2.1.40 Problem 42

Internal problem ID [9212]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 42
Date solved : Wednesday, March 05, 2025 at 07:37:53 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{x}^2{y}^{\prime \prime }-2x(x+2)y^{\prime }+(x^2+4x+6)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.068 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{0}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =0\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =0\end{array}$$

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 2.40: Necessary conditions for each Kovacic case

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

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

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 $infinity$ then the necessary conditions for case one are met. Therefore

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

Since $r=0$ 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

$$z_1(x)=1$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-2x^2-4x}{x^2}dx}\\ & =z_1{e}^{x+2\ln \left(x\right)}\\ & =z_1\left(x^2{\mathrm{e}}^x\right)\end{array}$$

Which simplifies to

$$y_1=x^2{\mathrm{e}}^x$$

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

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

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

$$y={\mathrm{e}}^x{x}^2(c_{2}x+c_1)$$

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

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)-2x(x+2)\left(\frac{d}{dx}y\left(x\right)\right)+(x^2+4x+6)y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=-\frac{(x^2+4x+6)y\left(x\right)}{x^2}+\frac{2(x+2)\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^2}{d{x}^2}y\left(x\right)-\frac{2(x+2)\left(\frac{d}{dx}y\left(x\right)\right)}{x}+\frac{(x^2+4x+6)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{2(x+2)}{x},P_3\left(x\right)=\frac{x^2+4x+6}{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}=-4\\ & \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^2}{d{x}^2}y\left(x\right)\right)-2x(x+2)\left(\frac{d}{dx}y\left(x\right)\right)+(x^2+4x+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}^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=1..2\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)x^{k+r}\\ & \circ & \text{Convert}{x}^2\cdot \left(\frac{d^2}{d{x}^2}y\left(x\right)\right)\text{to series expansion}\\ & & x^2\cdot \left(\frac{d^2}{d{x}^2}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(-2+r)(-3+r)x^r+(a_1(-1+r)(-2+r)-2a_0(-2+r))x^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k(k+r-2)(k+r-3)-2a_{k-1}(k+r-3)+a_{k-2})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (-2+r)(-3+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{2,3\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(-1+r)(-2+r)-2a_0(-2+r)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=\frac{2a_0}{-1+r}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_k(k+r-2)(k+r-3)-2a_{k-1}k-2a_{k-1}r+a_{k-2}+6a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+2}(k+r)(k+r-1)-2a_{k+1}(k+2)-2a_{k+1}r+a_k+6a_{k+1}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{2k{a}_{k+1}+2a_{k+1}r-a_k-2a_{k+1}}{(k+r)(k+r-1)}\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+2}=\frac{2k{a}_{k+1}-a_k+2a_{k+1}}{(k+2)(k+1)}\\ \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{2k{a}_{k+1}-a_k+2a_{k+1}}{(k+2)(k+1)},a_1=2a_0]\\ \bullet & & \text{Recursion relation for}r=3\\ & & a_{k+2}=\frac{2k{a}_{k+1}-a_k+4a_{k+1}}{(k+3)(k+2)}\\ \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{2k{a}_{k+1}-a_k+4a_{k+1}}{(k+3)(k+2)},a_1=a_0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+2})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k+3}),a_{k+2}=\frac{2k{a}_{k+1}-a_k+2a_{k+1}}{(k+2)(k+1)},a_1=2a_0,b_{k+2}=\frac{2k{b}_{k+1}-b_k+4b_{k+1}}{(k+3)(k+2)},b_1=b_0]\end{array}$$

✓ Mathematica. Time used: 0.049 (sec). Leaf size: 21

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

$$y(x)\to e^{x+2}{x}^2(c_{2}x+c_1)$$

✗ Sympy

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

False