2.1.744 Problem 764

Internal problem ID [9916]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 764
Date solved : Sunday, March 30, 2025 at 02:49:30 PM
CAS classification : [[_Emden, _Fowler]]

Solved as second order ode using Kovacic algorithm

Time used: 0.264 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

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

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =6\\ t& =x^2+1\end{array}$$

Therefore eq. (4) becomes

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

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 2.744: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-0\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=x^2+1$. There is a pole at $x=i$ of order $1$. There is a pole at $x=-i$ of order $1$. 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 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

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

Attempting to find a solution using case $n=1$.

Looking at poles of order 1. For the pole at $x=i$ of order 1 then

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

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{6}{x^2+1}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=6$. Hence

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

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

$$r=\frac{6}{x^2+1}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$i$	$1$	$0$	$0$	$1$

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

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 ${\alpha }_{\mathrm{\infty }}^{+}=3$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{-}\right)\\ & =3-\left(1\right)\\ & =2\end{array}$$

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

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

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=2$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x^2+a_{1}x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(2\right)+2\left(\frac{1}{x-i}\right)(2x+a_1)+((-\frac{1}{{(x-i)}^2})+{\left(\frac{1}{x-i}\right)}^2-\left(\frac{6}{x^2+1}\right))& =0\\ 2+\frac{-4x-2a_1}{-x+i}+\frac{-6x^2-6a_{1}x-6a_0}{x^2+1}=0\end{array}$$

Solving for the coefficients $a_i$ in the above using method of undetermined coefficients gives

$$\{a_0=0,a_1=i\}$$

Substituting these coefficients in $p(x)$ in eq. (2A) results in

$$\begin{array}{rl}p(x)& =x^2+ix\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x^2+ix){\mathrm{e}}^{\int \frac{1}{x-i}dx}\\ & =(x^2+ix){\mathrm{e}}^{\frac{\ln (x^2+1)}{2}+i\arctan \left(x\right)}\\ & =x(x+i)(ix+1)\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1\\ & =x(x+i)(ix+1)\end{array}$$

Which simplifies to

$$y_1=i{x}^3+ix$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dx\\ & =i{x}^3+ix\int \frac{1}{{(i{x}^3+ix)}^2}dx\\ & =i{x}^3+ix(\frac{1}{x}+\frac{x}{2x^2+2}+\frac{3\arctan \left(x\right)}{2})\end{array}$$

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.004 (sec). Leaf size: 31

ode:=(x^2+1)*diff(diff(y(x),x),x)-6*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{3x{c}_2(x^2+1)\arctan \left(x\right)}{2}+c_1{x}^3+\frac{3c_2{x}^2}{2}+c_{1}x+c_2$$

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) 
   Group is reducible, not completely reducible 
<- Kovacics algorithm successful
 

✓ Mathematica. Time used: 0.148 (sec). Leaf size: 33

ode=(x^2+1)*D[y[x],{x,2}]-6*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to (x^3+x)(c_2{\int }_1^x\frac{1}{{(K[1{]}^3+K[1])}^2}dK[1]+c_1)$$

✓ Sympy. Time used: 0.334 (sec). Leaf size: 70

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

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