left ( x-i\sqrt {2}\right ) }\right ) -\left ( -1\right ) \left ( \frac {1}{2\left ( x+i\sqrt {2}\right ) }\right ) \\ & =\frac {x}{x^{2}+2} \end {align*}

Since this is case 2 (\(n=2\)) then we need to ﬁrst ﬁnd \(P\left ( x\right ) \). The degree is \(d=0\). Hence constant. Say \(P=1\). But we need to verify this is valid. Setting up the equation\[ P^{\prime \prime \prime }+3\Theta P^{\prime \prime }+\left ( 3\Theta ^{2}+3\Theta ^{\prime }-4r\right ) P^{\prime }+\left ( \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }\right ) P=0 \] Which simpliﬁes to (since \(P=1\))\[ \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }=0 \] Using \(\Theta =\frac {x}{x^{2}+2},r=\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}\) the above reduces to \[ 0=0 \] Hence \(P\left ( x\right ) =1\) can be used. Now let \begin {align*} \phi & =\Theta +\frac {P^{\prime }}{P}\\ & =\frac {x}{x^{2}+2} \end {align*}

We now need to solve for \(\omega \) from (notice that original Kovacic paper has \(+\) and not \(-\) after ﬁrst term in the following equation. The \(+\) is from Smith paper. It seems to have been a typo in original paper as this version gives the correct solution).\begin {align*} \omega ^{2}-\phi \omega +\left ( \frac {1}{2}\phi ^{\prime }+\frac {1}{2}\phi ^{2}-r\right ) & =0\\ \omega ^{2}-\left ( \frac {x}{x^{2}+2}\right ) +\omega \left ( \frac {1}{2}\left ( \frac {x}{x^{2}+2}\right ) ^{\prime }+\frac {1}{2}\left ( \frac {x}{x^{2}+2}\right ) ^{2}-\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}\right ) & =0\\ \frac {4\omega ^{2}x^{4}-4\omega x^{3}+\left ( 16\omega ^{2}-7\right ) x^{2}-8x\omega +16\omega ^{2}-16}{4\left ( x^{2}+2\right ) ^{2}} & =0\\ 4\omega ^{2}x^{4}-4\omega x^{3}+\left ( 16\omega ^{2}-7\right ) x^{2}-8x\omega +16\omega ^{2}-16 & =0 \end {align*}

Solving for \(\omega \) (and picking ﬁrst root) gives \[ \omega =\frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }\] Before using this, we verify it satisﬁes \(\omega ^{\prime }+\omega ^{2}=r\)\begin {align*} \frac {d}{dx}\left ( \frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }\right ) +\left ( \frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }\right ) ^{2} & =\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}\\ \frac {7x^{2}+20}{4x^{4}+16x^{2}+16} & =\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}} \end {align*}

Veriﬁed OK. Hence solution is \begin {align*} z & =e^{\int \omega dx}\\ & =e^{\int \frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }dx}\\ & =\left ( x^{2}+2\right ) ^{\frac {1}{4}}e^{\sqrt {2}\operatorname {arcsinh}\left ( \frac {\sqrt {2}}{2}x\right ) } \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y_{1} & =ze^{\frac {-1}{2}\int adx}\\ & =\left ( x^{2}+2\right ) ^{\frac {1}{4}}e^{\sqrt {2}\operatorname {arcsinh}\left ( \frac {\sqrt {2}}{2}x\right ) }e^{-\frac {1}{2}\int \frac {3x}{\left ( x^{2}+2\right ) }dx}\\ & =\frac {e^{\sqrt {2}\operatorname {arcsinh}\left ( \frac {\sqrt {2}}{2}x\right ) }}{\sqrt {\left ( x^{2}+2\right ) }} \end {align*}

4.1.11 Example 11 case one

Solve \begin {align} x^{2}\left ( x^{2}+x+1\right ) y^{\prime \prime }-x\left ( -2x^{2}-4x+1\right ) y^{\prime }+y & =0\tag {1}\\ y^{\prime \prime }-\frac {x\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }y^{\prime }+\frac {1}{x^{2}\left ( x^{2}+x+1\right ) }y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =-\frac {x\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }\\ b & =\frac {1}{x^{2}\left ( x^{2}+x+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 {x\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( -\frac {x\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }\right ) -\frac {1}{x^{2}\left ( x^{2}+x+1\right ) }\nonumber \\ & =\frac {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+x+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 {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+x+1\right ) ^{2}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =10x^{2}-8x-1\\ t & =4x^{2}\left ( x^{2}+x+1\right ) ^{2}\\ & =\left ( x^{3}+x^{2}+x\right ) ^{2} \end {align*}

The free square factorization of \(t\) is \(t=\left [ 1,x^{3}+x^{2}+x\right ] \). 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^{3}+x^{2}+x \end {align*}

Now \begin {align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =6-2\\ & =4 \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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Using \(O\left ( \infty \right ) =2,t=\left ( x^{3}+x^{2}+x\right ) ^{2},t_{1}=1\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 6,2\right ) -6-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 2-6\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( \left ( x^{3}+x^{2}+x\right ) ^{2}\right ) }{\left ( x^{3}+x^{2}+x\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{2}\frac {3x^{2}+2x+1}{x^{3}+x^{2}+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 \(t_{2}=x^{3}+x^{2}+x\). In other words, the number of poles of \(r\) that are of order \(2\). There are three poles of order \(2.\) Hence \(k_{2}=3\). These poles at \(x=\left \{ 0,-\frac {1}{2}\pm \frac {1}{2}i\sqrt {3}\right \} \). The coeﬃcient of \(\frac {1}{\left ( x-c_{1}\right ) ^{2}}\) where \(c_{1}\) is ﬁrst pole is \(b_{1}=-\frac {1}{4}\). Hence \(e_{1}=\sqrt {1+4b}=\sqrt {1+4\left ( -\frac {1}{4}\right ) }=0\) and \(\theta _{1}=\frac {e_{1}}{x-c_{1}}=0\). The coeﬃcient of \(\frac {1}{\left ( x-c_{2}\right ) ^{2}}\) where \(c_{2}\) is second pole is \(b_{2}=\frac {9i\sqrt {3}+2}{3\left ( -1+i\sqrt {3}\right ) ^{2}}\). Hence \(e_{2}=\sqrt {1+4b}=\frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) }\) and \(\theta _{2}=\frac {e_{2}}{x-c_{2}}=\frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) }\). The coeﬃcient of \(\frac {1}{\left ( x-c_{3}\right ) ^{2}}\) where \(c_{3}\) is the third pole is \(b_{3}=\frac {-9i\sqrt {3}+2}{3\left ( 1+i\sqrt {3}\right ) ^{2}}\). Hence \(e_{3}=\sqrt {1+4b}=\frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\) and \(\theta _{3}=\frac {e_{3}}{x-c_{3}}=\frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\). Hence\begin {align*} e & =\left \{ 0,\frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) },\frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right \} \\ \theta & =\left \{ 0,\frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) },\frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\right \} \end {align*}

Now we need to ﬁnd \(e_{0},\theta _{0}\). Since this is case \(O\left ( \infty \right ) =4>2\) and since there are no poles or order \(4,6,8,\cdots \) then we do not need to handle case \(n=1\). Instead we use \begin {align*} e_{0} & =1\\ \theta _{0} & =0 \end {align*}

Hence now we have\begin {align*} e & =\left \{ 1,0,\frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) },\frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right \} \\ \theta & =\left \{ 0,0,\frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) },\frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\right \} \end {align*}

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 \(k_{2}=3\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=2^{4}=16\) sets \(s\) to try. The ﬁrst set \(s\) is\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \] Now we generate trial \(d\) using\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{k_{2}}s_{i}e_{i}\nonumber \end {equation} Since \(k_{2}=3\) then the above becomes\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1}-s_{2}e_{2}-s_{3}e_{3} \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}^{k_{2}}s_{i}\theta _{i} \tag {8} \end {equation} Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-1,\theta _{fixed}=\frac {1}{2}\frac {3x^{2}+2x+1}{x^{3}+x^{2}+x}\) gives\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( +1\right ) -\left ( \frac {-1}{2}\right ) \left ( 0\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) }\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right ) \\ & =-\frac {7}{6}i\sqrt {3}-\frac {3}{2} \end {align*}

Since not an integer, we try next set \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \). If we continue this process we will ﬁnd that set \(s=\left \{ \frac {1}{2},\frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \) works and generates\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {1}{2}\right ) \left ( +1\right ) -\left ( \frac {-1}{2}\right ) \left ( 0\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) }\right ) -\left ( \frac {1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right ) \\ & =0 \end {align*}

We can use this \(d.\ \)From Eq (8)\begin {align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}+s_{2}\theta _{2}+s_{3}\theta _{3}\\ & =\left ( 1\right ) \left ( \frac {1}{2}\frac {3x^{2}+2x+1}{x^{3}+x^{2}+x}\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) }\right ) +\left ( \frac {1}{2}\right ) \left ( \frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\right ) \\ & =\frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1} \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 \(d=0\) and \(\Theta =\frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}\) found from step 2 and also \(r=\frac {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+x+1\right ) ^{2}}\). Since degree \(d=0\), then let \(p\left ( x\right ) =1\). A constant. We need to verify\begin {align*} P^{\prime \prime }+2\Theta P^{\prime }+\left ( \Theta ^{\prime }+\Theta ^{2}-r\right ) P & =0\\ \Theta ^{\prime }+\Theta ^{2}-r & =0 \end {align*}

Substituting gives\begin {align*} \frac {d}{dx}\left ( \frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}\right ) +\left ( \frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}\right ) ^{2}-\frac {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+x+1\right ) ^{2}} & =0\\ 0 & =0 \end {align*}

Veriﬁed. The solution is\begin {align*} z & =P\left ( x\right ) e^{\int \Theta dx}\\ & =e^{\int \frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}dx}\\ & =\left ( x^{2}+x+1\right ) ^{\frac {1}{4}}\sqrt {x}e^{-\frac {7}{6}\sqrt {3}\arctan \left ( \frac {\left ( 2x+1\right ) \sqrt {3}}{3}\right ) } \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y_{1} & =ze^{\frac {-1}{2}\int adx}\\ & =\left ( x^{2}+x+1\right ) ^{\frac {1}{4}}\sqrt {x}e^{-\frac {7}{6}\sqrt {3}\arctan \left ( \frac {\left ( 2x+1\right ) \sqrt {3}}{3}\right ) }e^{\frac {-1}{2}\int \left ( \frac {1}{x^{2}\left ( x^{2}+x+1\right ) }\right ) dx}\\ & =\frac {xe^{-\frac {7}{3}\sqrt {3}\arctan \left ( \frac {\left ( 2x+1\right ) \sqrt {3}}{3}\right ) }}{\sqrt {x^{2}+x+4}} \end {align*}

4.1.12 Example 12 case one

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 {x^{4}-8x^{3}+24x^{2}-24x+12}{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 {x^{4}-8x^{3}+24x^{2}-24x+12}{4x^{2}\left ( x-2\right ) ^{2}}z \tag {5} \end {equation} Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =x^{4}-8x^{3}+24x^{2}-24x+12\\ t & =4x^{2}\left ( x-2\right ) ^{2} \end {align*}

The square free factorization of \(t\) is \(t=\left [ 1,x\left ( x-2\right ) \right ] \). Hence\begin {equation} m=2 \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} & =x\left ( x-2\right ) \end {align*}

Now \begin {align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =4-4\\ & =0 \end {align*}

There is one pole at \(x=0\) of order 2 and one pole at \(x=2\) also 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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Using \(O\left ( \infty \right ) =0,t=4x^{4},t_{1}=1\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 0,2\right ) -4-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 0-4\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4x^{2}\left ( x-2\right ) ^{2}\right ) }{4x^{2}\left ( x-2\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {x^{2}-3x+2}{x\left ( x-2\right ) ^{2}}\\ & =\frac {x-1}{x\left ( x-2\right ) } \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 \(t_{2}=x\left ( x-2\right ) \). In other words, the number of poles of \(r\) that are of order \(2\). There are two poles. Hence \(k_{2}=2\). These poles \(c_{i}\) where \(i=1,2\) at \(x=\left \{ 0,2\right \} \). For each \(c_{i}\) 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\begin {align*} r & =\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}} \end {align*}

The coeﬃcient of \(\frac {1}{\left ( x-0\right ) ^{2}}\) where \(c_{1}=0\) is ﬁrst pole 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 \(\theta _{1}=\frac {e_{1}}{x-c_{1}}=\frac {2}{x}\). The coeﬃcient of \(\frac {1}{\left ( x-c_{2}\right ) ^{2}}\) where \(c_{2}=2\) is second pole is \(b_{2}=\frac {3}{4}\). Hence \(e_{2}=\sqrt {1+4b}=\sqrt {1+4\left ( \frac {3}{4}\right ) }=2\) and \(\theta _{2}=\frac {2}{x-c_{2}}=\frac {2}{x-2}\). Therefore the lists \(e,\theta \) are\begin {align*} e & =\left \{ 2,2\right \} \\ \theta & =\left \{ \frac {2}{x},\frac {2}{x-2}\right \} \end {align*}

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.

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 the \(O\left ( \infty \right ) \) which is zero here. Hence \[ v=0 \] The following is sum of terms from the Laurent series expansion of \(\sqrt {r}\) at \(x=\infty \) which is \[ \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}-\frac {1}{x}+\frac {2}{x^{3}}+\frac {11}{x^{4}}+\cdots \] We want only terms for \(0\leq i\leq v\) but \(v=0\). Therefore only the constant term. Hence\[ \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}\] Then \(a\) is the coeﬃcient of \(x^{-\frac {v}{2}}=x^{0}\) or constant term. Hence \[ a=\frac {1}{2}\] 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\[ b=-1 \] Hence\begin {align*} e_{0} & =\frac {b}{a}=-2\\ \theta _{0} & =2\left [ \sqrt {r}\right ] _{\infty }=1 \end {align*}

Hence now we have\begin {align*} e & =\left \{ -2,2,2\right \} \\ \theta & =\left \{ 1,\frac {2}{x},\frac {2}{x-2}\right \} \end {align*}

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 \(k_{2}=2\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=2^{3}=8\) sets \(s\) to try. The ﬁrst set \(s\) is\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \] Now we generate trial \(d\) using\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{k_{2}}s_{i}e_{i}\nonumber \end {equation} Since \(k_{2}=2\) then the above becomes\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1}-s_{2}e_{2} \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}^{k_{2}}s_{i}\theta _{i} \tag {8} \end {equation} Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-1,\theta _{fixed}=\frac {x^{2}-3x+2}{x\left ( x-2\right ) ^{2}}\) gives\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( -2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) \\ & =2 \end {align*}

This will work. Let us ﬁnd all of the \(d\) 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 \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( -2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) \\ & =2 \end {align*}

Trying next set \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( -2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) -\left ( \frac {+1}{2}\right ) \left ( 2\right ) \\ & =0 \end {align*}

Trying next set \(s=\left \{ \frac {-1}{2},\frac {+1}{2},\frac {+1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( -2\right ) -\left ( \frac {+1}{2}\right ) \left ( 2\right ) -\left ( \frac {+1}{2}\right ) \left ( 2\right ) \\ & =-2 \end {align*}

Trying next set \(s=\left \{ \frac {+1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {+1}{2}\right ) \left ( -2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) \\ & =0 \end {align*}

Trying next set \(s=\left \{ \frac {+1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {+1}{2}\right ) \left ( -2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) -\left ( \frac {+1}{2}\right ) \left ( 2\right ) \\ & =-2 \end {align*}

Trying the next set \(s=\left \{ \frac {+1}{2},\frac {+1}{2},\frac {-1}{2}\right \} \) \begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {+1}{2}\right ) \left ( -2\right ) -\left ( \frac {+1}{2}\right ) \left ( 2\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) \\ & =-2 \end {align*}

Trying the next set \(s=\left \{ \frac {+1}{2},\frac {+1}{2},\frac {+1}{2}\right \} \) \begin {align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {+1}{2}\right ) \left ( -2\right ) -\left ( \frac {+1}{2}\right ) \left ( 2\right ) -\left ( \frac {+1}{2}\right ) \left ( 2\right ) \\ & =-4 \end {align*}

OK, we have all \(d\) values. We now try the ones which are \(d\geq 0\) and these are \(d-0,d=2\). Trying \(d=2\) ﬁrst which used the set \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \) gives \(\left \{ 1,\frac {2}{x},\frac {2}{x-2}\right \} \)\begin {align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}+s_{21}\theta _{1}\\ & =\left ( 1\right ) \left ( \frac {x^{2}-3x+2}{x\left ( x-2\right ) ^{2}}\right ) +\left ( \frac {-1}{2}\right ) \left ( 1\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {2}{x}\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {2}{x-2}\right ) \\ & =-\frac {1}{2x}\frac {x^{2}-2}{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 \(d=0\) and \(\Theta =-\frac {1}{2x}\frac {x^{2}-2}{x-2}\) found from step 2 and also \(r=\frac {x^{4}-8x^{3}+24x^{2}-24x+12}{4x^{2}\left ( x-2\right ) ^{2}}\) which comes from \(z^{\prime \prime }=rz\). Since degree \(d=2\), then let \(p\left ( x\right ) =x^{2}+a_{1}x+a_{2}\). Solving for \(p\left ( x\right ) \) from\[ P^{\prime \prime }+2\Theta P^{\prime }+\left ( \Theta ^{\prime }+\Theta ^{2}-r\right ) P=0 \] gives \(p\left ( x\right ) =x^{2}\) as solution. Hence the solution is\begin {align*} z & =P\left ( x\right ) e^{\int \Theta dx}\\ & =x^{2}e^{\int -\frac {1}{2x}\frac {x^{2}-2}{x-2}dx}\\ & =\frac {x^{\frac {3}{2}}}{\sqrt {x-2}}e^{-\frac {x}{2}} \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y_{1} & =ze^{\frac {-1}{2}\int adx}\\ & =\frac {x^{\frac {3}{2}}}{\sqrt {x-2}}e^{-\frac {x}{2}}e^{\frac {-1}{2}\int \frac {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }dx}\\ & =x^{2} \end {align*}

4.1.13 Example 13 case one

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

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}\nonumber \end {align}

Step 0 We need to ﬁnd which case it is. \(r=\frac {s}{t}\) where\begin {align*} s & =x^{2}-4x+6\\ t & =4\left ( x-1\right ) ^{2} \end {align*}

The square free factorization of \(t\) is \(t=\left [ 1,\left ( x-1\right ) \right ] \). Hence\begin {equation} m=2 \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} & =\left ( x-1\right ) \end {align*}

Now \begin {align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =2-2\\ & =0 \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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Using \(O\left ( \infty \right ) =0,t=4\left ( x-1\right ) ^{2},t_{1}=1\) the above gives\begin {align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 0,2\right ) -2-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 0-2\right ) \\ & =-\frac {1}{2}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4\left ( x-1\right ) ^{2}\right ) }{4\left ( x-1\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{2x-2} \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 \(t_{2}=\left ( x-1\right ) \). 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\[ \frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}=\frac {1}{4}+\frac {3}{4\left ( x-1\right ) ^{2}}-\frac {1}{2}\frac {1}{x-1}\] 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\begin {align*} e & =\left \{ 2\right \} \\ \theta & =\left \{ \frac {2}{x-1}\right \} \end {align*}

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.

\[ \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}-\frac {1}{2x}+\frac {1}{x^{3}}+\cdots \] We want only terms for \(0\leq i\leq v\) but \(v=0\). Therefore only the constant term. Hence\[ \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}\] Then \(a\) is the coeﬃcient of \(x^{-\frac {v}{2}}=x^{0}\) or constant term. Hence \[ a=\frac {1}{2}\] 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 \[ b=-\frac {1}{2}\] Therefore\begin {align*} e_{0} & =\frac {b}{a}=\frac {-\frac {1}{2}}{\frac {1}{2}}=-1\\ \theta _{0} & =2\left [ \sqrt {r}\right ] _{\infty }=1 \end {align*}

Hence now we have\begin {align*} e & =\left \{ -1,2\right \} \\ \theta & =\left \{ 1,\frac {2}{x-1}\right \} \end {align*}

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\[ s=\left \{ \frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \] Now we generate trial \(d\) using\begin {equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{k_{2}}s_{i}e_{i}\nonumber \end {equation} Since \(k_{2}=1\) 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}^{k_{2}}s_{i}\theta _{i} \tag {8} \end {equation} 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 \(\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( -1\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) \\ & =1 \end {align*}

This will work. The corresponding \(\Theta \) is from (8)\begin {align*} \Theta & =\left ( 1\right ) \left ( \frac {1}{2x-2}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}\\ & =\frac {1}{2x-2}-\frac {1}{2}\left ( 1\right ) -\frac {1}{2}\frac {2}{x-1}\\ & =-\frac {1}{2}\frac {x}{x-1} \end {align*}

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

We skip this \(d\) since negative. Next is \(s=\left \{ \frac {+1}{2},\frac {-1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {1}{2}\right ) \left ( -1\right ) -\left ( \frac {-1}{2}\right ) \left ( 2\right ) \\ & =0 \end {align*}

The corresponding \(\Theta \) is from (8) \begin {align*} \Theta & =\left ( 1\right ) \left ( \frac {1}{2x-2}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}\\ & =\frac {1}{2x-2}+\frac {1}{2}\left ( 1\right ) -\frac {1}{2}\frac {2}{x-1}\\ & =\frac {x-2}{2\left ( x-1\right ) } \end {align*}

The next set is \(\left \{ \frac {+1}{2},\frac {+1}{2}\right \} \)\begin {align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {1}{2}\right ) \left ( -1\right ) -\left ( \frac {1}{2}\right ) \left ( \allowbreak 2\right ) \\ & =-2 \end {align*}

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.

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\[ P^{\prime \prime }+2\Theta P^{\prime }+\left ( \Theta ^{\prime }+\Theta ^{2}-r\right ) P=0 \] gives \(p\left ( x\right ) =x\) as solution. Hence the solution is\begin {align*} z & =P\left ( x\right ) e^{\int \Theta dx}\\ & =xe^{\int -\frac {1}{2}\frac {x}{x-1}dx}\\ & =x\frac {e^{-\frac {1}{2}x}}{\sqrt {x-1}} \end {align*}

Hence ﬁrst solution to given ODE is\begin {align*} y_{1} & =x\frac {e^{-\frac {1}{2}x}}{\sqrt {x-1}}e^{\frac {-1}{2}\int adx}\\ & =x\frac {e^{-\frac {1}{2}x}}{\sqrt {x-1}}e^{\frac {-1}{2}\int \frac {x}{1-x}dx}\\ & =x \end {align*}

4.1 Worked examples

4.1.1 Example 1 case one

4.1.2 Example 2 case one

4.1.3 Example 3 case one

4.1.4 Example 4 case one

4.1.5 Example 5 case one

4.1.6 Example 6 case one

4.1.7 Example 7 case two

4.1.8 Example 8 case two

4.1.9 Example 9 case two

4.1.10 Example 10 case two

4.1.11 Example 11 case one

4.1.12 Example 12 case one

4.1.13 Example 13 case one