2.1.563 Problem 579

Internal problem ID [9735]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 579
Date solved : Sunday, March 30, 2025 at 02:45:28 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.372 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =2x^4+4x^2\\ \text{(3)}& B& =7x^{3}C& =3x^2+1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-3x^4-16}{16{(x^3+2x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-3x^4-16\\ t& =16{(x^3+2x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-3x^4-16}{16{(x^3+2x)}^2}\right)z(x)\end{array}$$

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 2.563: 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 $t=16{(x^3+2x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=i\sqrt{2}$ of order $2$. There is a pole at $x=-i\sqrt{2}$ 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

$$r=-\frac{1}{4x^2}-\frac{7}{64{(x-i\sqrt{2})}^2}-\frac{7}{64{(x+i\sqrt{2})}^2}-\frac{9i\sqrt{2}}{128(x-i\sqrt{2})}+\frac{9i\sqrt{2}}{128(x+i\sqrt{2})}$$

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 $x=i\sqrt{2}$ let $b$ be the coefficient of $\frac{1}{{(x-i\sqrt{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{7}{64}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{8}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{8}\end{array}$$

For the pole at $x=-i\sqrt{2}$ let $b$ be the coefficient of $\frac{1}{{(x+i\sqrt{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{7}{64}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{8}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{8}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-3x^4-16}{16{(x^3+2x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{3}{16}$. Hence

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

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

$$r=\frac{-3x^4-16}{16{(x^3+2x)}^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$

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 ${\alpha }_{\mathrm{\infty }}^{+}=\frac{3}{4}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-}+{\alpha }_{c_3}^{-})\\ & =\frac{3}{4}-\left(\frac{3}{4}\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+((-)[\sqrt{r}{]}_{c_3}+\frac{{\alpha }_{c_3}^{-}}{x-c_3})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{1}{2x}+\frac{1}{8x-8i\sqrt{2}}+\frac{1}{8x+8i\sqrt{2}}+\left(0\right)\\ & =\frac{1}{2x}+\frac{1}{8x-8i\sqrt{2}}+\frac{1}{8x+8i\sqrt{2}}\\ & =\frac{1}{2x}+\frac{x}{4x^2+8}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(\frac{1}{2x}+\frac{1}{8x-8i\sqrt{2}}+\frac{1}{8x+8i\sqrt{2}})\left(0\right)+((-\frac{1}{2x^2}-\frac{1}{8{(x-i\sqrt{2})}^2}-\frac{1}{8{(x+i\sqrt{2})}^2})+{(\frac{1}{2x}+\frac{1}{8x-8i\sqrt{2}}+\frac{1}{8x+8i\sqrt{2}})}^2-\left(\frac{-3x^4-16}{16{(x^3+2x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (\frac{1}{2x}+\frac{1}{8x-8i\sqrt{2}}+\frac{1}{8x+8i\sqrt{2}})dx}\\ & =\sqrt{x}{(x^2+2)}^{1/8}\end{array}$$

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

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

Which simplifies to

$$y_1=\frac{\sqrt{x}}{{(x^2+2)}^{3/4}}$$

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.176 (sec). Leaf size: 81

ode:=2*x^2*(x^2+2)*diff(diff(y(x),x),x)+7*x^3*diff(y(x),x)+(3*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{\sqrt{x}(2^{3/4}{c}_1+\ln (-\sqrt{2}{(2x^2+4)}^{1/4}+2)c_2-\ln (\sqrt{2}{(2x^2+4)}^{1/4}+2)c_2+2\arctan \left(\frac{\sqrt{2}{(2x^2+4)}^{1/4}}{2}\right)c_2)2^{1/4}}{2{(x^2+2)}^{3/4}}$$

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 
      -> Whittaker 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         <- heuristic approach successful 
         -> solution has integrals; searching for one without integrals... 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
            <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
         <- hypergeometric solution without integrals succesful 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved -\ 
 returning with no uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 2x^2(x^2+2)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+7x^3\left(\frac{d}{dx}y\left(x\right)\right)+(3x^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{(3x^2+1)y\left(x\right)}{2x^2(x^2+2)}-\frac{7x\left(\frac{d}{dx}y\left(x\right)\right)}{2(x^2+2)}\\ \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{7x\left(\frac{d}{dx}y\left(x\right)\right)}{2(x^2+2)}+\frac{(3x^2+1)y\left(x\right)}{2x^2(x^2+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{7x}{2(x^2+2)},P_3\left(x\right)=\frac{3x^2+1}{2x^2(x^2+2)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=0\\ & \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}{4}\\ & \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}\\ & & 2x^2(x^2+2)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+7x^3\left(\frac{d}{dx}y\left(x\right)\right)+(3x^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}^3\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x^3\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r+2}\\ & \circ & \text{Shift index using}k->k-2\\ & & x^3\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=2}}{a}_{k-2}(k-2+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+2r)}^2{x}^r+a_1{(1+2r)}^2{x}^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k{(2k+2r-1)}^2+a_{k-2}(2k+2r-1)(k+r-1))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & {(-1+2r)}^2=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r=\frac{1}{2}\\ \bullet & & \text{Each term must be 0}\\ & & a_1{(1+2r)}^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}\\ & & 4(k+r-\frac{1}{2})(\frac{a_{k-2}(k+r-1)}{2}+a_k(k+r-\frac{1}{2}))=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & 4(k+\frac{3}{2}+r)(\frac{a_k(k+r+1)}{2}+a_{k+2}(k+\frac{3}{2}+r))=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{a_k(k+r+1)}{2k+2r+3}\\ \bullet & & \text{Recursion relation for}r=\frac{1}{2}\\ & & a_{k+2}=-\frac{a_k(k+\frac{3}{2})}{2k+4}\\ \bullet & & \text{Solution for}r=\frac{1}{2}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{1}{2}},a_{k+2}=-\frac{a_k(k+\frac{3}{2})}{2k+4},a_1=0]\end{array}$$

✓ Mathematica. Time used: 0.429 (sec). Leaf size: 93

ode=2*x^2*(2+x^2)*D[y[x],{x,2}]+7*x^3*D[y[x],x]+(1+3*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{\exp ({\int }_1^x\frac{3K[1{]}^2+4}{4K[1{]}^3+8K[1]}dK[1])(c_2{\int }_1^x\exp (-2{\int }_1^{K[2]}\frac{3K[1{]}^2+4}{4K[1{]}^3+8K[1]}dK[1])dK[2]+c_1)}{{(x^2+2)}^{7/8}}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(7*x**3*Derivative(y(x), x) + 2*x**2*(x**2 + 2)*Derivative(y(x), (x, 2)) + (3*x**2 + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False