2.1.119 Problem 121

Internal problem ID [9291]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 121
Date solved : Sunday, March 30, 2025 at 02:31:37 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.192 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& 3x^2{(1+x)}^2{y}^{\prime \prime }+(11x^3+10x^2-x)y^{\prime }+(5x^2+1)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =3x^2{(1+x)}^2\\ \text{(3)}& B& =11x^3+10x^2-xC& =5x^2+1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-5}{36x^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-5\\ t& =36x^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(-\frac{5}{36x^2})z(x)\end{array}$$

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 2.119: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-0\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=36x^2$. There is a pole at $x=0$ 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

$$r=-\frac{5}{36x^2}$$

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 $b=-\frac{5}{36}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{6}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{6}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =-\frac{5}{36x^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{5}{36}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{6}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{6}\end{array}$$

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

$$r=-\frac{5}{36x^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{5}{6}$	$\frac{1}{6}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{5}{6}$	$\frac{1}{6}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{1}{6}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =\frac{1}{6}-\left(\frac{1}{6}\right)\\ & =0\end{array}$$

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

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

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

$$\begin{array}{rl}\left(0\right)+2\left(\frac{1}{6x}\right)\left(0\right)+((-\frac{1}{6x^2})+{\left(\frac{1}{6x}\right)}^2-(-\frac{5}{36x^2}))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int \frac{1}{6x}dx}\\ & =x^{1/6}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{11x^3+10x^2-x}{3x^2{(1+x)}^2}dx}\\ & =z_1{e}^{\frac{\ln \left(x\right)}{6}-2\ln (1+x)}\\ & =z_1\left(\frac{x^{1/6}}{{(1+x)}^2}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{x^{1/3}}{{(1+x)}^2}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{11x^3+10x^2-x}{3x^2{(1+x)}^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\frac{\ln \left(x\right)}{3}-4\ln (1+x)}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{3x^{1/3}{\mathrm{e}}^{\frac{\ln \left(x\right)}{3}-4\ln (1+x)}{(1+x)}^4}{2}\right)\end{array}$$

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.011 (sec). Leaf size: 19

ode:=3*x^2*(x+1)^2*diff(diff(y(x),x),x)-x*(-11*x^2-10*x+1)*diff(y(x),x)+(5*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_2{x}^{1/3}+c_{1}x}{{(1+x)}^2}$$

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) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 3x^2{(1+x)}^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-x(-11x^2-10x+1)\left(\frac{d}{dx}y\left(x\right)\right)+(5x^2+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{(5x^2+1)y\left(x\right)}{3x^2{(1+x)}^2}-\frac{\left(\frac{d}{dx}y\left(x\right)\right)(11x-1)}{3(1+x)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{\left(\frac{d}{dx}y\left(x\right)\right)(11x-1)}{3(1+x)x}+\frac{(5x^2+1)y\left(x\right)}{3x^2{(1+x)}^2}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{11x-1}{3(1+x)x},P_3\left(x\right)=\frac{5x^2+1}{3x^2{(1+x)}^2}]\\ & \circ & (1+x)\cdot P_2\left(x\right)\text{is analytic at}x=-1\\ & & ((1+x)\cdot P_2\left(x\right))|{}_{x=-1}=4\\ & \circ & {(1+x)}^2\cdot P_3\left(x\right)\text{is analytic at}x=-1\\ & & ({(1+x)}^2\cdot P_3\left(x\right))|{}_{x=-1}=2\\ & \circ & x=-1\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=-1\\ \bullet & & \text{Multiply by denominators}\\ & & 3x^2{(1+x)}^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x(1+x)(11x-1)\left(\frac{d}{dx}y\left(x\right)\right)+(5x^2+1)y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-1\text{so that the regular singular point is at}u=0\\ & & (3u^4-6u^3+3u^2)\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(11u^3-23u^2+12u)\left(\frac{d}{du}y\left(u\right)\right)+(5u^2-10u+6)y\left(u\right)=0\\ \bullet & & \text{Assume series solution for}y\left(u\right)\\ & & y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{u}^m\cdot y\left(u\right)\text{to series expansion for}m=0..2\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{u}^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=1..3\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)u^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=2..4\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)u^{k+r-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & 3a_0(2+r)(1+r)u^r+(3a_1(3+r)(2+r)-a_0(2+r)(5+6r))u^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(3a_k(k+r+2)(k+r+1)-a_{k-1}(k+r+1)(6k-1+6r)+a_{k-2}(3k-1+3r)(k+r-1))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & 3(2+r)(1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-2,-1\}\\ \bullet & & \text{Each term must be 0}\\ & & 3a_1(3+r)(2+r)-a_0(2+r)(5+6r)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=\frac{a_0(5+6r)}{3(3+r)}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 3(a_k+a_{k-2}-2a_{k-1})k^2+(6(a_k+a_{k-2}-2a_{k-1})r+9a_k-4a_{k-2}-5a_{k-1})k+3(a_k+a_{k-2}-2a_{k-1})r^2+(9a_k-4a_{k-2}-5a_{k-1})r+6a_k+a_{k-2}+a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & 3(a_{k+2}+a_k-2a_{k+1}){(k+2)}^2+(6(a_{k+2}+a_k-2a_{k+1})r+9a_{k+2}-4a_k-5a_{k+1})(k+2)+3(a_{k+2}+a_k-2a_{k+1})r^2+(9a_{k+2}-4a_k-5a_{k+1})r+6a_{k+2}+a_k+a_{k+1}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}+6kr{a}_k-12kr{a}_{k+1}+3r^2{a}_k-6r^2{a}_{k+1}+8k{a}_k-29k{a}_{k+1}+8r{a}_k-29r{a}_{k+1}+5a_k-33a_{k+1}}{3(k^2+2kr+r^2+7k+7r+12)}\\ \bullet & & \text{Recursion relation for}r=-2\\ & & a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}-4k{a}_k-5k{a}_{k+1}+a_k+a_{k+1}}{3(k^2+3k+2)}\\ \bullet & & \text{Solution for}r=-2\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k-2},a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}-4k{a}_k-5k{a}_{k+1}+a_k+a_{k+1}}{3(k^2+3k+2)},a_1=-\frac{7a_0}{3}]\\ \bullet & & \text{Revert the change of variables}u=1+x\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(1+x)}^{k-2},a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}-4k{a}_k-5k{a}_{k+1}+a_k+a_{k+1}}{3(k^2+3k+2)},a_1=-\frac{7a_0}{3}]\\ \bullet & & \text{Recursion relation for}r=-1\\ & & a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}+2k{a}_k-17k{a}_{k+1}-10a_{k+1}}{3(k^2+5k+6)}\\ \bullet & & \text{Solution for}r=-1\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k-1},a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}+2k{a}_k-17k{a}_{k+1}-10a_{k+1}}{3(k^2+5k+6)},a_1=-\frac{a_0}{6}]\\ \bullet & & \text{Revert the change of variables}u=1+x\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(1+x)}^{k-1},a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}+2k{a}_k-17k{a}_{k+1}-10a_{k+1}}{3(k^2+5k+6)},a_1=-\frac{a_0}{6}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(1+x)}^{k-2})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(1+x)}^{k-1}),a_{k+2}=-\frac{3k^2{a}_k-6k^2{a}_{k+1}-4k{a}_k-5k{a}_{k+1}+a_k+a_{k+1}}{3(k^2+3k+2)},a_1=-\frac{7a_0}{3},b_{k+2}=-\frac{3k^2{b}_k-6k^2{b}_{k+1}+2k{b}_k-17k{b}_{k+1}-10b_{k+1}}{3(k^2+5k+6)},b_1=-\frac{b_0}{6}]\end{array}$$

✓ Mathematica. Time used: 0.242 (sec). Leaf size: 58

ode=3*x^2*(1+x)^2*D[y[x],{x,2}]-x*(1-10*x-11*x^2)*D[y[x],x]+(1+5*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{1}{2}\sqrt[6]{x}(3c_2{x}^{2/3}+2c_1)\exp (-\frac{1}{2}{\int }_1^x(\frac{4}{K[1]+1}-\frac{1}{3K[1]})dK[1])$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(3*x**2*(x + 1)**2*Derivative(y(x), (x, 2)) - x*(-11*x**2 - 10*x + 1)*Derivative(y(x), x) + (5*x**2 + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False