problem Kovacic 1985 paper. page 25. section 5.2. Example 2

Internal problem ID [8331]
Internal ﬁle name [OUTPUT/7264_Sunday_June_05_2022_05_41_00_PM_16839255/index.tex]

Book: Collection of Kovacic problems
Section: section 3. Problems from Kovacic related papers
Problem number: Kovacic 1985 paper. page 25. section 5.2. Example 2.
ODE order: 2.
ODE degree: 1.

\[ \boxed {y^{\prime \prime }+\frac {\left (5 x^{2}+27\right ) y}{36 \left (x^{2}-1\right )^{2}}=0} \] Writing the ode as \begin {align*} y^{\prime \prime }+\frac {\left (5 x^{2}+27\right ) y}{36 \left (x^{2}-1\right )^{2}} &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end {align*}

Comparing (1) and (2) shows that \begin {align*} A &= 1 \\ B &= 0\tag {3} \\ C &= \frac {5 x^{2}+27}{36 \left (x^{2}-1\right )^{2}} \end {align*}

Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end {align*}

Where \(r\) is given by \begin {align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end {align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives \begin {align*} r &= \frac {-5 x^{2}-27}{36 \left (x^{2}-1\right )^{2}}\tag {6} \end {align*}

Comparing the above to (5) shows that \begin {align*} s &= -5 x^{2}-27\\ t &= 36 \left (x^{2}-1\right )^{2} \end {align*}

Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {-5 x^{2}-27}{36 \left (x^{2}-1\right )^{2}}\right ) z(x)\tag {7} \end {align*}

Equation (7) is now solved. After ﬁnding \(z(x)\) then \(y\) is found using the inverse transformation \begin {align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end {align*}

The ﬁrst step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.


Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)

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

2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisﬁed. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition

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

Table 841: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore \begin {align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 4 - 2 \\ &= 2 \end {align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=36 \left (x^{2}-1\right )^{2}\). There is a pole at \(x=1\) of order \(2\). There is a pole at \(x=-1\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then necessary conditions for case two are met. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore \begin {align*} L &= [1, 2, 4, 6, 12] \end {align*}

Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {11}{72 \left (x -1\right )}-\frac {2}{9 \left (x +1\right )^{2}}-\frac {11}{72 \left (x +1\right )}-\frac {2}{9 \left (x -1\right )^{2}} \] For the pole at \(x=1\) let \(b\) be the coeﬃcient of \(\frac {1}{ \left (x -1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b=-{\frac {2}{9}}\). Hence \begin {align*} E_c &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \end {align*}

Where \(n\) for case \(3\) is \(4,6\) or \(12\). For the current case \(n=4\). Hence the above becomes \begin {align*} E_c &= \{4, 5, 6, 7, 8\} \end {align*}

For the pole at \(x=-1\) let \(b\) be the coeﬃcient of \(\frac {1}{ \left (x +1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b=-{\frac {2}{9}}\). Hence \begin {align*} E_c &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \end {align*}

Where \(n\) for case \(3\) is \(4,6\) or \(12\). For the current case \(n=4\). Hence the above becomes \begin {align*} E_c &= \{4, 5, 6, 7, 8\} \end {align*}

Let \begin {align*} E_\infty &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \tag {B1} \end {align*}

Where \(b\) is the coeﬃcient of \(\frac {1}{x^{2}}\) in the Laurent series for \(r\) at \(\infty \) given by \[ r \approx -\frac {5}{36 x^{2}}-\frac {37}{36 x^{4}}-\frac {23}{12 x^{6}}-\frac {101}{36 x^{8}}-\frac {133}{36 x^{10}}-\frac {55}{12 x^{12}} + \cdots \] The above shows that \[ b = -{\frac {5}{36}} \] The value of \(n\) in eq. (B1) for case \(3\) is \(4,6\) or \(2\).For the current case \(n=4\). eq. (B1) simpliﬁes to the following, after removing any duplicate and non integer entries in the set. \begin {align*} E_\infty &= \{2, 4, 6, 8, 10\} \end {align*}

The following table summarizes the results found so far for poles and for the order of \(r\) at \(\infty \) for case 3 of Kovacic algorithm using \(n=4\).

Now that \(E_c\) sets for all poles are found and \(E_\infty \) set is found, the next step is to determine a non negative integer \(d\) using the following \begin {align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right ) \end {align*}

Where in the above \(e_c\) is a distinct element from each corresponding \(E_c\). This means all possible tuples \(\{e_{c_1},e_{c_2},\dots ,e_{c_n}\}\) are tried in the sum above, where \(e_{c_i}\) is one element of each \(E_c\) found earlier. Using the following family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=4,\hspace {3pt} e_2=4,\hspace {3pt} e_\infty =8 \] Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using \begin {align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {4}{12} \left ( 8 - \left (4+\left (4\right )\right )\right )\\ &= 0 \end {align*}

The following rational function is \begin {align*} \theta &= \frac {n}{12} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {4}{12} \left (\frac {4}{\left (x-\left (1\right )\right )}+\frac {4}{\left (x-\left (-1\right )\right )}\right ) \\ &= \frac {8 x}{3 x^{2}-3} \end {align*}

And \begin {align*} S &= \prod _{c\in \Gamma } (x-c) \\ &= \left (x -1\right ) \left (x +1\right ) \end {align*}

The polynomial \(p(x)\) is now determined. Since the degree of the polynomial is \(d=0\), then let \[ p(x) = 1 \] The following set of equations are set up in order to determine the coeﬃcients \(a_i\) (if any) of the above polynomial \begin {align*} P_n &= - p(x) \\ &= - 1 \\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \qquad i=n,n-1,\dots , 0 \tag {1A} \end {align*}

The coeﬃcients \(a_i\) are solved for from \[ P_{-1} = 0 \tag {2A} \] By using method of undetermined coeﬃcients. Carrying the above computation in eq. (1A) gives the following sequence of polynomials \(P_i\) (noting that \(n=4\) and \(r=\frac {-5 x^{2}-27}{36 \left (x^{2}-1\right )^{2}}\)). \begin{align*} P_{4} &= - p &= -1 \\ P_{3} &= \frac {8 x}{3} \\ P_{2} &= -5 x^{2}-\frac {1}{3} \\ P_{1} &= \frac {50}{9} x^{3}+\frac {14}{9} x \\ P_{0} &= -\frac {125}{54} x^{4}-\frac {67}{27} x^{2}+\frac {1}{18} \\ P_{-1} &= 0 \\ \end{align*} Because \(P_{-1} = 0\) then \(z=e^{\int \omega }\) is a solution. \(\omega \) is found by ﬁnding a solution to the equation generated by the following sum \begin {align*} \sum _{i=0}^{n} S^i \frac {P_i}{(n-i)!} \omega ^i &= 0 \\ \sum _{i=0}^{4} S^i \frac {P_i}{(4-i)!} \omega ^i &= 0 \end {align*}

Where the \(P_i\) are the polynomials found earlier. Computing the above sum gives \[ \tag{3A} \frac {\left (x -1\right )^{2} \left (x +1\right )^{2} \left (-5 x^{2}-\frac {1}{3}\right ) \omega ^{2}}{2}+\frac {8 \left (x +1\right )^{3} \left (x -1\right )^{3} x \,\omega ^{3}}{3}-\left (x -1\right )^{4} \left (x +1\right )^{4} \omega ^{4}-\frac {125 x^{4}}{1296}-\frac {67 x^{2}}{648}+\frac {1}{432}+\frac {25 \omega \,x^{5}}{27}-\frac {2 \omega \,x^{3}}{3}-\frac {7 \omega x}{27}=0 \] The solution \(\omega \) of eq. 3A is found as \begin{equation} \tag{4A} \omega =\frac {1}{6 x^{2}-6}\left (4 x +\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{\frac {2}{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}}}\right ) \end{equation} This \(\omega \) is used to ﬁnd a solution to \(z''=r z\). \[ z_1(x) = e^{ \int \omega \,dx}\tag {5A} \] Unable to integrate \(\int \omega dx\). Leaving the integral unevaluated. The ﬁrst solution to the original ode in \(y\) is found from \[ y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \] Since \(B=0\) then the above reduces to \begin{align*} y_1 &= z_1 \\ &= e^{\int \omega \,dx} \\ \end{align*} Where \(\omega \) given above. The second solution \(y_2\) to the original ode is found using reduction of order \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \\ &= e^{\int \omega \,dx}\int \frac { e^{\int -\frac {B}{A} \,dx}}{\left (e^{\int \omega \,dx}\right )^2} \,dx \\ \end{align*} Since \(B=0\) then the above reduces to \[ y_2 = e^{\int \omega \,dx} \int \left (e^{\int \omega \,dx}\right )^{-2} \,dx \] Therefore for kovacic case \(n=4\) the solution is

\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (e^{\int \omega \,dx}\right ) + c_{2} \left (e^{\int \omega \,dx} \int \left (e^{\int \omega \,dx}\right )^{-2} \,dx\right ) \\ \end{align*} Attempting to ﬁnd a solution using \(n=6\).

\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\right ) + c_{2} \left (\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\int \frac {4 x +\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{\frac {2}{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}}}}{6 x^{2}-6}d x}+c_{2} {\mathrm e}^{\frac {\left (\int \frac {4 x +\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{\frac {2}{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}}}}{x^{2}-1}d x \right )}{6}} \left (\int {\mathrm e}^{-\frac {\left (\int \frac {4 x +\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{\frac {2}{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}}}}{x^{2}-1}d x \right )}{3}}d x \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{\int \frac {4 x +\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{\frac {2}{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}}}}{6 x^{2}-6}d x}+c_{2} {\mathrm e}^{\frac {\left (\int \frac {4 x +\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{\frac {2}{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}}}}{x^{2}-1}d x \right )}{6}} \left (\int {\mathrm e}^{-\frac {\left (\int \frac {4 x +\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{\frac {2}{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{\frac {2}{3}}}}}}{x^{2}-1}d x \right )}{3}}d x \right ) \] Veriﬁed OK.

4.6.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {\left (5 x^{2}+27\right ) y}{36 \left (x^{2}-1\right )^{2}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=0, P_{3}\left (x \right )=\frac {5 x^{2}+27}{36 \left (x^{2}-1\right )^{2}}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\frac {2}{9} \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 36 y^{\prime \prime } \left (x^{2}-1\right )^{2}+\left (5 x^{2}+27\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (36 u^{4}-144 u^{3}+144 u^{2}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (5 u^{2}-10 u +32\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 16 a_{0} \left (-1+3 r \right ) \left (-2+3 r \right ) u^{r}+\left (16 a_{1} \left (2+3 r \right ) \left (1+3 r \right )-2 a_{0} \left (72 r^{2}-72 r +5\right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (16 a_{k} \left (3 k +3 r -1\right ) \left (3 k +3 r -2\right )-2 a_{k -1} \left (72 \left (k -1\right )^{2}+144 \left (k -1\right ) r +72 r^{2}-72 k +77-72 r \right )+a_{k -2} \left (6 k -13+6 r \right ) \left (6 k -17+6 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 16 \left (-1+3 r \right ) \left (-2+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{3}, \frac {2}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 16 a_{1} \left (2+3 r \right ) \left (1+3 r \right )-2 a_{0} \left (72 r^{2}-72 r +5\right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {a_{0} \left (72 r^{2}-72 r +5\right )}{8 \left (9 r^{2}+9 r +2\right )} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 36 \left (4 a_{k}+a_{k -2}-4 a_{k -1}\right ) k^{2}+36 \left (2 \left (4 a_{k}+a_{k -2}-4 a_{k -1}\right ) r -4 a_{k}-5 a_{k -2}+12 a_{k -1}\right ) k +36 \left (4 a_{k}+a_{k -2}-4 a_{k -1}\right ) r^{2}+36 \left (-4 a_{k}-5 a_{k -2}+12 a_{k -1}\right ) r +32 a_{k}+221 a_{k -2}-298 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 36 \left (4 a_{k +2}+a_{k}-4 a_{k +1}\right ) \left (k +2\right )^{2}+36 \left (2 \left (4 a_{k +2}+a_{k}-4 a_{k +1}\right ) r -4 a_{k +2}-5 a_{k}+12 a_{k +1}\right ) \left (k +2\right )+36 \left (4 a_{k +2}+a_{k}-4 a_{k +1}\right ) r^{2}+36 \left (-4 a_{k +2}-5 a_{k}+12 a_{k +1}\right ) r +32 a_{k +2}+221 a_{k}-298 a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}+72 k r a_{k}-288 k r a_{k +1}+36 r^{2} a_{k}-144 r^{2} a_{k +1}-36 k a_{k}-144 k a_{k +1}-36 r a_{k}-144 r a_{k +1}+5 a_{k}-10 a_{k +1}}{16 \left (9 k^{2}+18 k r +9 r^{2}+27 k +27 r +20\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}-12 k a_{k}-240 k a_{k +1}-3 a_{k}-74 a_{k +1}}{16 \left (9 k^{2}+33 k +30\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {1}{3}}, a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}-12 k a_{k}-240 k a_{k +1}-3 a_{k}-74 a_{k +1}}{16 \left (9 k^{2}+33 k +30\right )}, a_{1}=-\frac {11 a_{0}}{48}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +\frac {1}{3}}, a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}-12 k a_{k}-240 k a_{k +1}-3 a_{k}-74 a_{k +1}}{16 \left (9 k^{2}+33 k +30\right )}, a_{1}=-\frac {11 a_{0}}{48}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}+12 k a_{k}-336 k a_{k +1}-3 a_{k}-170 a_{k +1}}{16 \left (9 k^{2}+39 k +42\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2}{3}}, a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}+12 k a_{k}-336 k a_{k +1}-3 a_{k}-170 a_{k +1}}{16 \left (9 k^{2}+39 k +42\right )}, a_{1}=-\frac {11 a_{0}}{96}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +\frac {2}{3}}, a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}+12 k a_{k}-336 k a_{k +1}-3 a_{k}-170 a_{k +1}}{16 \left (9 k^{2}+39 k +42\right )}, a_{1}=-\frac {11 a_{0}}{96}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k +\frac {2}{3}}\right ), a_{k +2}=-\frac {36 k^{2} a_{k}-144 k^{2} a_{k +1}-12 k a_{k}-240 k a_{k +1}-3 a_{k}-74 a_{k +1}}{16 \left (9 k^{2}+33 k +30\right )}, a_{1}=-\frac {11 a_{0}}{48}, b_{k +2}=-\frac {36 k^{2} b_{k}-144 k^{2} b_{k +1}+12 k b_{k}-336 k b_{k +1}-3 b_{k}-170 b_{k +1}}{16 \left (9 k^{2}+39 k +42\right )}, b_{1}=-\frac {11 b_{0}}{96}\right ] \end {array} \]

Maple trace Kovacic algorithm successful

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Tetrahedral Galois group A4_SL2. 
<- Kovacics algorithm successful`

\[ y \left (x \right ) = c_{1} \left (x^{2}-1\right )^{\frac {1}{3}} {\mathrm e}^{\int \operatorname {RootOf}\left (-1+\left (432 x^{4}-864 x^{2}+432\right ) \textit {\_Z}^{4}+\left (-72 x^{2}+72\right ) \textit {\_Z}^{2}+16 x \textit {\_Z} , \operatorname {index} =1\right )d x}+c_{2} \left (x^{2}-1\right )^{\frac {1}{3}} {\mathrm e}^{\int \operatorname {RootOf}\left (-1+\left (432 x^{4}-864 x^{2}+432\right ) \textit {\_Z}^{4}+\left (-72 x^{2}+72\right ) \textit {\_Z}^{2}+16 x \textit {\_Z} , \operatorname {index} =2\right )d x} \]

\[ y(x)\to \sqrt {x^2-1} \left (c_1 P_{-\frac {1}{6}}^{\frac {1}{3}}(x)+c_2 Q_{-\frac {1}{6}}^{\frac {1}{3}}(x)\right ) \]


pole \(c\) location	pole order	set \(\{E_c\}\)

\(1\)	\(2\)	\(\{4, 5, 6, 7, 8\}\)

\(-1\)	\(2\)	\(\{4, 5, 6, 7, 8\}\)


Order of \(r\) at \(\infty \)	set \(\{E_\infty \}\)

\(2\)	\(\{2, 4, 6, 8, 10\}\)

4.6 problem Kovacic 1985 paper. page 25. section 5.2. Example 2

4.6.1 Maple step by step solution