2.1.271 Problem 274

Internal problem ID [9441]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 274
Date solved : Friday, April 25, 2025 at 06:08:24 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}4y^{\prime \prime }+\frac{3(-x^2+2)y}{{(-x^2+1)}^2}& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.235 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& 4y^{\prime \prime }+\frac{(-3x^2+6)y}{{(x^2-1)}^2}& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =4\\ \text{(3)}& B& =0C& =\frac{-3x^2+6}{{(x^2-1)}^2}\end{array}$$

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

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

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =3x^2-6\\ t& =4{(x^2-1)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{3x^2-6}{4{(x^2-1)}^2}\right)z(x)\end{array}$$

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 2.271: 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 $n=1$.

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=\frac{9}{16(x-1)}-\frac{3}{16{(x-1)}^2}-\frac{3}{16{(x+1)}^2}-\frac{9}{16(x+1)}$$

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

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{4}\end{array}$$

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

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{4}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is 2 then $[\sqrt{r}{]}_{\mathrm{\infty }}=0$. 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

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

Since the $\text{gcd}(s,t)=1$. This gives $b=\frac{3}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{2}\end{array}$$

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

$$r=\frac{3x^2-6}{4{(x^2-1)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$1$	$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$
$-1$	$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{+}=\frac{3}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})\\ & =\frac{3}{2}-\left(\frac{3}{2}\right)\\ & =0\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{x-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

Substituting the above values in the above results in

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{x-c_2})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{3}{4(x-1)}+\frac{3}{4(x+1)}+\left(0\right)\\ & =\frac{3}{4(x-1)}+\frac{3}{4(x+1)}\\ & =\frac{3x}{2x^2-2}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=0$ to solve the ode. The polynomial $p(x)$ needs to satisfy the equation

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

Let

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

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(\frac{3}{4(x-1)}+\frac{3}{4(x+1)})\left(0\right)+((-\frac{3}{4{(x-1)}^2}-\frac{3}{4{(x+1)}^2})+{(\frac{3}{4(x-1)}+\frac{3}{4(x+1)})}^2-\left(\frac{3x^2-6}{4{(x^2-1)}^2}\right))& =0\\ 0=0\end{array}$$

The equation is satisfied since both sides are zero. Therefore the first solution to the ode $z^{″}=rz$ is

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

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

$$y_1=z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}$$

Since $B=0$ then the above reduces to

$$\begin{array}{rl}{y}_1& =z_1\\ & ={(x^2-1)}^{3/4}\end{array}$$

Which simplifies to

$$y_1={(x^2-1)}^{3/4}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Since $B=0$ then the above becomes

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dx\\ & ={(x^2-1)}^{3/4}\int \frac{1}{{(x^2-1)}^{3/2}}dx\\ & ={(x^2-1)}^{3/4}(-\frac{(x-1)(x+1)x}{{(x^2-1)}^{3/2}})\end{array}$$

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.005 (sec). Leaf size: 24

ode:=4*diff(diff(y(x),x),x)+3*(-x^2+2)/(-x^2+1)^2*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1{(x^2-1)}^{3/4}+c_2{(x^2-1)}^{1/4}x$$

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 
   Reducible group (found an exponential solution) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 4\frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{3(-x^2+2)y\left(x\right)}{{(-x^2+1)}^2}=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{3(x^2-2)y\left(x\right)}{4{(x^2-1)}^2}\\ \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{3(x^2-2)y\left(x\right)}{4{(x^2-1)}^2}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=0,P_3\left(x\right)=-\frac{3(x^2-2)}{4{(x^2-1)}^2}]\\ & \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}=0\\ & \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}=\frac{3}{16}\\ & \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}\\ & & 4{(x^2-1)}^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(-3x^2+6)y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-1\text{so that the regular singular point is at}u=0\\ & & (4u^4-16u^3+16u^2)\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(-3u^2+6u+3)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 y\left(u\right)\text{to series expansion for}m=0..2\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{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=2..4\\ & & 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(-1+4r)(-3+4r)u^r+(a_1(3+4r)(1+4r)-2a_0(8r^2-8r-3))u^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k(4k+4r-1)(4k+4r-3)-2a_{k-1}(8{(k-1)}^2+16(k-1)r+8r^2-8k+5-8r)+a_{k-2}(2k+2r-3)(2k-7+2r))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (-1+4r)(-3+4r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{\frac{1}{4},\frac{3}{4}\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(3+4r)(1+4r)-2a_0(8r^2-8r-3)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=\frac{2a_0(8r^2-8r-3)}{16r^2+16r+3}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 4(4a_k+a_{k-2}-4a_{k-1})k^2+4(2(4a_k+a_{k-2}-4a_{k-1})r-4a_k-5a_{k-2}+12a_{k-1})k+4(4a_k+a_{k-2}-4a_{k-1})r^2+4(-4a_k-5a_{k-2}+12a_{k-1})r+3a_k+21a_{k-2}-26a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & 4(4a_{k+2}+a_k-4a_{k+1}){(k+2)}^2+4(2(4a_{k+2}+a_k-4a_{k+1})r-4a_{k+2}-5a_k+12a_{k+1})(k+2)+4(4a_{k+2}+a_k-4a_{k+1})r^2+4(-4a_{k+2}-5a_k+12a_{k+1})r+3a_{k+2}+21a_k-26a_{k+1}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}+8kr{a}_k-32kr{a}_{k+1}+4r^2{a}_k-16r^2{a}_{k+1}-4k{a}_k-16k{a}_{k+1}-4r{a}_k-16r{a}_{k+1}-3a_k+6a_{k+1}}{16k^2+32kr+16r^2+48k+48r+35}\\ \bullet & & \text{Recursion relation for}r=\frac{1}{4}\\ & & a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}-2k{a}_k-24k{a}_{k+1}-\frac{15}{4}{a}_k+a_{k+1}}{16k^2+56k+48}\\ \bullet & & \text{Solution for}r=\frac{1}{4}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+\frac{1}{4}},a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}-2k{a}_k-24k{a}_{k+1}-\frac{15}{4}{a}_k+a_{k+1}}{16k^2+56k+48},a_1=-\frac{9a_0}{8}]\\ \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}{4}},a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}-2k{a}_k-24k{a}_{k+1}-\frac{15}{4}{a}_k+a_{k+1}}{16k^2+56k+48},a_1=-\frac{9a_0}{8}]\\ \bullet & & \text{Recursion relation for}r=\frac{3}{4}\\ & & a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}+2k{a}_k-40k{a}_{k+1}-\frac{15}{4}{a}_k-15a_{k+1}}{16k^2+72k+80}\\ \bullet & & \text{Solution for}r=\frac{3}{4}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+\frac{3}{4}},a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}+2k{a}_k-40k{a}_{k+1}-\frac{15}{4}{a}_k-15a_{k+1}}{16k^2+72k+80},a_1=-\frac{3a_0}{8}]\\ \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}{4}},a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}+2k{a}_k-40k{a}_{k+1}-\frac{15}{4}{a}_k-15a_{k+1}}{16k^2+72k+80},a_1=-\frac{3a_0}{8}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1)}^{k+\frac{1}{4}})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(x+1)}^{k+\frac{3}{4}}),a_{k+2}=-\frac{4k^2{a}_k-16k^2{a}_{k+1}-2k{a}_k-24k{a}_{k+1}-\frac{15}{4}{a}_k+a_{k+1}}{16k^2+56k+48},a_1=-\frac{9a_0}{8},b_{k+2}=-\frac{4k^2{b}_k-16k^2{b}_{k+1}+2k{b}_k-40k{b}_{k+1}-\frac{15}{4}{b}_k-15b_{k+1}}{16k^2+72k+80},b_1=-\frac{3b_0}{8}]\end{array}$$

✓ Mathematica. Time used: 0.041 (sec). Leaf size: 51

ode=4*D[y[x],{x,2}]+3*(2-x^2)/(1-x^2)^2*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \sqrt{x^2-1}(c_2{Q}_{\frac{1}{2}}^{\frac{1}{2}}(x)+\frac{\sqrt{\frac{2}{\pi }}{c}_{1}x}{\sqrt[4]{1-x^2}})$$

✓ Sympy. Time used: 0.315 (sec). Leaf size: 48

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*Derivative(y(x), (x, 2)) + (6 - 3*x**2)*y(x)/(1 - x**2)**2,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=\frac{\sqrt[4]{x^2-1}(C_1\sqrt{x^2}{}_2{F}_1\left(\begin{array}{c}0,1\\ \frac{3}{2}\end{array}\right|x^2)+C_2{}_1{F}_0\left(\begin{array}{c}-\frac{1}{2}\end{array}\right|x^2))\sqrt[4]{x^2}}{\sqrt{x}}$$