2.1.10 Problem 10

Internal problem ID [9182]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 10
Date solved : Wednesday, March 05, 2025 at 07:37:30 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}(x^2-6x+10)y^{\prime \prime }-4(x-3)y^{\prime }+6y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.335 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (x^2-6x+10)y^{\prime \prime }+(-4x+12)y^{\prime }+6y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =x^2-6x+10\\ \text{(3)}& B& =-4x+12C& =6\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-8}{{(x^2-6x+10)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-8\\ t& ={(x^2-6x+10)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(-\frac{8}{{(x^2-6x+10)}^2})z(x)\end{array}$$

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 2.10: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-0\\ & =4\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t={(x^2-6x+10)}^2$. There is a pole at $x=3+i$ of order $2$. There is a pole at $x=3-i$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $4$ 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 $4$ 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

$$r=\frac{2}{{(x-3-i)}^2}+\frac{2}{{(x-3+i)}^2}+\frac{2i}{x-3-i}-\frac{2i}{x-3+i}$$

For the pole at $x=3+i$ let $b$ be the coefficient of $\frac{1}{{(x-3-i)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=2$. Hence

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

For the pole at $x=3-i$ let $b$ be the coefficient of $\frac{1}{{(x-3+i)}^2}$ 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 $4>2$ then

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

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

$$r=-\frac{8}{{(x^2-6x+10)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$3+i$	$2$	$0$	$2$	$-1$
$3-i$	$2$	$0$	$2$	$-1$

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

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

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})\\ & =1-\left(1\right)\\ & =0\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{x-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{x-c_2})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{x-3-i}+\frac{2}{x-3+i}+(-)\left(0\right)\\ & =-\frac{1}{x-3-i}+\frac{2}{x-3+i}\\ & =\frac{x-3-3i}{x^2-6x+10}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{x-3-i}+\frac{2}{x-3+i})\left(0\right)+((\frac{1}{{(x-3-i)}^2}-\frac{2}{{(x-3+i)}^2})+{(-\frac{1}{x-3-i}+\frac{2}{x-3+i})}^2-(-\frac{8}{{(x^2-6x+10)}^2}))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{1}{x-3-i}+\frac{2}{x-3+i})dx}\\ & =\frac{{(x^2-6x+10)}^2}{{(ix-3i+1)}^3}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-4x+12}{x^2-6x+10}dx}\\ & =z_1{e}^{\ln (x^2-6x+10)}\\ & =z_1(x^2-6x+10)\end{array}$$

Which simplifies to

$$y_1=\frac{{(x^2-6x+10)}^3}{{(ix-3i+1)}^3}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-4x+12}{x^2-6x+10}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{2\ln (x^2-6x+10)}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{x^2-6x+\frac{26}{3}}{{(x-3+i)}^3}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{{(x^2-6x+10)}^3}{{(ix-3i+1)}^3}\right)+c_2\left(\frac{{(x^2-6x+10)}^3}{{(ix-3i+1)}^3}\left(\frac{x^2-6x+\frac{26}{3}}{{(x-3+i)}^3}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.003 (sec). Leaf size: 31

ode:=(x^2-6*x+10)*diff(diff(y(x),x),x)-4*(x-3)*diff(y(x),x)+6*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1{x}^3+c_2{x}^2+6(-5c_1-c_2)x+60c_1+\frac{26c_2}{3}$$

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`
 

✓ Mathematica. Time used: 0.314 (sec). Leaf size: 84

ode=(x^2-6*x+10)*D[y[x],{x,2}]-4*(x-3)*D[y[x],x]+6*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to (x^2-6x+10)\exp ({\int }_1^x\frac{K[1]-(3-3i)}{(K[1]-6)K[1]+10}dK[1])(c_2{\int }_1^x\exp (-2{\int }_1^{K[2]}\frac{K[1]-(3-3i)}{(K[1]-6)K[1]+10}dK[1])dK[2]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((12 - 4*x)*Derivative(y(x), x) + (x**2 - 6*x + 10)*Derivative(y(x), (x, 2)) + 6*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False