2.1.90 Problem 92

Internal problem ID [9262]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 92
Date solved : Sunday, March 30, 2025 at 02:30:57 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.196 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{-3+16x}{16x^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-3+16x\\ t& =16x^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-3+16x}{16x^2}\right)z(x)\end{array}$$

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 2.90: 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

$$r=\frac{1}{x}-\frac{3}{16x^2}$$

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 $b=-\frac{3}{16}$. Hence

$$\begin{array}{rl}{E}_c& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{1,2,3\}\end{array}$$

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	$E_c$
$0$	$2$	$\{1,2,3\}$

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

$$\begin{array}{rl}d& =\frac{1}{2}(e_{\mathrm{\infty }}-\sum _{c\in \mathrm{\Gamma }}{e}_c)\\ & =\frac{1}{2}(1-\left(1\right))\\ & =0\end{array}$$

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 $d=0$ 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 $d=0$, then letting

$$\begin{array}{}\text{(2A)}& p=1\end{array}$$

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

$$0=0$$

And solving for $p$ gives

$$p=1$$

Now that $p(x)$ is found let

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

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

$$w^2-\frac{w}{2x}+\frac{1-16x}{16x^2}=0$$

Solving for $\omega$ gives

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

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

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

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

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

Which simplifies to

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

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.004 (sec). Leaf size: 25

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

$$y=\sqrt{x}(c_1\sinh \left(2\sqrt{x}\right)+c_2\cosh \left(2\sqrt{x}\right))$$

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)-x\left(\frac{d}{dx}y\left(x\right)\right)+(1-2x)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{(-1+2x)y\left(x\right)}{2x^2}+\frac{\frac{d}{dx}y\left(x\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{\frac{d}{dx}y\left(x\right)}{2x}-\frac{(-1+2x)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{1}{2x},P_3\left(x\right)=-\frac{-1+2x}{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{1}{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)-x\left(\frac{d}{dx}y\left(x\right)\right)+(1-2x)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+2r)(-1+r)x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_k(2k+2r-1)(k+r-1)-2a_{k-1})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (-1+2r)(-1+r)=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+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+r)}\\ \bullet & & \text{Recursion relation for}r=1\\ & & a_{k+1}=\frac{2a_k}{(2k+3)(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+3)(k+1)}]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{2}\\ & & a_{k+1}=\frac{2a_k}{(2k+2)(k+\frac{1}{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{1}{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+3)(k+1)},b_{k+1}=\frac{2b_k}{(2k+2)(k+\frac{1}{2})}]\end{array}$$

✓ Mathematica. Time used: 0.053 (sec). Leaf size: 41

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

$$y(x)\to \frac{1}{2}{e}^{-2\sqrt{x}}\sqrt{x}(2c_1{e}^{4\sqrt{x}}-c_2)$$

✓ Sympy. Time used: 0.200 (sec). Leaf size: 34

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

$$y\left(x\right)=x^{\frac{3}{4}}(C_1{J}_{\frac{1}{2}}\left(2i\sqrt{x}\right)+C_2{Y}_{\frac{1}{2}}\left(2i\sqrt{x}\right))$$