2.1.565 Problem 581

Internal problem ID [9737]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 581
Date solved : Sunday, March 30, 2025 at 02:45:31 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.464 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (4x^4+16x^2)y^{\prime \prime }+(9x^3+24x)y^{\prime }+(-9x^2+1)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =4x^4+16x^2\\ \text{(3)}& B& =9x^3+24xC& =-9x^2+1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{153x^4+704x^2-256}{64{(x^3+4x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =153x^4+704x^2-256\\ t& =64{(x^3+4x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{153x^4+704x^2-256}{64{(x^3+4x)}^2}\right)z(x)\end{array}$$

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 2.565: 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 $t=64{(x^3+4x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=2i$ of order $2$. There is a pole at $x=-2i$ 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

$$r=-\frac{1}{4x^2}-\frac{39}{256{(x-2i)}^2}-\frac{39}{256{(x+2i)}^2}-\frac{377i}{512(x-2i)}+\frac{377i}{512(x+2i)}$$

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 $x=2i$ let $b$ be the coefficient of $\frac{1}{{(x-2i)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{39}{256}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{13}{16}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{3}{16}\end{array}$$

For the pole at $x=-2i$ let $b$ be the coefficient of $\frac{1}{{(x+2i)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{39}{256}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{13}{16}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{3}{16}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{153x^4+704x^2-256}{64{(x^3+4x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=\frac{153}{64}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{17}{8}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{9}{8}\end{array}$$

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

$$r=\frac{153x^4+704x^2-256}{64{(x^3+4x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{1}{2}$	$\frac{1}{2}$
$2i$	$2$	$0$	$\frac{13}{16}$	$\frac{3}{16}$
$-2i$	$2$	$0$	$\frac{13}{16}$	$\frac{3}{16}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{17}{8}$	$-\frac{9}{8}$

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 ${\alpha }_{\mathrm{\infty }}^{+}=\frac{17}{8}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+}+{\alpha }_{c_3}^{+})\\ & =\frac{17}{8}-\left(\frac{17}{8}\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{x-c_2})+((+)[\sqrt{r}{]}_{c_3}+\frac{{\alpha }_{c_3}^{+}}{x-c_3})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{1}{2x}+\frac{13}{16(x-2i)}+\frac{13}{16(x+2i)}+\left(0\right)\\ & =\frac{1}{2x}+\frac{13}{16(x-2i)}+\frac{13}{16(x+2i)}\\ & =\frac{1}{2x}+\frac{13x}{8x^2+32}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(\frac{1}{2x}+\frac{13}{16(x-2i)}+\frac{13}{16(x+2i)})\left(0\right)+((-\frac{1}{2x^2}-\frac{13}{16{(x-2i)}^2}-\frac{13}{16{(x+2i)}^2})+{(\frac{1}{2x}+\frac{13}{16(x-2i)}+\frac{13}{16(x+2i)})}^2-\left(\frac{153x^4+704x^2-256}{64{(x^3+4x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (\frac{1}{2x}+\frac{13}{16(x-2i)}+\frac{13}{16(x+2i)})dx}\\ & ={(x^2+4)}^{13/16}\sqrt{x}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{9x^3+24x}{4x^4+16x^2}dx}\\ & =z_1{e}^{-\frac{3\ln (x^2+4)}{16}-\frac{3\ln \left(x\right)}{4}}\\ & =z_1\left(\frac{1}{{(x^2+4)}^{3/16}{x}^{3/4}}\right)\end{array}$$

Which simplifies to

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

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{9x^3+24x}{4x^4+16x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{3\ln (x^2+4)}{8}-\frac{3\ln \left(x\right)}{2}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{3\ln (x^2+4)}{8}-\frac{3\ln \left(x\right)}{2}}\sqrt{x}}{{(x^2+4)}^{5/4}}dx)\end{array}$$

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.116 (sec). Leaf size: 66

ode:=4*x^2*(x^2+4)*diff(diff(y(x),x),x)+3*x*(3*x^2+8)*diff(y(x),x)+(-9*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{(x^2\mathrm{hypergeom}([1,1,\frac{13}{8}],[2,2],-\frac{x^2}{4})-\frac{32\gamma }{5}+\frac{64\ln \left(2\right)}{5}-\frac{64\ln \left(x\right)}{5}-\frac{32\mathrm{\Psi }\left(\frac{5}{8}\right)}{5})c_2{(x^2+4)}^{5/8}{2}^{3/4}+c_1{(x^2+4)}^{5/8}-\frac{1024c_2}{25}}{x^{1/4}}$$

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 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Kummer 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         <- heuristic approach successful 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form is not straightforward to achieve - returning special \ 
function solution free of uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 4x^2(x^2+4)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x(3x^2+8)\left(\frac{d}{dx}y\left(x\right)\right)+(-9x^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{(9x^2-1)y\left(x\right)}{4x^2(x^2+4)}-\frac{3(3x^2+8)\left(\frac{d}{dx}y\left(x\right)\right)}{4x(x^2+4)}\\ \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{3(3x^2+8)\left(\frac{d}{dx}y\left(x\right)\right)}{4x(x^2+4)}-\frac{(9x^2-1)y\left(x\right)}{4x^2(x^2+4)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{3(3x^2+8)}{4x(x^2+4)},P_3\left(x\right)=-\frac{9x^2-1}{4x^2(x^2+4)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=\frac{3}{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}=\frac{1}{16}\\ & \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}\\ & & 4x^2(x^2+4)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x(3x^2+8)\left(\frac{d}{dx}y\left(x\right)\right)+(-9x^2+1)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..2\\ & & 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=1..3\\ & & 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=2..4\\ & & 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{(1+4r)}^2{x}^r+a_1{(5+4r)}^2{x}^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k{(4k+4r+1)}^2+a_{k-2}(4k+4r+1)(k-3+r))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & {(1+4r)}^2=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r=-\frac{1}{4}\\ \bullet & & \text{Each term must be 0}\\ & & a_1{(5+4r)}^2=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 16(\frac{a_{k-2}(k-3+r)}{4}+a_k(k+r+\frac{1}{4}))(k+r+\frac{1}{4})=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & 16(\frac{a_k(k+r-1)}{4}+a_{k+2}(k+\frac{9}{4}+r))(k+\frac{9}{4}+r)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{a_k(k+r-1)}{4k+4r+9}\\ \bullet & & \text{Recursion relation for}r=-\frac{1}{4}\\ & & a_{k+2}=-\frac{a_k(k-\frac{5}{4})}{4k+8}\\ \bullet & & \text{Solution for}r=-\frac{1}{4}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-\frac{1}{4}},a_{k+2}=-\frac{a_k(k-\frac{5}{4})}{4k+8},a_1=0]\end{array}$$

✓ Mathematica. Time used: 0.264 (sec). Leaf size: 118

ode=4*x^2*(4+x^2)*D[y[x],{x,2}]+3*x*(8+3*x^2)*D[y[x],x]+(1-9*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x\frac{17K[1{]}^2+16}{8K[1{]}^3+32K[1]}dK[1]-\frac{1}{2}{\int }_1^x\frac{9K[2{]}^2+24}{4K[2{]}^3+16K[2]}dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}\frac{17K[1{]}^2+16}{8K[1{]}^3+32K[1]}dK[1])dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*x**2*(x**2 + 4)*Derivative(y(x), (x, 2)) + 3*x*(3*x**2 + 8)*Derivative(y(x), x) + (1 - 9*x**2)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False