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

Internal problem ID [9420]
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
Date solved : Sunday, November 10, 2024 at 04:37:25 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }&=-\frac {\left (5 x^{2}+27\right ) y}{36 \left (x^{2}-1\right )^{2}} \end{align*}

Solved as second order ode using Kovacic algorithm

Time used: 89.451 (sec)

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*}

Then (2) becomes

\begin{align*} z''(x) = r z(x)\tag {4} \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 2.836: 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*}

Attempting to find a solution using case \(n=1\).

Unable to find solution using case one

Attempting to find a solution using case \(n=2\).

Unable to find solution using case two.

Attempting to find a solution using \(n=4\).

Looking at poles of order 2. The partial fractions decomposition of \(r\) is

\[ r = -\frac {11}{72 \left (x +1\right )}-\frac {2}{9 \left (x -1\right )^{2}}-\frac {2}{9 \left (x +1\right )^{2}}+\frac {11}{72 \left (x -1\right )} \]

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 \(n=4\).

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

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 )^{{2}/{3}}}+\sqrt {-\frac {2 \left (\left (-x^{2}+\frac {\left (x^{2}-1\right )^{{2}/{3}}}{2}+1\right ) \sqrt {x^{2}-1+\left (x^{2}-1\right )^{{2}/{3}}}+x^{3}-x \right )}{\sqrt {x^{2}-1+\left (x^{2}-1\right )^{{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 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*}

Will add steps showing solving for IC soon.

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {\left (5 x^{2}+27\right ) y \left (x \right )}{36 \left (x^{2}-1\right )^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )+\frac {\left (5 x^{2}+27\right ) y \left (x \right )}{36 \left (x^{2}-1\right )^{2}}=0 \\ \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 \left (x^{2}-1\right )^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (5 x^{2}+27\right ) y \left (x \right )=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 \left (x \right )=\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 \left (x \right )=\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 (x \right )=\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

`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`
 

Maple dsolve solution

Solving time : 0.013 (sec)
Leaf size : 25

dsolve(diff(diff(y(x),x),x) = -1/36*(5*x^2+27)/(x^2-1)^2*y(x), 
       y(x),singsol=all)
 

\[ y = \sqrt {x^{2}-1}\, \left (\operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, x\right ) c_{2} +\operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, x\right ) c_{1} \right ) \]

Mathematica DSolve solution

Solving time : 0.061 (sec)
Leaf size : 38

DSolve[{D[y[x],{x,2}]== -(5*x^2+27)/(36*(x^2-1)^2)*y[x],{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

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