2.1.266 Problem 269

Internal problem ID [9436]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 269
Date solved : Friday, April 25, 2025 at 06:08:16 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}2x^2{y}^{\prime \prime }-(3x+2)y^{\prime }+\frac{(2x-1)y}{x}& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.632 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& 2x^2{y}^{\prime \prime }+(-3x-2)y^{\prime }+(2-\frac{1}{x})y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{5x^2+36x+4}{16x^4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =5x^2+36x+4\\ t& =16x^4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{5x^2+36x+4}{16x^4}\right)z(x)\end{array}$$

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 2.266: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-2\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=16x^4$. There is a pole at $x=0$ of order $4$. 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. Therefore

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

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

Looking at higher order poles of order $2v$≥$4$ (must be even order for case one).Then for each pole $c$, $[\sqrt{r}{]}_c$ is the sum of terms $\frac{1}{(x-c{)}^i}$ for $2\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ expanded around each pole $c$. Hence

$$\begin{array}{}\text{(1B)}& [\sqrt{r}{]}_c& =\sum _2^v\frac{a_i}{(x-c{)}^i}\end{array}$$

Let $a$ be the coefficient of the term $\frac{1}{(x-c{)}^v}$ in the above where $v$ is the pole order divided by 2. Let $b$ be the coefficient of $\frac{1}{(x-c{)}^{v+1}}$ in $r$ minus the coefficient of $\frac{1}{(x-c{)}^{v+1}}$ in $[\sqrt{r}{]}_c$. Then

$$\begin{array}{rl}{\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)\\ {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)\end{array}$$

The partial fraction decomposition of $r$ is

$$r=\frac{9}{4x^3}+\frac{1}{4x^4}+\frac{5}{16x^2}$$

There is pole in $r$ at $x=0$ of order $4$, hence $v=2$. Expanding $\sqrt{r}$ as Laurent series about this pole $c=0$ gives

$$\begin{array}{}\text{(2B)}& [\sqrt{r}{]}_c\approx \frac{1}{2x^2}+\frac{9}{4x}-\frac{19}{4}+\frac{171x}{8}-\frac{475x^2}{4}+\frac{11799x^3}{16}+\dots \end{array}$$

Using eq. (1B), taking the sum up to $v=2$ the above becomes

$$\begin{array}{}\text{(3B)}& [\sqrt{r}{]}_c=\frac{1}{2x^2}\end{array}$$

The above shows that the coefficient of $\frac{1}{(x-0{)}^2}$ is

$$a=\frac{1}{2}$$

Now we need to find $b$. let $b$ be the coefficient of the term $\frac{1}{(x-c{)}^{v+1}}$ in $r$ minus the coefficient of the same term but in the sum $[\sqrt{r}{]}_c$ found in eq. (3B). Here $c$ is current pole which is $c=0$. This term becomes $\frac{1}{x^3}$. The coefficient of this term in the sum $[\sqrt{r}{]}_c$ is seen to be $0$ and the coefficient of this term $r$ is found from the partial fraction decomposition from above to be $\frac{9}{4}$. Therefore

$$\begin{array}{rl}b& =\left(\frac{9}{4}\right)-(0)\\ & =\frac{9}{4}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =\frac{1}{2x^2}\\ {\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)& & =\frac{1}{2}(\frac{\frac{9}{4}}{\frac{1}{2}}+2)& & =\frac{13}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)& & =\frac{1}{2}(-\frac{\frac{9}{4}}{\frac{1}{2}}+2)& & =-\frac{5}{4}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{5x^2+36x+4}{16x^4}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=\frac{5}{16}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{4}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{4}\end{array}$$

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

$$r=\frac{5x^2+36x+4}{16x^4}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$4$	$\frac{1}{2x^2}$	$\frac{13}{4}$	$-\frac{5}{4}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{5}{4}$	$-\frac{1}{4}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=-\frac{1}{4}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =-\frac{1}{4}-(-\frac{5}{4})\\ & =1\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{2x^2}-\frac{5}{4x}+(-)\left(0\right)\\ & =-\frac{1}{2x^2}-\frac{5}{4x}\\ & =\frac{-2-5x}{4x^2}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=1$ 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

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

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{2x^2}-\frac{5}{4x})\left(1\right)+((\frac{1}{x^3}+\frac{5}{4x^2})+{(-\frac{1}{2x^2}-\frac{5}{4x})}^2-\left(\frac{5x^2+36x+4}{16x^4}\right))& =0\\ \frac{-2+5a_0}{2x^2}=0\end{array}$$

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

$$\{a_0=\frac{2}{5}\}$$

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

$$\begin{array}{rl}p(x)& =x+\frac{2}{5}\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x+\frac{2}{5}){\mathrm{e}}^{\int (-\frac{1}{2x^2}-\frac{5}{4x})dx}\\ & =(x+\frac{2}{5}){\mathrm{e}}^{-\frac{5\ln \left(x\right)}{4}+\frac{1}{2x}}\\ & =\frac{(x+\frac{2}{5}){\mathrm{e}}^{\frac{1}{2x}}}{x^{5/4}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-3x-2}{2x^2}dx}\\ & =z_1{e}^{\frac{3\ln \left(x\right)}{4}-\frac{1}{2x}}\\ & =z_1\left(x^{3/4}{\mathrm{e}}^{-\frac{1}{2x}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{2+5x}{5\sqrt{x}}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-3x-2}{2x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\frac{3\ln \left(x\right)}{2}-\frac{1}{x}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{25{\mathrm{e}}^{\frac{3\ln \left(x\right)}{2}-\frac{1}{x}}x}{{(2+5x)}^2}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{2+5x}{5\sqrt{x}}\right)+c_2\left(\frac{2+5x}{5\sqrt{x}}(\int \frac{25{\mathrm{e}}^{\frac{3\ln \left(x\right)}{2}-\frac{1}{x}}x}{{(2+5x)}^2}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.059 (sec). Leaf size: 35

ode:=2*x^2*diff(diff(y(x),x),x)-(2+3*x)*diff(y(x),x)+(-1+2*x)/x*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_1(5x+2)+c_2{\mathrm{e}}^{-\frac{1}{x}}\mathrm{hypergeom}(\left[2\right],[-\frac{1}{2}],\frac{1}{x})x^{5/2}}{\sqrt{x}}$$

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 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      <- Kummer successful 
   <- special function solution successful 
      Solution using Kummer functions still has integrals. Trying a hypergeomet\ 
ric solution. 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved -\ 
 returning with no uncomputed integrals 
   <- Kovacics algorithm successful
 

✓ Mathematica. Time used: 0.509 (sec). Leaf size: 65

ode=2*x^2*D[y[x],{x,2}]-(3*x+2)*D[y[x],x]+(2*x-1)/x*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{\sqrt{e}(5x+2)(c_2{\int }_1^x\frac{25e^{-\frac{5}{2}-\frac{1}{K[1]}}K[1{]}^{5/2}}{(5K[1]+2{)}^2}dK[1]+c_1)}{5\sqrt{x}}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**2*Derivative(y(x), (x, 2)) - (3*x + 2)*Derivative(y(x), x) + (2*x - 1)*y(x)/x,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) - (2*x**3*Derivative(y(x), (x, 2)) + 2*x*y(x) - y(x))/(x*(3*x + 2)) cannot be solved by the factorable group method