2.2.17 problem 18

Internal problem ID [8816]
Book : Second order enumerated odes
Section : section 2
Problem number : 18
Date solved : Thursday, December 12, 2024 at 09:51:56 AM
CAS classification : [_Gegenbauer, [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

Solve

\begin{align*} \left (-x^{2}+1\right ) y^{\prime \prime }-x y^{\prime }-c^{2} y&=0 \end{align*}

Solved as second order ode using change of variable on x method 2

Time used: 0.484 (sec)

In normal form the ode

\begin{align*} \left (-x^{2}+1\right ) y^{\prime \prime }-x y^{\prime }-c^{2} y&=0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {x}{x^{2}-1}\\ q \left (x \right )&=\frac {c^{2}}{x^{2}-1} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \frac {x}{x^{2}-1}d x}d x\\ &= \int e^{-\frac {\ln \left (x -1\right )}{2}-\frac {\ln \left (x +1\right )}{2}} \,dx\\ &= \int \frac {1}{\sqrt {x -1}\, \sqrt {x +1}}d x\\ &= \frac {\sqrt {\left (x -1\right ) \left (x +1\right )}\, \ln \left (x +\sqrt {x^{2}-1}\right )}{\sqrt {x -1}\, \sqrt {x +1}}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {c^{2}}{x^{2}-1}}{\frac {1}{\left (x -1\right ) \left (x +1\right )}}\\ &= c^{2}\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right )&=0 \end{align*}

The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \]

Where in the above \(A=1, B=0, C=c^{2}\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }+c^{2} {\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ c^{2}+\lambda ^{2} = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=c^{2}\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (c^{2}\right )}\\ &= \pm \sqrt {-c^{2}} \end{align*}

Hence

\begin{align*} \lambda _1 &= + \sqrt {-c^{2}} \\ \lambda _2 &= - \sqrt {-c^{2}} \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= \sqrt {-c^{2}} \\ \lambda _2 &= -\sqrt {-c^{2}} \\ \end{align*}

Since roots are real and distinct, then the solution is

\begin{align*} y \left (\tau \right ) &= c_1 e^{\lambda _1 \tau } + c_2 e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_1 e^{\left (\sqrt {-c^{2}}\right )\tau } +c_2 e^{\left (-\sqrt {-c^{2}}\right )\tau } \\ \end{align*}

Or

\[ y \left (\tau \right ) =c_1 \,{\mathrm e}^{\tau \sqrt {-c^{2}}}+c_2 \,{\mathrm e}^{-\tau \sqrt {-c^{2}}} \]

Will add steps showing solving for IC soon.

The above solution is now transformed back to \(y\) using (6) which results in

\[ y = c_1 \,{\mathrm e}^{\frac {\sqrt {\left (x -1\right ) \left (x +1\right )}\, \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {-c^{2}}}{\sqrt {x -1}\, \sqrt {x +1}}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {\left (x -1\right ) \left (x +1\right )}\, \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {-c^{2}}}{\sqrt {x -1}\, \sqrt {x +1}}} \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \,{\mathrm e}^{\frac {\sqrt {\left (x -1\right ) \left (x +1\right )}\, \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {-c^{2}}}{\sqrt {x -1}\, \sqrt {x +1}}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {\left (x -1\right ) \left (x +1\right )}\, \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {-c^{2}}}{\sqrt {x -1}\, \sqrt {x +1}}} \\ \end{align*}

Solved as second order ode using change of variable on x method 1

Time used: 0.161 (sec)

In normal form the ode

\begin{align*} \left (-x^{2}+1\right ) y^{\prime \prime }-x y^{\prime }-c^{2} y&=0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {x}{x^{2}-1}\\ q \left (x \right )&=\frac {c^{2}}{x^{2}-1} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) results

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5)

\begin{align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {\frac {c^{2}}{x^{2}-1}}}{c}\tag {6} \\ \tau '' &= -\frac {c^{2} x}{c \sqrt {\frac {c^{2}}{x^{2}-1}}\, \left (x^{2}-1\right )^{2}} \end{align*}

Substituting the above into (4) results in

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {-\frac {c^{2} x}{c \sqrt {\frac {c^{2}}{x^{2}-1}}\, \left (x^{2}-1\right )^{2}}+\frac {x}{x^{2}-1}\frac {\sqrt {\frac {c^{2}}{x^{2}-1}}}{c}}{\left (\frac {\sqrt {\frac {c^{2}}{x^{2}-1}}}{c}\right )^2} \\ &=0 \end{align*}

Therefore ode (3) now becomes

\begin{align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end{align*}

The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give

\begin{align*} y \left (\tau \right ) &= c_1 \cos \left (c \tau \right )+c_2 \sin \left (c \tau \right ) \end{align*}

Now from (6)

\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {\frac {c^{2}}{x^{2}-1}}d x}{c}\\ &= \frac {\sqrt {\frac {c^{2}}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right )}{c} \end{align*}

Substituting the above into the solution obtained gives

\[ y = c_1 \cos \left (\sqrt {\frac {c^{2}}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right )\right )+c_2 \sin \left (\sqrt {\frac {c^{2}}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right )\right ) \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \cos \left (\sqrt {\frac {c^{2}}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right )\right )+c_2 \sin \left (\sqrt {\frac {c^{2}}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right )\right ) \\ \end{align*}

Solved as second order ode using Kovacic algorithm

Time used: 0.320 (sec)

Writing the ode as

\begin{align*} \left (-x^{2}+1\right ) y^{\prime \prime }-x y^{\prime }-c^{2} 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 &= -x\tag {3} \\ C &= -c^{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 {-4 c^{2} x^{2}+4 c^{2}-x^{2}-2}{4 \left (x^{2}-1\right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= -4 c^{2} x^{2}+4 c^{2}-x^{2}-2\\ t &= 4 \left (x^{2}-1\right )^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {-4 c^{2} x^{2}+4 c^{2}-x^{2}-2}{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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.35: 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 \(t=4 \left (x^{2}-1\right )^{2}\). 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

\[ r = -\frac {3}{16 \left (x -1\right )^{2}}+\frac {\frac {1}{16}-\frac {c^{2}}{2}}{x -1}-\frac {3}{16 \left (x +1\right )^{2}}+\frac {\frac {c^{2}}{2}-\frac {1}{16}}{x +1} \]

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*}

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 c^{2} x^{2}+4 c^{2}-x^{2}-2}{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	\(E_c\)
\(1\)	\(2\)	\(\{1, 2, 3\}\)
\(-1\)	\(2\)	\(\{1, 2, 3\}\)

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 )\\ &= 0 \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 x -2}+\frac {1}{2 x +2} \end{align*}

Now we search for a monic polynomial \(p(x)\) of degree \(d=0\) 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 \(d=0\), then letting

\[ p = 1\tag {2A} \]

Substituting \(p\) and \(\theta \) into Eq. (1A) gives

\[ 0 = 0 \]

And solving for \(p\) gives

\[ p = 1 \]

Now that \(p(x)\) is found let

\begin{align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {1}{2 x -2}+\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

\[ w^{2}-\left (\frac {1}{2 x -2}+\frac {1}{2 x +2}\right ) w +\frac {4 c^{2} x^{2}-4 c^{2}+x^{2}}{4 \left (x^{2}-1\right )^{2}} = 0 \]

Solving for \(\omega \) gives

\begin{align*} \omega &= \frac {x +2 c \sqrt {-x^{2}+1}}{2 \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 {x +2 c \sqrt {-x^{2}+1}}{2 \left (x -1\right ) \left (x +1\right )}d x}\\ &= \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \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 {-x}{-x^{2}+1} \,dx} \\ &= z_1 e^{-\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}} \\ &= z_1 \left (\frac {1}{\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = \frac {\left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}}{\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}}} \]

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 {-x}{-x^{2}+1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\frac {\ln \left (x -1\right )}{2}-\frac {\ln \left (x +1\right )}{2}}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {{\mathrm e}^{-\frac {\ln \left (x -1\right )}{2}-\frac {\ln \left (x +1\right )}{2}} \sqrt {x -1}\, \sqrt {x +1}\, {\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (\frac {\left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}}{\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}}}\right ) + c_2 \left (\frac {\left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}}{\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}}}\left (\int \frac {{\mathrm e}^{-\frac {\ln \left (x -1\right )}{2}-\frac {\ln \left (x +1\right )}{2}} \sqrt {x -1}\, \sqrt {x +1}\, {\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {c_1 \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}}{\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}}}+\frac {c_2 \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x}{\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}}} \\ \end{align*}

Solved as second order ode adjoint method

Time used: 0.891 (sec)

In normal form the ode

\begin{align*} \left (-x^{2}+1\right ) y^{\prime \prime }-x y^{\prime }-c^{2} y = 0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {x}{x^{2}-1}\\ q \left (x \right )&=\frac {c^{2}}{x^{2}-1}\\ r \left (x \right )&=0 \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {x \xi \left (x \right )}{x^{2}-1}\right )' + \left (\frac {c^{2} \xi \left (x \right )}{x^{2}-1}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\frac {x \xi ^{\prime }\left (x \right )}{x^{2}-1}+\frac {\left (x^{2}+1+c^{2} \left (x^{2}-1\right )\right ) \xi \left (x \right )}{\left (x^{2}-1\right )^{2}}&= 0 \end{align*}

Which is solved for \(\xi (x)\). Writing the ode as

\begin{align*} \xi ^{\prime \prime }-\frac {x \xi ^{\prime }}{x^{2}-1}+\frac {\left (x^{2}+1+c^{2} \left (x^{2}-1\right )\right ) \xi }{\left (x^{2}-1\right )^{2}} &= 0 \tag {1} \\ A \xi ^{\prime \prime } + B \xi ^{\prime } + C \xi &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= 1 \\ B &= -\frac {x}{x^{2}-1}\tag {3} \\ C &= \frac {x^{2}+1+c^{2} \left (x^{2}-1\right )}{\left (x^{2}-1\right )^{2}} \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= \xi 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 c^{2} x^{2}+4 c^{2}-x^{2}-2}{4 \left (x^{2}-1\right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= -4 c^{2} x^{2}+4 c^{2}-x^{2}-2\\ t &= 4 \left (x^{2}-1\right )^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {-4 c^{2} x^{2}+4 c^{2}-x^{2}-2}{4 \left (x^{2}-1\right )^{2}}\right ) z(x)\tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(x)\) then \(\xi \) is found using the inverse transformation

\begin{align*} \xi &= 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 2.36: 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 \(t=4 \left (x^{2}-1\right )^{2}\). 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

\[ r = -\frac {3}{16 \left (x -1\right )^{2}}+\frac {\frac {1}{16}-\frac {c^{2}}{2}}{x -1}-\frac {3}{16 \left (x +1\right )^{2}}+\frac {\frac {c^{2}}{2}-\frac {1}{16}}{x +1} \]

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*}

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 c^{2} x^{2}+4 c^{2}-x^{2}-2}{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	\(E_c\)
\(1\)	\(2\)	\(\{1, 2, 3\}\)
\(-1\)	\(2\)	\(\{1, 2, 3\}\)

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 )\\ &= 0 \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 x -2}+\frac {1}{2 x +2} \end{align*}

Now we search for a monic polynomial \(p(x)\) of degree \(d=0\) 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 \(d=0\), then letting

\[ p = 1\tag {2A} \]

Substituting \(p\) and \(\theta \) into Eq. (1A) gives

\[ 0 = 0 \]

And solving for \(p\) gives

\[ p = 1 \]

Now that \(p(x)\) is found let

\begin{align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {1}{2 x -2}+\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

\[ w^{2}-\left (\frac {1}{2 x -2}+\frac {1}{2 x +2}\right ) w +\frac {4 c^{2} x^{2}-4 c^{2}+x^{2}}{4 \left (x^{2}-1\right )^{2}} = 0 \]

Solving for \(\omega \) gives

\begin{align*} \omega &= \frac {x +2 c \sqrt {-x^{2}+1}}{2 \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 {x +2 c \sqrt {-x^{2}+1}}{2 \left (x -1\right ) \left (x +1\right )}d x}\\ &= \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \end{align*}

The first solution to the original ode in \(\xi \) is found from

\begin{align*} \xi _1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-\frac {x}{x^{2}-1}}{1} \,dx} \\ &= z_1 e^{\frac {\ln \left (x -1\right )}{4}+\frac {\ln \left (x +1\right )}{4}} \\ &= z_1 \left (\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}}\right ) \\ \end{align*}

Which simplifies to

\[ \xi _1 = \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \]

The second solution \(\xi _2\) to the original ode is found using reduction of order

\[ \xi _2 = \xi _1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{\xi _1^2} \,dx \]

Substituting gives

\begin{align*} \xi _2 &= \xi _1 \int \frac { e^{\int -\frac {-\frac {x}{x^{2}-1}}{1} \,dx}}{\left (\xi _1\right )^2} \,dx \\ &= \xi _1 \int \frac { e^{\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}}}{\left (\xi _1\right )^2} \,dx \\ &= \xi _1 \left (\int \frac {{\mathrm e}^{\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}} {\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x -1}\, \sqrt {x +1}\, \sqrt {x^{2}-1}}d x\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} \xi &= c_1 \xi _1 + c_2 \xi _2 \\ &= c_1 \left (\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}\right ) + c_2 \left (\left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}\left (\int \frac {{\mathrm e}^{\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}} {\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x -1}\, \sqrt {x +1}\, \sqrt {x^{2}-1}}d x\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}

Or

\begin{align*} y^{\prime }+y \left (\frac {x}{x^{2}-1}-\frac {\frac {c_1 \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}}{4 \left (x -1\right )^{{3}/{4}}}+\frac {c_1 \left (x -1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}}{4 \left (x +1\right )^{{3}/{4}}}+\frac {c_1 \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} x}{2 \left (x^{2}-1\right )^{{3}/{4}}}-\frac {c_1 \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} c \,{\mathrm e}^{-c \arcsin \left (x \right )}}{\sqrt {-x^{2}+1}}+\frac {c_2 \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x}{4 \left (x -1\right )^{{3}/{4}}}+\frac {c_2 \left (x -1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x}{4 \left (x +1\right )^{{3}/{4}}}+\frac {c_2 \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x x}{2 \left (x^{2}-1\right )^{{3}/{4}}}-\frac {c_2 \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} c \,{\mathrm e}^{-c \arcsin \left (x \right )} \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x}{\sqrt {-x^{2}+1}}+\frac {c_2 \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} {\mathrm e}^{2 c \arcsin \left (x \right )}}{\left (x^{2}-1\right )^{{1}/{4}}}}{c_1 \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )}+c_2 \left (x -1\right )^{{1}/{4}} \left (x +1\right )^{{1}/{4}} \left (x^{2}-1\right )^{{1}/{4}} {\mathrm e}^{-c \arcsin \left (x \right )} \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x}\right )&=0 \end{align*}

Which is now a first order ode. This is now solved for \(y\). The ode \(y^{\prime } = -\frac {\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c \right ) y}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}\) is separable as it can be written as

\begin{align*} y^{\prime }&= -\frac {\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c \right ) y}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= -\frac {\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}\\ g(y) &= y \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{y}\,dy} &= \int { -\frac {\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )} \,dx}\\ \ln \left (y\right )&=\int -\frac {\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}d x +2 c_3 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is zero, since we had to divide by this above. Solving \(g(y)=0\) or \(y=0\) for \(y\) gives

\begin{align*} y&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \ln \left (y\right ) = \int -\frac {\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}d x +2 c_3\\ y = 0 \end{align*}

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} \ln \left (y\right ) &= \int -\frac {\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}d x +2 c_3 \\ y &= 0 \\ \end{align*}

The constants can be merged to give

\[ \ln \left (y\right ) = \int -\frac {\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}d x +2 \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} \ln \left (y\right ) &= \int -\frac {\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 c \,x^{2}-c_2 \,{\mathrm e}^{2 c \arcsin \left (x \right )} \sqrt {x^{2}-1}\, \sqrt {-x^{2}+1}+c_1 c \,x^{2}-c_2 c \int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x -c_1 c}{\left (\int \frac {{\mathrm e}^{2 c \arcsin \left (x \right )}}{\sqrt {x^{2}-1}}d x c_2 +c_1 \right ) \sqrt {-x^{2}+1}\, \left (x -1\right ) \left (x +1\right )}d x +2 \\ y &= 0 \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-x \left (\frac {d}{d x}y \left (x \right )\right )-c^{2} y \left (x \right )=0 \\ \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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {c^{2} y \left (x \right )}{x^{2}-1}-\frac {\left (\frac {d}{d x}y \left (x \right )\right ) x}{x^{2}-1} \\ \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 (\frac {d}{d x}y \left (x \right )\right ) x}{x^{2}-1}+\frac {c^{2} y \left (x \right )}{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 {x}{x^{2}-1}, P_{3}\left (x \right )=\frac {c^{2}}{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 ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+c^{2} 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 (u^{2}-2 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (u -1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+c^{2} 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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +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 (-1+2 r \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 (c^{2}+k^{2}+2 k r +r^{2}\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+2 r \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 +1+r \right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}+a_{k} \left (c^{2}+k^{2}+2 k r +r^{2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (c^{2}+k^{2}+2 k r +r^{2}\right )}{\left (k +1+r \right ) \left (2 k +1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (c^{2}+k^{2}\right )}{\left (k +1\right ) \left (2 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 (c^{2}+k^{2}\right )}{\left (k +1\right ) \left (2 k +1\right )}\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}, a_{k +1}=\frac {a_{k} \left (c^{2}+k^{2}\right )}{\left (k +1\right ) \left (2 k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (c^{2}+k^{2}+k +\frac {1}{4}\right )}{\left (k +\frac {3}{2}\right ) \left (2 k +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 (c^{2}+k^{2}+k +\frac {1}{4}\right )}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )}\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}{2}}, a_{k +1}=\frac {a_{k} \left (c^{2}+k^{2}+k +\frac {1}{4}\right )}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )}\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}\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 (c^{2}+k^{2}\right )}{\left (k +1\right ) \left (2 k +1\right )}, b_{k +1}=\frac {b_{k} \left (c^{2}+k^{2}+k +\frac {1}{4}\right )}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )}\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)] 
<- linear_1 successful`
 

Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 35

dsolve((-x^2+1)*diff(diff(y(x),x),x)-diff(y(x),x)*x-c^2*y(x) = 0, 
       y(x),singsol=all)
 

\[ y = c_{1} \left (x +\sqrt {x^{2}-1}\right )^{i c}+c_{2} \left (x +\sqrt {x^{2}-1}\right )^{-i c} \]

Mathematica DSolve solution

Solving time : 0.086 (sec)
Leaf size : 42

DSolve[{(1-x^2)*D[y[x],{x,2}]-x*D[y[x],x]-c^2*y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_1 \cos \left (c \text {arctanh}\left (\frac {x}{\sqrt {x^2-1}}\right )\right )+c_2 \sin \left (c \text {arctanh}\left (\frac {x}{\sqrt {x^2-1}}\right )\right ) \]