2.2.17 Problem 18

Internal problem ID [9140]
Book : Second order enumerated odes
Section : section 2
Problem number : 18
Date solved : Sunday, March 30, 2025 at 02:23:15 PM
CAS classification : [_Gegenbauer, [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

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

Time used: 0.373 (sec)

Solve

$$\begin{array}{rl}(-x^2+1)y^{\prime \prime }-x{y}^{\prime }-c^{2}y& =0\end{array}$$

In normal form the ode

$$\begin{array}{}\text{(1)}& (-x^2+1)y^{\prime \prime }-x{y}^{\prime }-c^{2}y& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

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

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

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $p_1=0$. Eq (4) simplifies to

$$\begin{array}{rl}{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)& =0\end{array}$$

This ode is solved resulting in

$$\begin{array}{rl}\tau & =\int {\mathrm{e}}^{-\int p\left(x\right)dx}dx\\ & =\int {\mathrm{e}}^{-\int \frac{x}{x^2-1}dx}dx\\ & =\int e^{-\frac{\ln (x-1)}{2}-\frac{\ln (x+1)}{2}}dx\\ & =\int \frac{1}{\sqrt{x-1}\sqrt{x+1}}dx\\ \text{(6)}& & =\frac{\sqrt{(x-1)(x+1)}\ln (x+\sqrt{x^2-1})}{\sqrt{x-1}\sqrt{x+1}}\end{array}$$

Using (6) to evaluate $q_1$ from (5) gives

$$\begin{array}{rl}{q}_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{\frac{c^2}{x^2-1}}{\frac{1}{(x+1)(x-1)}}\\ \text{(7)}& & =c^2\end{array}$$

Substituting the above in (3) and noting that now $p_1=0$ results in

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

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

$$A{y}^{″}(\tau )+B{y}^{\prime }(\tau )+Cy(\tau )=0$$

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

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{\tau \lambda }+c^2{\mathrm{e}}^{\tau \lambda }=0\end{array}$$

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

$$\begin{array}{}\text{(2)}& c^2+{\lambda }^2=0\end{array}$$

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

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=c^2$ into the above gives

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

Hence

$$\begin{array}{rl}{\lambda }_1& =+\sqrt{-c^2}\\ {\lambda }_2& =-\sqrt{-c^2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =ic\\ {\lambda }_2& =-ic\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =c$. Therefore the final solution, when using Euler relation, can be written as

$$y=e^{\alpha \tau }(c_1\cos (\beta \tau )+c_2\sin (\beta \tau ))$$

Which becomes

$$y=e^0(c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right))$$

Or

$$y=c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right)$$

Will add steps showing solving for IC soon.

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

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

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

Time used: 0.173 (sec)

Solve

$$\begin{array}{rl}(-x^2+1)y^{\prime \prime }-x{y}^{\prime }-c^{2}y& =0\end{array}$$

In normal form the ode

$$\begin{array}{}\text{(1)}& (-x^2+1)y^{\prime \prime }-x{y}^{\prime }-c^{2}y& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

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

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

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

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

$$\begin{array}{rl}{\tau }^{\prime }& =\frac{1}{c}\sqrt{q}\\ \text{(6)}& & =\frac{\sqrt{\frac{c^2}{x^2-1}}}{c}{\tau }^{″}& =-\frac{c^{2}x}{c\sqrt{\frac{c^2}{x^2-1}}{(x^2-1)}^2}\end{array}$$

Substituting the above into (4) results in

$$\begin{array}{rl}{p}_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{-\frac{c^{2}x}{c\sqrt{\frac{c^2}{x^2-1}}{(x^2-1)}^2}+\frac{x}{x^2-1}\frac{\sqrt{\frac{c^2}{x^2-1}}}{c}}{{\left(\frac{\sqrt{\frac{c^2}{x^2-1}}}{c}\right)}^2}\\ & =0\end{array}$$

Therefore ode (3) now becomes

$$\begin{array}{rl}y{\left(\tau \right)}^{″}+p_{1}y{\left(\tau \right)}^{\prime }+q_{1}y\left(\tau \right)& =0\\ \text{(7)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+c^{2}y\left(\tau \right)& =0\end{array}$$

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

$$\begin{array}{rl}y\left(\tau \right)& =c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right)\end{array}$$

Now from (6)

$$\begin{array}{rl}\tau & =\int \frac{1}{c}\sqrt{q}dx\\ & =\frac{\int \sqrt{\frac{c^2}{x^2-1}}dx}{c}\\ & =\frac{\sqrt{\frac{c^2}{x^2-1}}\sqrt{x^2-1}\ln (x+\sqrt{x^2-1})}{c}\end{array}$$

Substituting the above into the solution obtained gives

$$y=c_1\cos (\sqrt{\frac{c^2}{x^2-1}}\sqrt{x^2-1}\ln (x+\sqrt{x^2-1}))+c_2\sin (\sqrt{\frac{c^2}{x^2-1}}\sqrt{x^2-1}\ln (x+\sqrt{x^2-1}))$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos (\sqrt{\frac{c^2}{x^2-1}}\sqrt{x^2-1}\ln (x+\sqrt{x^2-1}))+c_2\sin (\sqrt{\frac{c^2}{x^2-1}}\sqrt{x^2-1}\ln (x+\sqrt{x^2-1}))\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.526 (sec)

Solve

$$\begin{array}{rl}(-x^2+1)y^{\prime \prime }-x{y}^{\prime }-c^{2}y& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& (-x^2+1)y^{\prime \prime }-x{y}^{\prime }-c^{2}y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =-x^2+1\\ \text{(3)}& B& =-xC& =-c^2\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =y{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{-4c^2{x}^2+4c^2-x^2-2}{4{(x^2-1)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-4c^2{x}^2+4c^2-x^2-2\\ t& =4{(x^2-1)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-4c^2{x}^2+4c^2-x^2-2}{4{(x^2-1)}^2}\right)z(x)\end{array}$$

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

$$\begin{array}{rl}y& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

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 $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
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	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.34: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-2\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4{(x^2-1)}^2$. 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 $\mathrm{\infty }$ 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 $\mathrm{\infty }$ is $2$ then the necessary conditions for case three are met. Therefore

$$\begin{array}{rl}L& =[1,2,4,6,12]\end{array}$$

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

$$r=-\frac{3}{16{(x+1)}^2}+\frac{-\frac{1}{16}+\frac{c^2}{2}}{x+1}-\frac{3}{16{(x-1)}^2}+\frac{\frac{1}{16}-\frac{c^2}{2}}{x-1}$$

For the pole at $x=1$ let $b$ be the coefficient of $\frac{1}{{(x-1)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{3}{16}$. Hence

$$\begin{array}{rl}{E}_c& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{1,2,3\}\end{array}$$

For the pole at $x=-1$ let $b$ be the coefficient of $\frac{1}{{(x+1)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{3}{16}$. Hence

$$\begin{array}{rl}{E}_c& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{1,2,3\}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is 2 then let $b$ be the coefficient of $\frac{1}{x^2}$ in the Laurent series expansion of $r$ at $\mathrm{\infty }$. which can be found by dividing the leading coefficient of $s$ by the leading coefficient of $t$ from

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-4c^2{x}^2+4c^2-x^2-2}{4{(x^2-1)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-1$. Hence

$$\begin{array}{rl}{E}_{\mathrm{\infty }}& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{2\}\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ 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 $\mathrm{\infty }$	$E_{\mathrm{\infty }}$
$2$	$\{2\}$

Using the family $\{e_1,e_2,\dots ,e_{\mathrm{\infty }}\}$ given by

$$e_1=1,e_2=1,e_{\mathrm{\infty }}=2$$

Gives a non negative integer $d$ (the degree of the polynomial $p(x)$), which is generated using

$$\begin{array}{rl}d& =\frac{1}{2}(e_{\mathrm{\infty }}-\sum _{c\in \mathrm{\Gamma }}{e}_c)\\ & =\frac{1}{2}(2-(1+\left(1\right)))\\ & =0\end{array}$$

We now form the following rational function

$$\begin{array}{rl}\theta & =\frac{1}{2}\sum _{c\in \mathrm{\Gamma }}\frac{e_c}{x-c}\\ & =\frac{1}{2}(\frac{1}{(x-\left(1\right))}+\frac{1}{(x-(-1))})\\ & =\frac{1}{2x-2}+\frac{1}{2x+2}\end{array}$$

Now we search for a monic polynomial $p(x)$ of degree $d=0$ such that

$$\begin{array}{}\text{(1A)}& p^{‴}+3\theta p^{″}+(3{\theta }^2+3{\theta }^{\prime }-4r)p^{\prime }+({\theta }^{″}+3\theta {\theta }^{\prime }+{\theta }^3-4r\theta -2r^{\prime })p=0\end{array}$$

Since $d=0$, then letting

$$\begin{array}{}\text{(2A)}& p=1\end{array}$$

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

$$0=0$$

And solving for $p$ gives

$$p=1$$

Now that $p(x)$ is found let

$$\begin{array}{rl}\varphi & =\theta +\frac{p^{\prime }}{p}\\ & =\frac{1}{2x-2}+\frac{1}{2x+2}\end{array}$$

Let $\omega$ be the solution of

$$\begin{array}{rl}{\omega }^2-\varphi \omega +(\frac{1}{2}{\varphi }^{\prime }+\frac{1}{2}{\varphi }^2-r)& =0\end{array}$$

Substituting the values for $\varphi$ and $r$ into the above equation gives

$$w^2-(\frac{1}{2x-2}+\frac{1}{2x+2})w+\frac{4c^2{x}^2-4c^2+x^2}{4{(x^2-1)}^2}=0$$

Solving for $\omega$ gives

$$\begin{array}{rl}\omega & =\frac{x+2c\sqrt{-x^2+1}}{2(x-1)(x+1)}\end{array}$$

Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =e^{\int \omega dx}\\ & ={\mathrm{e}}^{\int \frac{x+2c\sqrt{-x^2+1}}{2(x-1)(x+1)}dx}\\ & ={(x^2-1)}^{1/4}{\mathrm{e}}^{-c\arcsin \left(x\right)}\end{array}$$

The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-x}{-x^2+1}dx}\\ & =z_1{e}^{-\frac{\ln (x-1)}{4}-\frac{\ln (x+1)}{4}}\\ & =z_1\left(\frac{1}{{(x-1)}^{1/4}{(x+1)}^{1/4}}\right)\end{array}$$

Which simplifies to

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

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-x}{-x^2+1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{\ln (x-1)}{2}-\frac{\ln (x+1)}{2}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{\ln (x-1)}{2}-\frac{\ln (x+1)}{2}}\sqrt{x-1}\sqrt{x+1}{\mathrm{e}}^{2c\arcsin \left(x\right)}}{\sqrt{x^2-1}}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{{(x^2-1)}^{1/4}{\mathrm{e}}^{-c\arcsin \left(x\right)}}{{(x-1)}^{1/4}{(x+1)}^{1/4}}\right)+c_2\left(\frac{{(x^2-1)}^{1/4}{\mathrm{e}}^{-c\arcsin \left(x\right)}}{{(x-1)}^{1/4}{(x+1)}^{1/4}}(\int \frac{{\mathrm{e}}^{-\frac{\ln (x-1)}{2}-\frac{\ln (x+1)}{2}}\sqrt{x-1}\sqrt{x+1}{\mathrm{e}}^{2c\arcsin \left(x\right)}}{\sqrt{x^2-1}}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

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

✓ Maple. Time used: 0.002 (sec). Leaf size: 35

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

$$y=c_1{(x+\sqrt{x^2-1})}^{ic}+c_2{(x+\sqrt{x^2-1})}^{-ic}$$

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 step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (-x^2+1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-x\left(\frac{d}{dx}y\left(x\right)\right)-c^{2}y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=-\frac{c^{2}y\left(x\right)}{x^2-1}-\frac{x\left(\frac{d}{dx}y\left(x\right)\right)}{x^2-1}\\ \bullet & & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{x\left(\frac{d}{dx}y\left(x\right)\right)}{x^2-1}+\frac{c^{2}y\left(x\right)}{x^2-1}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{x}{x^2-1},P_3\left(x\right)=\frac{c^2}{x^2-1}]\\ & \circ & (x+1)\cdot P_2\left(x\right)\text{is analytic at}x=-1\\ & & ((x+1)\cdot P_2\left(x\right))|{}_{x=-1}=\frac{1}{2}\\ & \circ & {(x+1)}^2\cdot P_3\left(x\right)\text{is analytic at}x=-1\\ & & ({(x+1)}^2\cdot P_3\left(x\right))|{}_{x=-1}=0\\ & \circ & x=-1\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=-1\\ \bullet & & \text{Multiply by denominators}\\ & & (x^2-1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x\left(\frac{d}{dx}y\left(x\right)\right)+c^{2}y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-1\text{so that the regular singular point is at}u=0\\ & & (u^2-2u)\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(u-1)\left(\frac{d}{du}y\left(u\right)\right)+c^{2}y\left(u\right)=0\\ \bullet & & \text{Assume series solution for}y\left(u\right)\\ & & y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=0..1\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)u^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=1..2\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)u^{k+r-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & -a_{0}r(-1+2r)u^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(-a_{k+1}(k+1+r)(2k+1+2r)+a_k(c^2+k^2+2kr+r^2))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -r(-1+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{1}{2}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & -2(k+1+r)(k+\frac{1}{2}+r)a_{k+1}+a_k(c^2+k^2+2kr+r^2)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k(c^2+k^2+2kr+r^2)}{(k+1+r)(2k+1+2r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=\frac{a_k(c^2+k^2)}{(k+1)(2k+1)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^k,a_{k+1}=\frac{a_k(c^2+k^2)}{(k+1)(2k+1)}]\\ \bullet & & \text{Revert the change of variables}u=x+1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1)}^k,a_{k+1}=\frac{a_k(c^2+k^2)}{(k+1)(2k+1)}]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{2}\\ & & a_{k+1}=\frac{a_k(c^2+k^2+k+\frac{1}{4})}{(k+\frac{3}{2})(2k+2)}\\ \bullet & & \text{Solution for}r=\frac{1}{2}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+\frac{1}{2}},a_{k+1}=\frac{a_k(c^2+k^2+k+\frac{1}{4})}{(k+\frac{3}{2})(2k+2)}]\\ \bullet & & \text{Revert the change of variables}u=x+1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1)}^{k+\frac{1}{2}},a_{k+1}=\frac{a_k(c^2+k^2+k+\frac{1}{4})}{(k+\frac{3}{2})(2k+2)}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1)}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(x+1)}^{k+\frac{1}{2}}),a_{k+1}=\frac{a_k(c^2+k^2)}{(k+1)(2k+1)},b_{k+1}=\frac{b_k(c^2+k^2+k+\frac{1}{4})}{(k+\frac{3}{2})(2k+2)}]\end{array}$$

✓ Mathematica. Time used: 0.045 (sec). Leaf size: 42

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

$$y(x)\to c_1\cos (c\log (\sqrt{x^2-1}+x))+c_2\sin (c\log (\sqrt{x^2-1}+x))$$

✗ Sympy

from sympy import * 
x = symbols("x") 
c = symbols("c") 
y = Function("y") 
ode = Eq(-c**2*y(x) - x*Derivative(y(x), x) + (1 - x**2)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False