2.1.248 Problem 251

Internal problem ID [9418]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 251
Date solved : Friday, April 25, 2025 at 06:07:50 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.443 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

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

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 2.248: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =1-1\\ & =0\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=x$. There is a pole at $x=0$ of order $1$. 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

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

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

Looking at poles of order 1. For the pole at $x=0$ 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 $O_r(\mathrm{\infty })=0$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{0}{2}& & =0\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $x^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{x}^i\\ \text{(8)}& & =\sum _{i=0}^0{a}_i{x}^i\end{array}$$

Let $a$ be the coefficient of $x^v=x^0$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx 1+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{2x^3}-\frac{5}{8x^4}+\frac{7}{8x^5}-\frac{21}{16x^6}+\frac{33}{16x^7}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=1$$

From Eq. (9) the sum up to $v=0$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^0{a}_i{x}^i\\ \text{(10)}& & =1\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^{-1}=\frac{1}{x}$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=1$$

This shows that the coefficient of $\frac{1}{x}$ in the above is $0$. Now we need to find the coefficient of $\frac{1}{x}$ in $r$. How this is done depends on if $v=0$ or not. Since $v=0$ then starting from $r=\frac{s}{t}$ and doing long division in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $\frac{1}{x}$ in $r$ will be the coefficient in $R$ of the term in $x$ of degree of $t$ minus one, divided by the leading coefficient in $t$. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{2+x}{x}\\ & =Q+\frac{R}{x}\\ & =\left(1\right)+\left(\frac{2}{x}\right)\\ & =1+\frac{2}{x}\end{array}$$

Since the degree of $t$ is $1$, then we see that the coefficient of the term $1$ in the remainder $R$ is $2$. Dividing this by leading coefficient in $t$ which is $1$ gives $2$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(2\right)-\left(0\right)\\ & =2\end{array}$$

Hence

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

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

$$r=\frac{2+x}{x}$$

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

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

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 ${\alpha }_{\mathrm{\infty }}^{+}=1$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{-}\right)\\ & =1-\left(1\right)\\ & =0\end{array}$$

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

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

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

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

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

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

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

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

Which simplifies to

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

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{2x^2+4x}{x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-2x-4\ln \left(x\right)}}{{\left(y_1\right)}^2}dx\\ & =y_1(-\frac{{\mathrm{e}}^{-2x}}{3x}+\frac{{\mathrm{e}}^{-2x}}{3}-\frac{2x{\mathrm{e}}^{-2x}}{3}+\frac{4x^2{\mathrm{Ei}}_1\left(2x\right)}{3}-\frac{4{\mathrm{Ei}}_1\left(2x\right)x^3-2{\mathrm{e}}^{-2x}{x}^2+x{\mathrm{e}}^{-2x}-6x{\mathrm{Ei}}_1\left(2x\right)+2{\mathrm{e}}^{-2x}}{3x})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{1}{x}\right)+c_2\left(\frac{1}{x}(-\frac{{\mathrm{e}}^{-2x}}{3x}+\frac{{\mathrm{e}}^{-2x}}{3}-\frac{2x{\mathrm{e}}^{-2x}}{3}+\frac{4x^2{\mathrm{Ei}}_1\left(2x\right)}{3}-\frac{4{\mathrm{Ei}}_1\left(2x\right)x^3-2{\mathrm{e}}^{-2x}{x}^2+x{\mathrm{e}}^{-2x}-6x{\mathrm{Ei}}_1\left(2x\right)+2{\mathrm{e}}^{-2x}}{3x})\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.004 (sec). Leaf size: 28

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

$$y=\frac{2{\mathrm{Ei}}_1\left(2x\right)c_{2}x-{\mathrm{e}}^{-2x}{c}_2+c_{1}x}{x^2}$$

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 
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: 0.239 (sec). Leaf size: 33

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

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

✗ 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) + (2*x + 2)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False