2.1.321 Problem 328

Internal problem ID [9491]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 328
Date solved : Friday, April 25, 2025 at 06:09:41 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.331 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

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

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 2.321: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-1\\ & =1\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=16x^2$. There is a pole at $x=0$ of order $2$. Since there is a pole of order $2$ then necessary conditions for case two are met. Therefore

$$\begin{array}{rl}L& =[2]\end{array}$$

Attempting to find a solution using case $n=2$.

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=\frac{5}{16x^2}-\frac{1}{x}$$

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 $b=\frac{5}{16}$. Hence

$$\begin{array}{rl}{E}_c& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{-1,2,5\}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $1<2$ then

$$\begin{array}{r}{E}_{\mathrm{\infty }}=\{1\}\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ for case 2 of Kovacic algorithm.

pole $c$ location	pole order	$E_c$
$0$	$2$	$\{-1,2,5\}$

Order of $r$ at $\mathrm{\infty }$	$E_{\mathrm{\infty }}$
$1$	$\{1\}$

Using the family $\{e_1,e_2,\dots ,e_{\mathrm{\infty }}\}$ given by

$$e_1=-1,e_{\mathrm{\infty }}=1$$

Gives a non negative integer $d$ (the degree of the polynomial $p(x)$), which is generated using

$$\begin{array}{rl}d& =\frac{1}{2}(e_{\mathrm{\infty }}-\sum _{c\in \mathrm{\Gamma }}{e}_c)\\ & =\frac{1}{2}(1-(-1))\\ & =1\end{array}$$

We now form the following rational function

$$\begin{array}{rl}\theta & =\frac{1}{2}\sum _{c\in \mathrm{\Gamma }}\frac{e_c}{x-c}\\ & =\frac{1}{2}\left(\frac{-1}{(x-\left(0\right))}\right)\\ & =-\frac{1}{2x}\end{array}$$

Now we search for a monic polynomial $p(x)$ of degree $d=1$ such that

$$\begin{array}{}\text{(1A)}& p^{‴}+3\theta p^{″}+(3{\theta }^2+3{\theta }^{\prime }-4r)p^{\prime }+({\theta }^{″}+3\theta {\theta }^{\prime }+{\theta }^3-4r\theta -2r^{\prime })p=0\end{array}$$

Since $d=1$, then letting

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

Substituting $p$ and $\theta$ into Eq. (1A) gives

$$\frac{1-4a_0}{x^2}=0$$

And solving for $p$ gives

$$p=x+\frac{1}{4}$$

Now that $p(x)$ is found let

$$\begin{array}{rl}\varphi & =\theta +\frac{p^{\prime }}{p}\\ & =\frac{1}{x+\frac{1}{4}}-\frac{1}{2x}\end{array}$$

Let $\omega$ be the solution of

$$\begin{array}{rl}{\omega }^2-\varphi \omega +(\frac{1}{2}{\varphi }^{\prime }+\frac{1}{2}{\varphi }^2-r)& =0\end{array}$$

Substituting the values for $\varphi$ and $r$ into the above equation gives

$$w^2-(\frac{1}{x+\frac{1}{4}}-\frac{1}{2x})w+\frac{64x^2-12x+1}{64x^3+16x^2}=0$$

Solving for $\omega$ gives

$$\begin{array}{rl}\omega & =\frac{16x\sqrt{-x}+4x-1}{4(4x+1)x}\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =e^{\int \omega dx}\\ & ={\mathrm{e}}^{\int \frac{16x\sqrt{-x}+4x-1}{4(4x+1)x}dx}\\ & =\frac{(2\sqrt{-x}-1){\mathrm{e}}^{2\sqrt{-x}}}{{(-x)}^{1/4}}\end{array}$$

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

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

Which simplifies to

$$y_1=\frac{(2\sqrt{-x}-1){\mathrm{e}}^{2\sqrt{-x}}}{x^{3/4}{(-x)}^{1/4}}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{3x}{2x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{3\ln \left(x\right)}{2}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\frac{{\mathrm{e}}^{-4\sqrt{-x}}}{8}+\frac{{\mathrm{e}}^{-4\sqrt{-x}}}{8\sqrt{-x}-4})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{(2\sqrt{-x}-1){\mathrm{e}}^{2\sqrt{-x}}}{x^{3/4}{(-x)}^{1/4}}\right)+c_2\left(\frac{(2\sqrt{-x}-1){\mathrm{e}}^{2\sqrt{-x}}}{x^{3/4}{(-x)}^{1/4}}(\frac{{\mathrm{e}}^{-4\sqrt{-x}}}{8}+\frac{{\mathrm{e}}^{-4\sqrt{-x}}}{8\sqrt{-x}-4})\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.242 (sec). Leaf size: 73

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

$$y=\frac{c_1\sqrt{\frac{(2\sqrt{x}+i)(4x+1)}{-2\sqrt{x}+i}}{\mathrm{e}}^{2i\sqrt{x}}+c_2\sqrt{\frac{(-2\sqrt{x}+i)(4x+1)}{2\sqrt{x}+i}}{\mathrm{e}}^{-2i\sqrt{x}}}{x}$$

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 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 2x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x\left(\frac{d}{dx}y\left(x\right)\right)+(2x-1)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{(2x-1)y\left(x\right)}{2x^2}-\frac{3\left(\frac{d}{dx}y\left(x\right)\right)}{2x}\\ \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{3\left(\frac{d}{dx}y\left(x\right)\right)}{2x}+\frac{(2x-1)y\left(x\right)}{2x^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{3}{2x},P_3\left(x\right)=\frac{2x-1}{2x^2}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=\frac{3}{2}\\ & \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}=-\frac{1}{2}\\ & \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}\\ & & 2x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x\left(\frac{d}{dx}y\left(x\right)\right)+(2x-1)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..1\\ & & 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(1+r)(-1+2r)x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_k(k+r+1)(2k+2r-1)+2a_{k-1})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (1+r)(-1+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-1,\frac{1}{2}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 2(k+r-\frac{1}{2})(k+r+1)a_k+2a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & 2(k+\frac{1}{2}+r)(k+2+r)a_{k+1}+2a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{2a_k}{(2k+1+2r)(k+2+r)}\\ \bullet & & \text{Recursion relation for}r=-1\\ & & a_{k+1}=-\frac{2a_k}{(2k-1)(k+1)}\\ \bullet & & \text{Solution for}r=-1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-1},a_{k+1}=-\frac{2a_k}{(2k-1)(k+1)}]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{2}\\ & & a_{k+1}=-\frac{2a_k}{(2k+2)(k+\frac{5}{2})}\\ \bullet & & \text{Solution for}r=\frac{1}{2}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{1}{2}},a_{k+1}=-\frac{2a_k}{(2k+2)(k+\frac{5}{2})}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-1})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k+\frac{1}{2}}),a_{k+1}=-\frac{2a_k}{(2k-1)(k+1)},b_{k+1}=-\frac{2b_k}{(2k+2)(k+\frac{5}{2})}]\end{array}$$

✓ Mathematica. Time used: 0.211 (sec). Leaf size: 77

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

$$y(x)\to \frac{e^{2i\sqrt{x}}(2\sqrt{x}+i)(c_2{\int }_1^x\frac{e^{-4i\sqrt{K[1]}}\sqrt{K[1]}}{{(2\sqrt{K[1]}+i)}^2}dK[1]+c_1)}{x}$$

✓ Sympy. Time used: 0.242 (sec). Leaf size: 31

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

$$y\left(x\right)=\frac{C_1{J}_{\frac{3}{2}}\left(2\sqrt{x}\right)+C_2{Y}_{\frac{3}{2}}\left(2\sqrt{x}\right)}{\sqrt[4]{x}}$$