2.1.587 Problem 603

Internal problem ID [9757]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 603
Date solved : Friday, April 25, 2025 at 06:18:59 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{x}^2(-2x^2+1)y^{\prime \prime }+x(-13x^2+7)y^{\prime }-14x^{2}y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.371 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (-2x^4+x^2)y^{\prime \prime }+(-13x^3+7x)y^{\prime }-14x^{2}y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =-2x^4+x^2\\ \text{(3)}& B& =-13x^3+7xC& =-14x^2\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{5x^4-68x^2+35}{4{(2x^3-x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =5x^4-68x^2+35\\ t& =4{(2x^3-x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{5x^4-68x^2+35}{4{(2x^3-x)}^2}\right)z(x)\end{array}$$

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 2.587: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =6-4\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4{(2x^3-x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=\frac{\sqrt{2}}{2}$ of order $2$. There is a pole at $x=-\frac{\sqrt{2}}{2}$ 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

$$r=\frac{9}{64{(x-\frac{\sqrt{2}}{2})}^2}+\frac{9}{64{(x+\frac{\sqrt{2}}{2})}^2}-\frac{279\sqrt{2}}{64(x-\frac{\sqrt{2}}{2})}+\frac{279\sqrt{2}}{64(x+\frac{\sqrt{2}}{2})}+\frac{35}{4x^2}$$

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 $b=\frac{35}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{5}{2}\end{array}$$

For the pole at $x=\frac{\sqrt{2}}{2}$ let $b$ be the coefficient of $\frac{1}{{(x-\frac{\sqrt{2}}{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{9}{64}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{9}{8}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{8}\end{array}$$

For the pole at $x=-\frac{\sqrt{2}}{2}$ let $b$ be the coefficient of $\frac{1}{{(x+\frac{\sqrt{2}}{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{9}{64}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{9}{8}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{8}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{5x^4-68x^2+35}{4{(2x^3-x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=\frac{5}{16}$. Hence

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

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

$$r=\frac{5x^4-68x^2+35}{4{(2x^3-x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{7}{2}$	$-\frac{5}{2}$
$\frac{\sqrt{2}}{2}$	$2$	$0$	$\frac{9}{8}$	$-\frac{1}{8}$
$-\frac{\sqrt{2}}{2}$	$2$	$0$	$\frac{9}{8}$	$-\frac{1}{8}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{5}{4}$	$-\frac{1}{4}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=-\frac{1}{4}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+}+{\alpha }_{c_3}^{+})\\ & =-\frac{1}{4}-(-\frac{1}{4})\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{x-c_2})+((+)[\sqrt{r}{]}_{c_3}+\frac{{\alpha }_{c_3}^{+}}{x-c_3})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{5}{2x}+\frac{9}{8(x-\frac{\sqrt{2}}{2})}+\frac{9}{8(x+\frac{\sqrt{2}}{2})}+(-)\left(0\right)\\ & =-\frac{5}{2x}+\frac{9}{8(x-\frac{\sqrt{2}}{2})}+\frac{9}{8(x+\frac{\sqrt{2}}{2})}\\ & =\frac{-x^2+5}{4x^3-2x}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{5}{2x}+\frac{9}{8(x-\frac{\sqrt{2}}{2})}+\frac{9}{8(x+\frac{\sqrt{2}}{2})})\left(0\right)+((\frac{5}{2x^2}-\frac{9}{8{(x-\frac{\sqrt{2}}{2})}^2}-\frac{9}{8{(x+\frac{\sqrt{2}}{2})}^2})+{(-\frac{5}{2x}+\frac{9}{8(x-\frac{\sqrt{2}}{2})}+\frac{9}{8(x+\frac{\sqrt{2}}{2})})}^2-\left(\frac{5x^4-68x^2+35}{4{(2x^3-x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{5}{2x}+\frac{9}{8(x-\frac{\sqrt{2}}{2})}+\frac{9}{8(x+\frac{\sqrt{2}}{2})})dx}\\ & =\frac{{(2x-\sqrt{2})}^{9/8}{(2x+\sqrt{2})}^{9/8}}{x^{5/2}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-13x^3+7x}{-2x^4+x^2}dx}\\ & =z_1{e}^{\frac{\ln (2x^2-1)}{8}-\frac{7\ln \left(x\right)}{2}}\\ & =z_1\left(\frac{{(2x^2-1)}^{1/8}}{x^{7/2}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{2{(2x^2-1)}^{5/4}{2}^{1/8}}{x^6}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-13x^3+7x}{-2x^4+x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\frac{\ln (2x^2-1)}{4}-7\ln \left(x\right)}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{(5x^4-20x^2+8)x^7{\mathrm{e}}^{\frac{\ln (2x^2-1)}{4}-7\ln \left(x\right)}{2}^{3/4}}{120{(2x^2-1)}^{3/2}}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{2{(2x^2-1)}^{5/4}{2}^{1/8}}{x^6}\right)+c_2\left(\frac{2{(2x^2-1)}^{5/4}{2}^{1/8}}{x^6}\left(\frac{(5x^4-20x^2+8)x^7{\mathrm{e}}^{\frac{\ln (2x^2-1)}{4}-7\ln \left(x\right)}{2}^{3/4}}{120{(2x^2-1)}^{3/2}}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.016 (sec). Leaf size: 35

ode:=x^2*(-2*x^2+1)*diff(diff(y(x),x),x)+x*(-13*x^2+7)*diff(y(x),x)-14*x^2*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_1{(2x^2-1)}^{5/4}+5c_2{x}^4-20c_2{x}^2+8c_2}{x^6}$$

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

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2(-2x^2+1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x(-13x^2+7)\left(\frac{d}{dx}y\left(x\right)\right)-14x^{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{14y\left(x\right)}{2x^2-1}-\frac{(13x^2-7)\left(\frac{d}{dx}y\left(x\right)\right)}{x(2x^2-1)}\\ \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{(13x^2-7)\left(\frac{d}{dx}y\left(x\right)\right)}{x(2x^2-1)}+\frac{14y\left(x\right)}{2x^2-1}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{13x^2-7}{x(2x^2-1)},P_3\left(x\right)=\frac{14}{2x^2-1}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=7\\ & \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\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & x(2x^2-1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(13x^2-7)\left(\frac{d}{dx}y\left(x\right)\right)+14y\left(x\right)x=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\cdot y\left(x\right)\text{to series expansion}\\ & & x\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r+1}\\ & \circ & \text{Shift index using}k->k-1\\ & & x\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=1}}{a}_{k-1}{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..2\\ & & 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}^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=1..3\\ & & x^m\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-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & x^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & -a_{0}r(6+r)x^{-1+r}-a_1(1+r)(7+r)x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(-a_{k+1}(k+r+1)(k+7+r)+a_{k-1}(2k+5+2r)(k+r+1))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -r(6+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-6,0\}\\ \bullet & & \text{Each term must be 0}\\ & & -a_1(1+r)(7+r)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 2((k+r+\frac{5}{2})a_{k-1}-\frac{a_{k+1}(k+7+r)}{2})(k+r+1)=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & 2((k+\frac{7}{2}+r)a_k-\frac{a_{k+2}(k+8+r)}{2})(k+r+2)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{(2k+2r+7)a_k}{k+8+r}\\ \bullet & & \text{Recursion relation for}r=-6\\ & & a_{k+2}=\frac{(2k-5)a_k}{k+2}\\ \bullet & & \text{Solution for}r=-6\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-6},a_{k+2}=\frac{(2k-5)a_k}{k+2},5a_1=0]\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=\frac{(2k+7)a_k}{k+8}\\ \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+7)a_k}{k+8},-7a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-6})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^k),a_{k+2}=\frac{(2k-5)a_k}{k+2},5a_1=0,b_{k+2}=\frac{(2k+7)b_k}{k+8},-7b_1=0]\end{array}$$

✓ Mathematica. Time used: 0.214 (sec). Leaf size: 116

ode=x^2*(1-2*x^2)*D[y[x],{x,2}]+x*(7-13*x^2)*D[y[x],x]-14*x^2*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x\frac{5-K[1{]}^2}{4K[1{]}^3-2K[1]}dK[1]-\frac{1}{2}{\int }_1^x\frac{7-13K[2{]}^2}{K[2]-2K[2{]}^3}dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}\frac{5-K[1{]}^2}{4K[1{]}^3-2K[1]}dK[1])dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(1 - 2*x**2)*Derivative(y(x), (x, 2)) - 14*x**2*y(x) + x*(7 - 13*x**2)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False