Internal problem ID [8323]
Internal file name [OUTPUT/7256_Sunday_June_05_2022_05_39_54_PM_73652588/index.tex]
Book: Collection of Kovacic problems
Section: section 2. Solution found using all possible Kovacic cases
Problem number: 7.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (-x^{2}+1\right ) y^{\prime \prime }+y^{\prime }+y=0} \] Writing the ode as \begin {align*} \left (-x^{2}+1\right ) y^{\prime \prime }+y^{\prime }+y &= 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 &= -x^{2}+1 \\ B &= 1\tag {3} \\ C &= 1 \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 {4 x^{2}+4 x -3}{4 \left (x^{2}-1\right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 4 x^{2}+4 x -3\\ t &= 4 \left (x^{2}-1\right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {4 x^{2}+4 x -3}{4 \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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
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 \(t=4 \left (x^{2}-1\right )^{2}\). 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 \(n=2\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {7}{16 \left (x -1\right )}-\frac {3}{16 \left (x +1\right )^{2}}-\frac {7}{16 \left (x +1\right )}+\frac {5}{16 \left (x -1\right )^{2}} \] 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. Therefore \(b={\frac {5}{16}}\). Hence \begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{-1, 2, 5\} \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. Therefore \(b=-{\frac {3}{16}}\). Hence \begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{1, 2, 3\} \end {align*}
Since the order of \(r\) at \(\infty \) is 2 then let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\) from \begin {alignat*} {2} r &= \frac {s}{t} &&= \frac {4 x^{2}+4 x -3}{4 \left (x^{2}-1\right )^{2}} \end {alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=1\). Hence \begin {align*} E_\infty &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{2\} \end {align*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) for case 2 of Kovacic algorithm.
| pole \(c\) location | pole order | \(E_c\) |
| \(1\) | \(2\) | \(\{-1, 2, 5\}\) |
| \(-1\) | \(2\) | \(\{1, 2, 3\}\) |
| Order of \(r\) at \(\infty \) | \(E_\infty \) |
| \(2\) | \(\{2\}\) |
Using the family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=-1,\hspace {3pt} e_2=1,\hspace {3pt} e_\infty =2 \] Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using \begin {align*} d &= \frac {1}{2} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {1}{2} \left ( 2 - \left (-1+\left (1\right )\right )\right )\\ &= 1 \end {align*}
We now form the following rational function \begin {align*} \theta &= \frac {1}{2} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {1}{2} \left (\frac {-1}{\left (x-\left (1\right )\right )}+\frac {1}{\left (x-\left (-1\right )\right )}\right ) \\ &= -\frac {1}{2 \left (x -1\right )}+\frac {1}{2 x +2} \end {align*}
Now we search for a monic polynomial \(p(x)\) of degree \(d=1\) such that \[ p'''+3 \theta p'' + \left (3 \theta ^2 + 3 \theta ' - 4 r\right )p' + \left (\theta '' + 3 \theta \theta ' + \theta ^3 - 4 r \theta - 2 r' \right ) p = 0 \tag {1A} \] Since \(d=1\), then letting \[ p = x +a_{0}\tag {2A} \] Substituting \(p\) and \(\theta \) into Eq. (1A) gives \[ \frac {4 a_{0}+6}{\left (x -1\right )^{2} \left (x +1\right )} = 0 \] And solving for \(p\) gives \[ p = x -\frac {3}{2} \] Now that \(p(x)\) is found let \begin {align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {1}{x -\frac {3}{2}}-\frac {1}{2 \left (x -1\right )}+\frac {1}{2 x +2} \end {align*}
Let \(\omega \) be the solution of \begin {align*} \omega ^2 - \phi \omega + \left ( \frac {1}{2} \phi ' + \frac {1}{2} \phi ^2 - r \right ) &= 0 \end {align*}
Substituting the values for \(\phi \) and \(r\) into the above equation gives \[ w^{2}-\left (\frac {1}{x -\frac {3}{2}}-\frac {1}{2 \left (x -1\right )}+\frac {1}{2 x +2}\right ) w +\frac {-8 x^{3}+4 x^{2}+10 x -7}{4 \left (x^{2}-1\right )^{2} \left (2 x -3\right )} = 0 \] Solving for \(\omega \) gives \begin {align*} \omega &= \frac {2 \sqrt {5}\, \sqrt {\left (x -1\right ) \left (x +1\right )}\, x -2 \sqrt {5}\, \sqrt {\left (x -1\right ) \left (x +1\right )}+2 x^{2}-2 x +1}{2 \left (2 x -3\right ) \left (x -1\right ) \left (x +1\right )} \end {align*}
Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {2 \sqrt {5}\, \sqrt {\left (x -1\right ) \left (x +1\right )}\, x -2 \sqrt {5}\, \sqrt {\left (x -1\right ) \left (x +1\right )}+2 x^{2}-2 x +1}{2 \left (2 x -3\right ) \left (x -1\right ) \left (x +1\right )}d x}\\ &= \frac {\left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}\, \left (x +\sqrt {x^{2}-1}\right )^{\frac {\sqrt {5}}{2}} 5^{\frac {1}{4}}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {5 \sqrt {x^{2}-1}+\left (3 x -2\right ) \sqrt {5}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}} \end {align*}
The first solution to the original ode in \(y\) is found from \begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {1}{-x^{2}+1} \,dx} \\ &= z_1 e^{-\frac {\operatorname {arctanh}\left (x \right )}{2}} \\ &= z_1 \left (\frac {1}{\sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {\left (x +\sqrt {x^{2}-1}\right )^{\frac {\sqrt {5}}{2}} \sqrt {2 x -3}\, \left (5 x +5\right )^{\frac {1}{4}}}{\sqrt {\frac {i \left (3 \sqrt {5}\, x -2 \sqrt {5}+5 \sqrt {x^{2}-1}\right )}{2 x -3}}\, \left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {1}{-x^{2}+1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\operatorname {arctanh}\left (x \right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {i \left (x +\sqrt {x^{2}-1}\right )^{-\sqrt {5}} \left (3 \sqrt {5}\, x -2 \sqrt {5}+5 \sqrt {x^{2}-1}\right ) \sqrt {x -1}}{\left (2 x -3\right )^{2} \sqrt {5 x +5}}d x\right ) \\ \end{align*} Therefore for kovacic case \(n=2\) the solution is \begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{\frac {\sqrt {5}}{2}} \sqrt {2 x -3}\, \left (5 x +5\right )^{\frac {1}{4}}}{\sqrt {\frac {i \left (3 \sqrt {5}\, x -2 \sqrt {5}+5 \sqrt {x^{2}-1}\right )}{2 x -3}}\, \left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\right ) + c_{2} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{\frac {\sqrt {5}}{2}} \sqrt {2 x -3}\, \left (5 x +5\right )^{\frac {1}{4}}}{\sqrt {\frac {i \left (3 \sqrt {5}\, x -2 \sqrt {5}+5 \sqrt {x^{2}-1}\right )}{2 x -3}}\, \left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\left (\int \frac {i \left (x +\sqrt {x^{2}-1}\right )^{-\sqrt {5}} \left (3 \sqrt {5}\, x -2 \sqrt {5}+5 \sqrt {x^{2}-1}\right ) \sqrt {x -1}}{\left (2 x -3\right )^{2} \sqrt {5 x +5}}d x\right )\right ) \\ \end{align*} Attempting to find a solution using \(n=4\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {7}{16 \left (x -1\right )}-\frac {3}{16 \left (x +1\right )^{2}}-\frac {7}{16 \left (x +1\right )}+\frac {5}{16 \left (x -1\right )^{2}} \] 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 \(b={\frac {5}{16}}\). 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 &= \{-3, 6, 15\} \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 \(b=-{\frac {3}{16}}\). 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 &= \{3, 6, 9\} \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 \[ r \approx \frac {1}{x^{2}}+\frac {1}{x^{3}}+\frac {5}{4 x^{4}}+\frac {2}{x^{5}}+\frac {3}{2 x^{6}}+\frac {3}{x^{7}} + \cdots \] The above shows that \[ b = 1 \] 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 &= \{6\} \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 \(\{E_c\}\) |
| \(1\) | \(2\) | \(\{-3, 6, 15\}\) |
| \(-1\) | \(2\) | \(\{3, 6, 9\}\) |
| Order of \(r\) at \(\infty \) | set \(\{E_\infty \}\) |
| \(2\) | \(\{6\}\) |
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 \[ e_1=-3,\hspace {3pt} e_2=3,\hspace {3pt} e_\infty =6 \] 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 ( 6 - \left (-3+\left (3\right )\right )\right )\\ &= 2 \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 {-3}{\left (x-\left (1\right )\right )}+\frac {3}{\left (x-\left (-1\right )\right )}\right ) \\ &= -\frac {2}{x^{2}-1} \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 \(d=2\), then let \[ p(x) = x^{2}+a_{1} x +a_{0} \] 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) \\ &= - \left (x^{2}+a_{1} x +a_{0}\right ) \\ 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 \(r=\frac {4 x^{2}+4 x -3}{4 \left (x^{2}-1\right )^{2}}\)). \begin{align*} P_{4} &= - p &= -x^{2}-a_{1} x -a_{0} \\ P_{3} &= 2 x^{3}+\left (a_{1}-2\right ) x^{2}+\left (-2 a_{1}-2\right ) x -2 a_{0}-a_{1} \\ P_{2} &= 2 x^{4}+\left (4 a_{1}+8\right ) x^{3}+\left (4 a_{0}+4 a_{1}-3\right ) x^{2}+\left (-7 a_{1}-8\right ) x -7 a_{0}-4 a_{1}-2 \\ P_{1} &= -12-12 x^{5}+2 \left (6-a_{1}\right ) x^{4}+\left (51+8 a_{0}+22 a_{1}\right ) x^{3}+\frac {\left (40 a_{0}+43 a_{1}-6\right ) x^{2}}{2}+\left (-39-8 a_{0}-25 a_{1}\right ) x -23 a_{0}-\frac {39 a_{1}}{2} \\ P_{0} &= -72-24 x^{6}+4 \left (-15-7 a_{1}\right ) x^{5}+2 \left (48+7 a_{1}\right ) x^{4}+2 \left (120+36 a_{0}+79 a_{1}\right ) x^{3}+\frac {\left (168 a_{0}+136 a_{1}-3\right ) x^{2}}{2}+\frac {\left (-144 a_{0}-263 a_{1}-360\right ) x}{2}-\frac {171 a_{0}}{2}-82 a_{1} \\ P_{-1} &= -2 \left (x -1\right ) \left (38 x^{2} a_{1}+114 x^{2}+16 x a_{0}-36 a_{1} x -144 x -156 a_{0}-177 a_{1}-180\right ) \left (x +1\right )^{3} \\ \end{align*} Using method of undetermined coefficient to solve for \(a_i\) from the last equation above \(P_{-1} = 0 \) gives the following solution for the coefficients \(a_i\) \begin {align*} a_{0}&={\frac {9}{4}}\\ a_{1}&=-3\\ \end {align*}
Substituting these in the the polynomial \(p(x)\) results in \begin {align*} P(x) &= x^{2}-3 x +\frac {9}{4} \end {align*}
\(\omega \) 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 \[ \tag{3A} -3-x^{6}+\frac {\left (-15-7 a_{1}\right ) x^{5}}{6}+\frac {\left (48+7 a_{1}\right ) x^{4}}{12}+\frac {\left (120+36 a_{0}+79 a_{1}\right ) x^{3}}{12}+\frac {\left (168 a_{0}+136 a_{1}-3\right ) x^{2}}{48}+\frac {\left (-144 a_{0}-263 a_{1}-360\right ) x}{48}-\frac {57 a_{0}}{16}-\frac {41 a_{1}}{12}-2 \omega \left (x +1\right ) \left (x^{5}+\left (\frac {a_{1}}{6}-1\right ) x^{4}+\left (-\frac {2 a_{0}}{3}-\frac {11 a_{1}}{6}-\frac {17}{4}\right ) x^{3}+\left (-\frac {5 a_{0}}{3}-\frac {43 a_{1}}{24}+\frac {1}{4}\right ) x^{2}+\left (\frac {2 a_{0}}{3}+\frac {25 a_{1}}{12}+\frac {13}{4}\right ) x +\frac {23 a_{0}}{12}+\frac {13 a_{1}}{8}+1\right ) \left (x -1\right )+\omega ^{2} \left (x +1\right )^{2} \left (x^{4}+\left (2 a_{1}+4\right ) x^{3}+\left (2 a_{0}+2 a_{1}-\frac {3}{2}\right ) x^{2}+\left (-\frac {7 a_{1}}{2}-4\right ) x -\frac {7 a_{0}}{2}-2 a_{1}-1\right ) \left (x -1\right )^{2}+\left (x +1\right )^{3} \left (x -1\right )^{3} \left (2 x^{3}+\left (a_{1}-2\right ) x^{2}+\left (-2 a_{1}-2\right ) x -2 a_{0}-a_{1}\right ) \omega ^{3}-\left (x -1\right )^{4} \left (x +1\right )^{4} \left (x^{2}+a_{1} x +a_{0}\right ) \omega ^{4}=0 \] The solution \(\omega \) of eq. 3A is found as \[ \omega =\frac {1}{4 x^{3}-6 x^{2}-4 x +6}\left (-2 \sqrt {5}\, \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}+2 x^{2}-2 x +1\right ) \] This \(\omega \) is used to find a solution to \(z''=r z\). \[ z_1(x) = e^{ \int \omega \,dx}\tag {5A} \] Doing the integration gives in eq. (4A) gives \begin{align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {-2 \sqrt {5}\, \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}+2 x^{2}-2 x +1}{4 x^{3}-6 x^{2}-4 x +6}d x} \\ &= \frac {\left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}\, \left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}\, \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}}{\sqrt {x^{2}-1}\, \left (2 x -2\right )}} {\mathrm e}^{\frac {\operatorname {arctanh}\left (\frac {\left (3 x -2\right ) \sqrt {5}}{5 \sqrt {x^{2}-1}}\right ) \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}}{\sqrt {x^{2}-1}\, \left (2 x -2\right )}}}{\left (x -1\right )^{\frac {1}{4}}} \\ \end{align*} Which simplifies to \[ z_1(x) = \frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}}} \] The first solution to the original ode in \(y\) is found from \begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {1}{-x^{2}+1} \,dx} \\ &= z_1 e^{-\frac {\operatorname {arctanh}\left (x \right )}{2}} \\ &= z_1 \left (\frac {1}{\sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {1}{-x^{2}+1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\operatorname {arctanh}\left (x \right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {\left (x +\sqrt {x^{2}-1}\right )^{\sqrt {5}} \sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}\, \sqrt {x -1}}{\sqrt {x +1}\, \left (3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}\right ) \left (2 x -3\right )}d x\right ) \\ \end{align*} Therefore for kovacic case \(n=4\) the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\right ) + c_{2} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\left (\int \frac {\left (x +\sqrt {x^{2}-1}\right )^{\sqrt {5}} \sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}\, \sqrt {x -1}}{\sqrt {x +1}\, \left (3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}\right ) \left (2 x -3\right )}d x\right )\right ) \\ \end{align*} Attempting to find a solution using \(n=6\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {7}{16 \left (x -1\right )}-\frac {3}{16 \left (x +1\right )^{2}}-\frac {7}{16 \left (x +1\right )}+\frac {5}{16 \left (x -1\right )^{2}} \] 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 \(b={\frac {5}{16}}\). 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 \(n=6\). Hence the above becomes \begin {align*} E_c &= \{-3, 0, 3, 6, 9, 12, 15\} \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 \(b=-{\frac {3}{16}}\). 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 \(n=6\). Hence the above becomes \begin {align*} E_c &= \{3, 4, 5, 6, 7, 8, 9\} \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 \[ r \approx \frac {1}{x^{2}}+\frac {1}{x^{3}}+\frac {5}{4 x^{4}}+\frac {2}{x^{5}}+\frac {3}{2 x^{6}}+\frac {3}{x^{7}} + \cdots \] The above shows that \[ b = 1 \] The value of \(n\) in eq. (B1) for case \(3\) is \(4,6\) or \(2\).For the current case \(n=6\). eq. (B1) simplifies to the following, after removing any duplicate and non integer entries in the set. \begin {align*} E_\infty &= \{6\} \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 \(n=6\).
| pole \(c\) location | pole order | set \(\{E_c\}\) |
| \(1\) | \(2\) | \(\{-3, 0, 3, 6, 9, 12, 15\}\) |
| \(-1\) | \(2\) | \(\{3, 4, 5, 6, 7, 8, 9\}\) |
| Order of \(r\) at \(\infty \) | set \(\{E_\infty \}\) |
| \(2\) | \(\{6\}\) |
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 \[ e_1=-3,\hspace {3pt} e_2=3,\hspace {3pt} e_\infty =6 \] 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 {6}{12} \left ( 6 - \left (-3+\left (3\right )\right )\right )\\ &= 3 \end {align*}
The following rational function is \begin {align*} \theta &= \frac {n}{12} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {6}{12} \left (\frac {-3}{\left (x-\left (1\right )\right )}+\frac {3}{\left (x-\left (-1\right )\right )}\right ) \\ &= -\frac {3}{x^{2}-1} \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 \(d=3\), then let \[ p(x) = x^{3}+a_{2} x^{2}+a_{1} x +a_{0} \] 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) \\ &= - \left (x^{3}+a_{2} x^{2}+a_{1} x +a_{0}\right ) \\ 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 \(n=6\) and \(r=\frac {4 x^{2}+4 x -3}{4 \left (x^{2}-1\right )^{2}}\)). \begin{align*} P_{6} &= - p &= -x^{3}-a_{2} x^{2}-a_{1} x -a_{0} \\ P_{5} &= 3 x^{4}+\left (2 a_{2}-3\right ) x^{3}+\left (a_{1}-3 a_{2}-3\right ) x^{2}+\left (-3 a_{1}-2 a_{2}\right ) x -3 a_{0}-a_{1} \\ P_{4} &= 2 \left (9+2 a_{2}\right ) x^{4}+\frac {3 \left (-1+4 a_{1}+8 a_{2}\right ) x^{3}}{2}+\frac {\left (12 a_{0}+12 a_{1}-19 a_{2}-36\right ) x^{2}}{2}+\frac {3 \left (-4-9 a_{1}-8 a_{2}\right ) x}{2}-\frac {27 a_{0}}{2}-6 a_{1}-2 a_{2} \\ P_{3} &= -30 x^{6}-20 x^{5} a_{2}+\left (-4 a_{1}+34 a_{2}+135\right ) x^{4}+\left (12 a_{0}+50 a_{1}+98 a_{2}+39\right ) x^{3}+\left (48 a_{0}+43 a_{1}-31 a_{2}-99\right ) x^{2}+\left (-12 a_{0}-65 a_{1}-78 a_{2}-54\right ) x -63 a_{0}-39 a_{1}-18 a_{2}-6 \\ P_{2} &= -72+34 \left (-9-2 a_{2}\right ) x^{6}+4 \left (-27-20 a_{1}-46 a_{2}\right ) x^{5}+2 \left (459-18 a_{0}-3 a_{1}+151 a_{2}\right ) x^{4}+\frac {3 \left (357+104 a_{0}+300 a_{1}+448 a_{2}\right ) x^{3}}{2}+\frac {\left (672 a_{0}+484 a_{1}-213 a_{2}-1080\right ) x^{2}}{2}+\frac {\left (-312 a_{0}-785 a_{1}-976 a_{2}-900\right ) x}{2}-\frac {645 a_{0}}{2}-236 a_{1}-150 a_{2} \\ P_{1} &= -720+360 x^{8}+4 \left (-63+26 a_{2}\right ) x^{7}+12 \left (-261-16 a_{1}-77 a_{2}\right ) x^{6}+4 \left (-144-72 a_{0}-204 a_{1}-379 a_{2}\right ) x^{5}+\frac {\left (-96 a_{0}+1360 a_{1}+5696 a_{2}+13653\right ) x^{4}}{2}+\frac {\left (4032 a_{0}+6984 a_{1}+9550 a_{2}+8451\right ) x^{3}}{2}+\frac {\left (3432 a_{0}+1927 a_{1}-1693 a_{2}-6669\right ) x^{2}}{2}+\frac {3 \left (-2280-1152 a_{0}-1799 a_{1}-2242 a_{2}\right ) x}{2}-\frac {3381 a_{0}}{2}-\frac {2903 a_{1}}{2}-1100 a_{2} \\ P_{0} &= -6120+720 x^{9}+8 \left (423+124 a_{2}\right ) x^{8}+8 \left (-783+4 a_{1}-108 a_{2}\right ) x^{7}+\left (-24435-1080 a_{0}-3796 a_{1}-11314 a_{2}\right ) x^{6}+2 \left (2547-1176 a_{0}-2280 a_{1}-2062 a_{2}\right ) x^{5}+\left (54108+7338 a_{0}+15439 a_{1}+30999 a_{2}\right ) x^{4}+\frac {\left (75816 a_{0}+103136 a_{1}+119720 a_{2}+108891\right ) x^{3}}{4}+\frac {\left (11796 a_{0}-11280 a_{1}-51601 a_{2}-107748\right ) x^{2}}{4}+\frac {\left (-66408 a_{0}-85069 a_{1}-99768 a_{2}-107136\right ) x}{4}-\frac {36873 a_{0}}{4}-8823 a_{1}-7788 a_{2} \\ P_{-1} &= 3344 \left (x -1\right ) \left (x +1\right )^{4} \left (\left (a_{2}+\frac {9}{2}\right ) x^{4}+\left (\frac {82 a_{1}}{209}-\frac {807 a_{2}}{418}-\frac {9477}{836}\right ) x^{3}+\left (-\frac {27 a_{0}}{19}-\frac {2571 a_{1}}{418}-\frac {2172 a_{2}}{209}-\frac {8397}{836}\right ) x^{2}+\left (-\frac {39 a_{0}}{38}+\frac {3987 a_{1}}{836}+\frac {2678 a_{2}}{209}+\frac {4599}{209}\right ) x +\frac {6483 a_{0}}{418}+\frac {13567 a_{1}}{836}+\frac {6657 a_{2}}{418}+\frac {3024}{209}\right ) \\ \end{align*} Using method of undetermined coefficient to solve for \(a_i\) from the last equation above \(P_{-1} = 0 \) gives the following solution for the coefficients \(a_i\) \begin {align*} a_{0}&=-{\frac {27}{8}}\\ a_{1}&={\frac {27}{4}}\\ a_{2}&=-{\frac {9}{2}}\\ \end {align*}
Substituting these in the the polynomial \(p(x)\) results in \begin {align*} P(x) &= x^{3}-\frac {9}{2} x^{2}+\frac {27}{4} x -\frac {27}{8} \end {align*}
\(\omega \) 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}^{6} S^i \frac {P_i}{(6-i)!} \omega ^i &= 0 \end {align*}
Where the \(P_i\) are the polynomials found earlier. Computing the above sum gives \[ \tag{3A} \text {Expression too large to display}=0 \] The solution \(\omega \) of eq. 3A is found as \[ \omega =\frac {1}{4 x^{3}-6 x^{2}-4 x +6}\left (-2 \sqrt {5}\, \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}+2 x^{2}-2 x +1\right ) \] This \(\omega \) is used to find a solution to \(z''=r z\). \[ z_1(x) = e^{ \int \omega \,dx}\tag {5A} \] Doing the integration gives in eq. (4A) gives \begin{align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {-2 \sqrt {5}\, \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}+2 x^{2}-2 x +1}{4 x^{3}-6 x^{2}-4 x +6}d x} \\ &= \frac {\left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}\, \left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}\, \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}}{\sqrt {x^{2}-1}\, \left (2 x -2\right )}} {\mathrm e}^{\frac {\operatorname {arctanh}\left (\frac {\left (3 x -2\right ) \sqrt {5}}{5 \sqrt {x^{2}-1}}\right ) \sqrt {\left (x +1\right ) \left (x -1\right )^{3}}}{\sqrt {x^{2}-1}\, \left (2 x -2\right )}}}{\left (x -1\right )^{\frac {1}{4}}} \\ \end{align*} Which simplifies to \[ z_1(x) = \frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}}} \] The first solution to the original ode in \(y\) is found from \begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {1}{-x^{2}+1} \,dx} \\ &= z_1 e^{-\frac {\operatorname {arctanh}\left (x \right )}{2}} \\ &= z_1 \left (\frac {1}{\sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {1}{-x^{2}+1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\operatorname {arctanh}\left (x \right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {\left (x +\sqrt {x^{2}-1}\right )^{\sqrt {5}} \sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}\, \sqrt {x -1}}{\sqrt {x +1}\, \left (3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}\right ) \left (2 x -3\right )}d x\right ) \\ \end{align*} Therefore for kovacic case \(n=6\) the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\right ) + c_{2} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\left (\int \frac {\left (x +\sqrt {x^{2}-1}\right )^{\sqrt {5}} \sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}\, \sqrt {x -1}}{\sqrt {x +1}\, \left (3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}\right ) \left (2 x -3\right )}d x\right )\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\right ) + c_{2} \left (\frac {\left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}\left (\int \frac {\left (x +\sqrt {x^{2}-1}\right )^{\sqrt {5}} \sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}\, \sqrt {x -1}}{\sqrt {x +1}\, \left (3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}\right ) \left (2 x -3\right )}d x\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}+\frac {c_{2} \left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}\, \left (\int \frac {\left (x +\sqrt {x^{2}-1}\right )^{\sqrt {5}} \sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}\, \sqrt {x -1}}{\sqrt {x +1}\, \left (3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}\right ) \left (2 x -3\right )}d x \right )}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}}+\frac {c_{2} \left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{2}} \sqrt {\frac {3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}}}\, \left (x +1\right )^{\frac {1}{4}} \sqrt {2 x -3}\, \left (\int \frac {\left (x +\sqrt {x^{2}-1}\right )^{\sqrt {5}} \sqrt {x^{2}-1}\, \sqrt {-\frac {\left (2 x -3\right )^{2}}{x^{2}-1}}\, \sqrt {x -1}}{\sqrt {x +1}\, \left (3 x -2+\sqrt {5}\, \sqrt {x^{2}-1}\right ) \left (2 x -3\right )}d x \right )}{\left (x -1\right )^{\frac {1}{4}} \sqrt {\frac {x +1}{\sqrt {-x^{2}+1}}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right ) y^{\prime \prime }+y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {y^{\prime }}{x^{2}-1}+\frac {y}{x^{2}-1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-\frac {y^{\prime }}{x^{2}-1}-\frac {y}{x^{2}-1}=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 )=-\frac {1}{x^{2}-1}, P_{3}\left (x \right )=-\frac {1}{x^{2}-1}\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}}}=\frac {1}{2} \\ {} & \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}}}=0 \\ {} & \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} \\ {} & {} & \left (x^{2}-1\right ) y^{\prime \prime }-y^{\prime }-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 (u^{2}-2 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )-\frac {d}{d u}y \left (u \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} \frac {d}{d u}y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) 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 =1..2 \\ {} & {} & 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} \\ {} & {} & -a_{0} r \left (2 r -1\right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+a_{k} \left (k^{2}+2 k r +r^{2}-k -r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (2 r -1\right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +\frac {1}{2}+r \right ) \left (k +1+r \right ) a_{k +1}+a_{k} \left (k^{2}+\left (2 r -1\right ) k +r^{2}-r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}+2 k r +r^{2}-k -r -1\right )}{\left (2 k +1+2 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}-k -1\right )}{\left (2 k +1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {a_{k} \left (k^{2}-k -1\right )}{\left (2 k +1\right ) \left (k +1\right )}\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}, a_{k +1}=\frac {a_{k} \left (k^{2}-k -1\right )}{\left (2 k +1\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}-\frac {5}{4}\right )}{\left (2 k +2\right ) \left (k +\frac {3}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {1}{2}}, a_{k +1}=\frac {a_{k} \left (k^{2}-\frac {5}{4}\right )}{\left (2 k +2\right ) \left (k +\frac {3}{2}\right )}\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}{2}}, a_{k +1}=\frac {a_{k} \left (k^{2}-\frac {5}{4}\right )}{\left (2 k +2\right ) \left (k +\frac {3}{2}\right )}\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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k +\frac {1}{2}}\right ), a_{k +1}=\frac {a_{k} \left (k^{2}-k -1\right )}{\left (2 k +1\right ) \left (k +1\right )}, b_{k +1}=\frac {b_{k} \left (k^{2}-\frac {5}{4}\right )}{\left (2 k +2\right ) \left (k +\frac {3}{2}\right )}\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 Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 177
dsolve((1-x^2)*diff(y(x),x$2)+diff(y(x),x)+y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \sqrt {-3+2 x}\, {\left (\frac {3 \sqrt {5}\, x -2 \sqrt {5}-5 \sqrt {x^{2}-1}}{3 \sqrt {5}\, x -2 \sqrt {5}+5 \sqrt {x^{2}-1}}\right )}^{\frac {1}{4}} \left (x +\sqrt {x^{2}-1}\right )^{\frac {3 \sqrt {5}}{10}} \left (x +\sqrt {x^{2}-1}\right )^{\frac {\sqrt {5}}{5}}+c_{2} \sqrt {-3+2 x}\, {\left (\frac {3 \sqrt {5}\, x -2 \sqrt {5}+5 \sqrt {x^{2}-1}}{3 \sqrt {5}\, x -2 \sqrt {5}-5 \sqrt {x^{2}-1}}\right )}^{\frac {1}{4}} \left (x +\sqrt {x^{2}-1}\right )^{-\frac {3 \sqrt {5}}{10}} \left (x +\sqrt {x^{2}-1}\right )^{-\frac {\sqrt {5}}{5}} \]
✓ Solution by Mathematica
Time used: 36.335 (sec). Leaf size: 171
DSolve[(1-x^2)*y''[x]+y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {\sqrt [4]{1-x} \left (\sqrt {5} \sqrt {x-1}-\sqrt {x+1}\right ) e^{2 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {x+1}+\sqrt {2}}{\sqrt {x-1}}\right )} \left (c_2 \int _1^x\frac {2 e^{-4 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {K[1]+1}+\sqrt {2}}{\sqrt {K[1]-1}}\right )} \sqrt {\frac {K[1]-1}{K[1]+1}}}{\left (\sqrt {K[1]+1}-\sqrt {5} \sqrt {K[1]-1}\right )^2}dK[1]+c_1\right )}{\sqrt {2} \sqrt [4]{x-1}} \]