2.1.571 Problem 587

Internal problem ID [9743]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 587
Date solved : Sunday, March 30, 2025 at 02:45:39 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.248 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (-x^3+x^2)y^{\prime \prime }+(5x^2-3x)y^{\prime }+(4-5x)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =-x^3+x^2\\ \text{(3)}& B& =5x^2-3xC& =4-5x\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{15x^2-6x-1}{4{(x^2-x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =15x^2-6x-1\\ t& =4{(x^2-x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{15x^2-6x-1}{4{(x^2-x)}^2}\right)z(x)\end{array}$$

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 2.571: 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 $t=4{(x^2-x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=1$ of order $2$. 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. Since there is a pole of order $2$ then necessary conditions for case two are met. Since pole order is not larger than $2$ and the order at $\mathrm{\infty }$ is $2$ then the necessary conditions for case three are met. Therefore

$$\begin{array}{rl}L& =[1,2,4,6,12]\end{array}$$

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

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

$$r=\frac{2}{{(-1+x)}^2}-\frac{2}{x}+\frac{2}{-1+x}-\frac{1}{4x^2}$$

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 $b=-\frac{1}{4}$. Hence

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

For the pole at $x=1$ let $b$ be the coefficient of $\frac{1}{{(-1+x)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=2$. Hence

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

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{15x^2-6x-1}{4{(x^2-x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=\frac{15}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{3}{2}\end{array}$$

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

$$r=\frac{15x^2-6x-1}{4{(x^2-x)}^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{5}{2}$	$-\frac{3}{2}$

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 ${\alpha }_{\mathrm{\infty }}^{+}=\frac{5}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})\\ & =\frac{5}{2}-\left(\frac{5}{2}\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{x-c_2})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{1}{2x}+\frac{2}{-1+x}+\left(0\right)\\ & =\frac{1}{2x}+\frac{2}{-1+x}\\ & =\frac{-1+5x}{2x(-1+x)}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(\frac{1}{2x}+\frac{2}{-1+x})\left(0\right)+((-\frac{1}{2x^2}-\frac{2}{{(-1+x)}^2})+{(\frac{1}{2x}+\frac{2}{-1+x})}^2-\left(\frac{15x^2-6x-1}{4{(x^2-x)}^2}\right))& =0\\ 0=0\end{array}$$

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

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

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{5x^2-3x}{-x^3+x^2}dx}\\ & =z_1{e}^{\frac{3\ln \left(x\right)}{2}+\ln (-1+x)}\\ & =z_1\left(x^{3/2}(-1+x)\right)\end{array}$$

Which simplifies to

$$y_1=x^2{(-1+x)}^3$$

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.005 (sec). Leaf size: 47

ode:=x^2*(1-x)*diff(diff(y(x),x),x)-x*(3-5*x)*diff(y(x),x)+(4-5*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=x^2(c_1{(-1+x)}^3+c_2(-{(-1+x)}^3\ln (-1+x)+{(-1+x)}^3\ln \left(x\right)-x^2+\frac{5x}{2}-\frac{11}{6}))$$

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: 0.244 (sec). Leaf size: 104

ode=x^2*(1-x)*D[y[x],{x,2}]-x*(3-5*x)*D[y[x],x]+(4-5*x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x(\frac{1}{2K[1]}+\frac{2}{K[1]-1})dK[1]-\frac{1}{2}{\int }_1^x(-\frac{3}{K[2]}-\frac{2}{K[2]-1})dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}\frac{1-5K[1]}{2K[1]-2K[1{]}^2}dK[1])dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(1 - x)*Derivative(y(x), (x, 2)) - x*(3 - 5*x)*Derivative(y(x), x) + (4 - 5*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False