2.1.294 Problem 297

Internal problem ID [9464]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 297
Date solved : Friday, April 25, 2025 at 06:08:58 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 1.053 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{4x^2-4x-3}{4{(x^2-1)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =4x^2-4x-3\\ t& =4{(x^2-1)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{4x^2-4x-3}{4{(x^2-1)}^2}\right)z(x)\end{array}$$

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 2.294: 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

$$r=-\frac{3}{16{(x-1)}^2}+\frac{5}{16{(x+1)}^2}+\frac{7}{16(x-1)}-\frac{7}{16(x+1)}$$

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 $b=\frac{5}{16}$. Hence

$$\begin{array}{rl}{E}_c& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{-1,2,5\}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{4x^2-4x-3}{4{(x^2-1)}^2}\end{array}$$

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	$E_c$
$1$	$2$	$\{1,2,3\}$
$-1$	$2$	$\{-1,2,5\}$

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

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

We now form the following rational function

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

Now we search for a monic polynomial $p(x)$ of degree $d=1$ 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 $d=1$, then letting

$$\begin{array}{}\text{(2A)}& p=x+a_0\end{array}$$

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

$$\frac{4a_0-6}{{(x+1)}^2(x-1)}=0$$

And solving for $p$ gives

$$p=x+\frac{3}{2}$$

Now that $p(x)$ is found let

$$\begin{array}{rl}\varphi & =\theta +\frac{p^{\prime }}{p}\\ & =\frac{1}{x+\frac{3}{2}}+\frac{1}{2x-2}-\frac{1}{2(x+1)}\end{array}$$

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

$$w^2-(\frac{1}{x+\frac{3}{2}}+\frac{1}{2x-2}-\frac{1}{2(x+1)})w+\frac{-8x^3-4x^2+10x+7}{4{(x^2-1)}^2(2x+3)}=0$$

Solving for $\omega$ gives

$$\begin{array}{rl}\omega & =\frac{2\sqrt{5}\sqrt{(x-1)(x+1)}x+2\sqrt{5}\sqrt{(x-1)(x+1)}+2x^2+2x+1}{2(2x+3)(x-1)(x+1)}\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =e^{\int \omega dx}\\ & ={\mathrm{e}}^{\int \frac{2\sqrt{5}\sqrt{(x-1)(x+1)}x+2\sqrt{5}\sqrt{(x-1)(x+1)}+2x^2+2x+1}{2(2x+3)(x-1)(x+1)}dx}\\ & =\frac{\sqrt{2x+3}{(x-1)}^{1/4}{(x+\sqrt{x^2-1})}^{\frac{\sqrt{5}}{2}}{5}^{1/4}}{{(x+1)}^{1/4}\sqrt{\frac{5\sqrt{x^2-1}+(2+3x)\sqrt{5}}{\sqrt{x^2-1}\sqrt{-\frac{{(2x+3)}^2}{x^2-1}}}}}\end{array}$$

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

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

Which simplifies to

$$y_1=\frac{\sqrt{\frac{x+1}{\sqrt{-x^2+1}}}{(x+\sqrt{x^2-1})}^{\frac{\sqrt{5}}{2}}\sqrt{2x+3}{(5x-5)}^{1/4}}{\sqrt{\frac{i(3\sqrt{5}x+5\sqrt{x^2-1}+2\sqrt{5})}{2x+3}}{(x+1)}^{1/4}}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-1}{-x^2+1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\mathrm{arctanh}\left(x\right)}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{i\sqrt{x+1}{(x+\sqrt{x^2-1})}^{-\sqrt{5}}(3\sqrt{5}x+5\sqrt{x^2-1}+2\sqrt{5})}{{(2x+3)}^2\sqrt{5x-5}}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{\sqrt{\frac{x+1}{\sqrt{-x^2+1}}}{(x+\sqrt{x^2-1})}^{\frac{\sqrt{5}}{2}}\sqrt{2x+3}{(5x-5)}^{1/4}}{\sqrt{\frac{i(3\sqrt{5}x+5\sqrt{x^2-1}+2\sqrt{5})}{2x+3}}{(x+1)}^{1/4}}\right)+c_2\left(\frac{\sqrt{\frac{x+1}{\sqrt{-x^2+1}}}{(x+\sqrt{x^2-1})}^{\frac{\sqrt{5}}{2}}\sqrt{2x+3}{(5x-5)}^{1/4}}{\sqrt{\frac{i(3\sqrt{5}x+5\sqrt{x^2-1}+2\sqrt{5})}{2x+3}}{(x+1)}^{1/4}}(\int \frac{i\sqrt{x+1}{(x+\sqrt{x^2-1})}^{-\sqrt{5}}(3\sqrt{5}x+5\sqrt{x^2-1}+2\sqrt{5})}{{(2x+3)}^2\sqrt{5x-5}}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.024 (sec). Leaf size: 66

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

$$y=c_1\mathrm{hypergeom}([-\frac{1}{2}-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}-\frac{1}{2}],[-\frac{1}{2}],\frac{1}{2}+\frac{x}{2})+2c_2\sqrt{2+2x}\mathrm{hypergeom}([1-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}+1],\left[\frac{5}{2}\right],\frac{1}{2}+\frac{x}{2})(x+1)$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Group is reducible or imprimitive 
<- Kovacics algorithm 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)-\frac{d}{dx}y\left(x\right)+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{y\left(x\right)}{x^2-1}-\frac{\frac{d}{dx}y\left(x\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{\frac{d}{dx}y\left(x\right)}{x^2-1}-\frac{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{1}{x^2-1},P_3\left(x\right)=-\frac{1}{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)+\frac{d}{dx}y\left(x\right)-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)+\frac{d}{du}y\left(u\right)-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}\frac{d}{du}y\left(u\right)\text{to series expansion}\\ & & \frac{d}{du}y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & \frac{d}{du}y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+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(-3+2r)u^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(-a_{k+1}(k+1+r)(2k-1+2r)+a_k(k^2+2kr+r^2-k-r-1))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -r(-3+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{3}{2}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & -2(k-\frac{1}{2}+r)(k+1+r)a_{k+1}+a_k(k^2+(2r-1)k+r^2-r-1)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k(k^2+2kr+r^2-k-r-1)}{(2k-1+2r)(k+1+r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=\frac{a_k(k^2-k-1)}{(2k-1)(k+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(k^2-k-1)}{(2k-1)(k+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(k^2-k-1)}{(2k-1)(k+1)}]\\ \bullet & & \text{Recursion relation for}r=\frac{3}{2}\\ & & a_{k+1}=\frac{a_k(k^2+2k-\frac{1}{4})}{(2k+2)(k+\frac{5}{2})}\\ \bullet & & \text{Solution for}r=\frac{3}{2}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+\frac{3}{2}},a_{k+1}=\frac{a_k(k^2+2k-\frac{1}{4})}{(2k+2)(k+\frac{5}{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{3}{2}},a_{k+1}=\frac{a_k(k^2+2k-\frac{1}{4})}{(2k+2)(k+\frac{5}{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{3}{2}}),a_{k+1}=\frac{a_k(k^2-k-1)}{(2k-1)(k+1)},b_{k+1}=\frac{b_k(k^2+2k-\frac{1}{4})}{(2k+2)(k+\frac{5}{2})}]\end{array}$$

✓ Mathematica. Time used: 6.1 (sec). Leaf size: 210

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

$$y(x)\to {(\sqrt{x-1}-\sqrt{x+1})}^{-\frac{1}{2}-\frac{\sqrt{5}}{2}}{(\sqrt{x-1}+\sqrt{x+1})}^{\frac{1}{2}(\sqrt{5}-1)}(\sqrt{x-1}-\sqrt{5}\sqrt{x+1})(c_2{\int }_1^x-\frac{2e^{\text{arctanh}(K[2])}{(\sqrt{K[2]-1}-\sqrt{K[2]+1})}^{\sqrt{5}}{(\sqrt{K[2]-1}+\sqrt{K[2]+1})}^{-\sqrt{5}}}{{(\sqrt{K[2]-1}-\sqrt{5}\sqrt{K[2]+1})}^2}dK[2]+c_1)\exp (-\frac{1}{2}{\int }_1^x\frac{1}{K[1{]}^2-1}dK[1]-\frac{\text{arctanh}(x)}{2})$$

✓ Sympy. Time used: 0.612 (sec). Leaf size: 122

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

$$y\left(x\right)=\frac{\sqrt[4]{x-1}\sqrt[4]{x^2-1}{(x^2-1)}^{\frac{\sqrt{5}}{4}}(C_1{\left(\frac{x+1}{x-1}\right)}^{\frac{3}{2}}{}_2{F}_1\left(\begin{array}{c}1-\frac{\sqrt{5}}{2},\frac{3}{2}-\frac{\sqrt{5}}{2}\\ \frac{5}{2}\end{array}\right|\frac{x+1}{x-1})+C_2{}_2{F}_1\left(\begin{array}{c}-\frac{\sqrt{5}}{2},-\frac{\sqrt{5}}{2}-\frac{1}{2}\\ -\frac{1}{2}\end{array}\right|\frac{x+1}{x-1}))e^{\frac{\sqrt{5}(\log (x-1)-\log (x+1))}{4}}}{\sqrt[4]{x+1}}$$