2.1.771 Problem 793

Internal problem ID [9941]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 793
Date solved : Friday, April 25, 2025 at 06:23:33 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.098 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

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

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 2.771: 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

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

Which simplifies to

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

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.003 (sec). Leaf size: 14

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

$$y={\mathrm{e}}^x(c_1+\frac{c_2}{x})$$

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) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(2-2x)\left(\frac{d}{dx}y\left(x\right)\right)+(x-2)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{(x-2)y\left(x\right)}{x}+\frac{2(-1+x)\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}{dx}\frac{d}{dx}y\left(x\right)-\frac{2(-1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x}+\frac{(x-2)y\left(x\right)}{x}=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(-1+x)}{x},P_3\left(x\right)=\frac{x-2}{x}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=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}=0\\ & \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\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(2-2x)\left(\frac{d}{dx}y\left(x\right)\right)+(x-2)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}^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=0..1\\ & & 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\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\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-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(1+r)x^{-1+r}+(a_1(1+r)(2+r)-2a_0(1+r))x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+1}(k+1+r)(k+2+r)-2a_k(k+1+r)+a_{k-1})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-1,0\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(1+r)(2+r)-2a_0(1+r)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+1+r)(k+2+r)-2a_{k}k-2a_{k}r-2a_k+a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & a_{k+2}(k+2+r)(k+3+r)-2a_{k+1}(k+1)-2r{a}_{k+1}-2a_{k+1}+a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{2k{a}_{k+1}+2r{a}_{k+1}-a_k+4a_{k+1}}{(k+2+r)(k+3+r)}\\ \bullet & & \text{Recursion relation for}r=-1\\ & & a_{k+2}=\frac{2k{a}_{k+1}-a_k+2a_{k+1}}{(k+1)(k+2)}\\ \bullet & & \text{Solution for}r=-1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-1},a_{k+2}=\frac{2k{a}_{k+1}-a_k+2a_{k+1}}{(k+1)(k+2)},0=0]\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=\frac{2k{a}_{k+1}-a_k+4a_{k+1}}{(k+2)(k+3)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=\frac{2k{a}_{k+1}-a_k+4a_{k+1}}{(k+2)(k+3)},2a_1-2a_0=0]\\ \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),a_{k+2}=\frac{2k{a}_{k+1}-a_k+2a_{k+1}}{(k+1)(k+2)},0=0,b_{k+2}=\frac{2k{b}_{k+1}-b_k+4b_{k+1}}{(k+2)(k+3)},2b_1-2b_0=0]\end{array}$$

✓ Mathematica. Time used: 0.027 (sec). Leaf size: 19

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

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

✗ Sympy

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

False