2.1.772 Problem 794

Internal problem ID [9944]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 794
Date solved : Sunday, March 30, 2025 at 02:50:06 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.198 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =x^2\\ \text{(3)}& B& =6xC& =4x^2+6\end{array}$$

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

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

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-4\\ t& =1\end{array}$$

Therefore eq. (4) becomes

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

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 2.772: Necessary conditions for each Kovacic case

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

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

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=-4$ is not a function of $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)=\cos \left(2x\right)$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{6x}{x^2}dx}\\ & =z_1{e}^{-3\ln \left(x\right)}\\ & =z_1\left(\frac{1}{x^3}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{\cos \left(2x\right)}{x^3}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{6x}{x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-6\ln \left(x\right)}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{\tan \left(2x\right)}{2}\right)\end{array}$$

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.003 (sec). Leaf size: 21

ode:=x^2*diff(diff(y(x),x),x)+6*diff(y(x),x)*x+(4*x^2+6)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_1\sin \left(2x\right)+c_2\cos \left(2x\right)}{x^3}$$

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\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+6x\left(\frac{d}{dx}y\left(x\right)\right)+(4x^2+6)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{2(2x^2+3)y\left(x\right)}{x^2}-\frac{6\left(\frac{d}{dx}y\left(x\right)\right)}{x}\\ \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{6\left(\frac{d}{dx}y\left(x\right)\right)}{x}+\frac{2(2x^2+3)y\left(x\right)}{x^2}=0\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{6}{x},P_3\left(x\right)=\frac{2(2x^2+3)}{x^2}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=6\\ & \circ & x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & (x^2\cdot P_3\left(x\right))|{}_{x=0}=6\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+6x\left(\frac{d}{dx}y\left(x\right)\right)+(4x^2+6)y\left(x\right)=0\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{x}^m\cdot y\left(x\right)\text{to series expansion for}m=0..2\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{x}^{k+r}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r}\\ & \circ & \text{Convert}{x}^2\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x^2\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0(3+r)(2+r)x^r+a_1(4+r)(3+r)x^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k(k+r+3)(k+r+2)+4a_{k-2})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (3+r)(2+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-3,-2\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(4+r)(3+r)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_k(k+r+3)(k+r+2)+4a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+2}(k+5+r)(k+4+r)+4a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{4a_k}{(k+5+r)(k+4+r)}\\ \bullet & & \text{Recursion relation for}r=-3\\ & & a_{k+2}=-\frac{4a_k}{(k+2)(k+1)}\\ \bullet & & \text{Solution for}r=-3\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-3},a_{k+2}=-\frac{4a_k}{(k+2)(k+1)},a_1=0]\\ \bullet & & \text{Recursion relation for}r=-2\\ & & a_{k+2}=-\frac{4a_k}{(k+3)(k+2)}\\ \bullet & & \text{Solution for}r=-2\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-2},a_{k+2}=-\frac{4a_k}{(k+3)(k+2)},a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-3})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k-2}),a_{k+2}=-\frac{4a_k}{(k+2)(k+1)},a_1=0,b_{k+2}=-\frac{4b_k}{(k+3)(k+2)},b_1=0]\end{array}$$

✓ Mathematica. Time used: 0.032 (sec). Leaf size: 37

ode=x^2*D[y[x],{x,2}]+6*x*D[y[x],x]+(4*x^2+6)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{4c_1{e}^{-2ix}-i{c}_2{e}^{2ix}}{4x^3}$$

✓ Sympy. Time used: 0.237 (sec). Leaf size: 24

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

$$y\left(x\right)=\frac{C_1{J}_{\frac{1}{2}}\left(2x\right)+C_2{Y}_{\frac{1}{2}}\left(2x\right)}{x^{\frac{5}{2}}}$$