2.1.821 Problem 845

Internal problem ID [9993]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 845
Date solved : Sunday, March 30, 2025 at 02:51:09 PM
CAS classification : [[_Emden, _Fowler]]

Solved as second order ode using Kovacic algorithm

Time used: 0.124 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

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

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 2.821: 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 $t=x^2$. There is a pole at $x=0$ 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

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

For the pole at $x=0$ let $b$ be the coefficient of $\frac{1}{x^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=6$. Hence

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

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

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

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

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

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

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 ${\alpha }_{\mathrm{\infty }}^{-}=-2$ then

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

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

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

The above gives

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

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

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

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

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

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

$$\begin{array}{rl}{y}_1& =z_1\\ & =\frac{1}{x^2}\end{array}$$

Which simplifies to

$$y_1=\frac{1}{x^2}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dx\\ & =\frac{1}{x^2}\int \frac{1}{\frac{1}{x^4}}dx\\ & =\frac{1}{x^2}\left(\frac{x^5}{5}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{1}{x^2}\right)+c_2\left(\frac{1}{x^2}\left(\frac{x^5}{5}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.003 (sec). Leaf size: 15

ode:=diff(diff(y(x),x),x) = 6/x^2*y(x); 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_2{x}^5+c_1}{x^2}$$

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 
<- LODE of Euler type successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\frac{6y\left(x\right)}{x^2}\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \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{6y\left(x\right)}{x^2}=0\\ \bullet & & \text{Multiply by denominators of the ODE}\\ & & \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)x^2-6y\left(x\right)=0\\ \bullet & & \text{Make a change of variables}\\ & & t=\ln \left(x\right)\\ ◻& & \text{Substitute the change of variables back into the ODE}\\ & \circ & \text{Calculate the}\text{1st}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}y\left(t\right)\right)\left(\frac{d}{dx}t\left(x\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}y\left(t\right)}{x}\\ & \circ & \text{Calculate the}\text{2nd}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right){\left(\frac{d}{dx}t\left(x\right)\right)}^2+\left(\frac{d}{dx}\frac{d}{dx}t\left(x\right)\right)\left(\frac{d}{dt}y\left(t\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & (\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2})x^2-6y\left(t\right)=0\\ \bullet & & \text{Simplify}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)-\frac{d}{dt}y\left(t\right)-6y\left(t\right)=0\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2-r-6=0\\ \bullet & & \text{Factor the characteristic polynomial}\\ & & (r+2)(r-3)=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-2,3)\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(t\right)={\mathrm{e}}^{-2t}\\ \bullet & & \text{2nd solution of the ODE}\\ & & y_2\left(t\right)={\mathrm{e}}^{3t}\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{y}_1\left(t\right)+\mathit{C}\mathit{2}{y}_2\left(t\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-2t}+\mathit{C}\mathit{2}{\mathrm{e}}^{3t}\\ \bullet & & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=\frac{\mathit{C}\mathit{1}}{x^2}+\mathit{C}\mathit{2}{x}^3\\ \bullet & & \text{Simplify}\\ & & y\left(x\right)=\frac{\mathit{C}\mathit{1}}{x^2}+\mathit{C}\mathit{2}{x}^3\end{array}$$

✓ Mathematica. Time used: 0.012 (sec). Leaf size: 18

ode=D[y[x],{x,2}]==((4*(5/2)^2-1)/(4*x^2))*y[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{c_2{x}^5+c_1}{x^2}$$

✓ Sympy. Time used: 0.049 (sec). Leaf size: 12

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

$$y\left(x\right)=\frac{C_1}{x^2}+C_2{x}^3$$