2.1.575 Problem 591

Internal problem ID [9745]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 591
Date solved : Friday, April 25, 2025 at 06:18:41 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}4x^2{y}^{\prime \prime }+2x(-x^2+4)y^{\prime }+(7x^2+1)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.776 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& 4x^2{y}^{\prime \prime }+(-2x^3+8x)y^{\prime }+(7x^2+1)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =4x^2\\ \text{(3)}& B& =-2x^3+8xC& =7x^2+1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{x^4-40x^2-4}{16x^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =x^4-40x^2-4\\ t& =16x^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{x^4-40x^2-4}{16x^2}\right)z(x)\end{array}$$

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 2.575: Necessary conditions for each Kovacic case

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

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

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=16x^2$. 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. Therefore

$$\begin{array}{rl}L& =[1,2]\end{array}$$

Attempting to find a solution using case $n=1$.

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=\frac{x^2}{16}-\frac{5}{2}-\frac{1}{4x^2}$$

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 $b=-\frac{1}{4}$. Hence

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

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-2$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{2}{2}& & =1\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $x^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{x}^i\\ \text{(8)}& & =\sum _{i=0}^1{a}_i{x}^i\end{array}$$

Let $a$ be the coefficient of $x^v=x^1$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{x}{4}-\frac{5}{x}-\frac{101}{2x^3}-\frac{1010}{x^5}-\frac{50601}{2x^7}-\frac{710030}{x^9}-\frac{21351501}{x^{11}}-\frac{672670100}{x^{13}}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=\frac{1}{4}$$

From Eq. (9) the sum up to $v=1$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^1{a}_i{x}^i\\ \text{(10)}& & =\frac{x}{4}\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^0=1$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{x^2}{16}$$

This shows that the coefficient of $1$ in the above is $0$. Now we need to find the coefficient of $1$ in $r$. How this is done depends on if $v=0$ or not. Since $v=1$ which is not zero, then starting $r=\frac{s}{t}$, we do long division and write this in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $1$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^4-40x^2-4}{16x^2}\\ & =Q+\frac{R}{16x^2}\\ & =(\frac{x^2}{16}-\frac{5}{2})+(-\frac{1}{4x^2})\\ & =\frac{x^2}{16}-\frac{5}{2}-\frac{1}{4x^2}\end{array}$$

We see that the coefficient of the term $x$ in the quotient is $-\frac{5}{2}$. Now $b$ can be found.

$$\begin{array}{rl}b& =(-\frac{5}{2})-\left(0\right)\\ & =-\frac{5}{2}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{x}{4}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-\frac{5}{2}}{\frac{1}{4}}-1)& & =-\frac{11}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-\frac{5}{2}}{\frac{1}{4}}-1)& & =\frac{9}{2}\end{array}$$

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

$$r=\frac{x^4-40x^2-4}{16x^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-2$	$\frac{x}{4}$	$-\frac{11}{2}$	$\frac{9}{2}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{9}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{+}\right)\\ & =\frac{9}{2}-\left(\frac{1}{2}\right)\\ & =4\end{array}$$

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

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

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=4$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}(12x^2+6x{a}_3+2a_2)+2(\frac{1}{2x}-\frac{x}{4})(4x^3+3a_3{x}^2+2a_{2}x+a_1)+((-\frac{1}{2x^2}-\frac{1}{4})+{(\frac{1}{2x}-\frac{x}{4})}^2-\left(\frac{x^4-40x^2-4}{16x^2}\right))& =0\\ \frac{x^4{a}_3+2(16+a_2)x^3+3(a_1+6a_3)x^2+4(a_0+2a_2)x+2a_1}{2x}=0\end{array}$$

Solving for the coefficients $a_i$ in the above using method of undetermined coefficients gives

$$\{a_0=32,a_1=0,a_2=-16,a_3=0\}$$

Substituting these coefficients in $p(x)$ in eq. (2A) results in

$$\begin{array}{rl}p(x)& =x^4-16x^2+32\end{array}$$

Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x^4-16x^2+32){\mathrm{e}}^{\int (\frac{1}{2x}-\frac{x}{4})dx}\\ & =(x^4-16x^2+32){\mathrm{e}}^{-\frac{x^2}{8}+\frac{\ln \left(x\right)}{2}}\\ & =(x^4-16x^2+32)\sqrt{x}{\mathrm{e}}^{-\frac{x^2}{8}}\end{array}$$

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

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

Which simplifies to

$$y_1=\frac{x^4-16x^2+32}{\sqrt{x}}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-2x^3+8x}{4x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\frac{x^2}{4}-2\ln \left(x\right)}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{\frac{x^2}{4}-2\ln \left(x\right)}x}{{(x^4-16x^2+32)}^2}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{x^4-16x^2+32}{\sqrt{x}}\right)+c_2\left(\frac{x^4-16x^2+32}{\sqrt{x}}(\int \frac{{\mathrm{e}}^{\frac{x^2}{4}-2\ln \left(x\right)}x}{{(x^4-16x^2+32)}^2}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.040 (sec). Leaf size: 24

ode:=4*x^2*diff(diff(y(x),x),x)+2*x*(-x^2+4)*diff(y(x),x)+(7*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{(x^4-16x^2+32)(c_1+2c_2)}{32\sqrt{x}}$$

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 successful 
   <- special function solution successful 
      Solution using Kummer functions still has integrals. Trying a hypergeomet\ 
ric solution. 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form could result into a too large expression - returning s\ 
pecial function form of solution, free of uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 4x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+2x(-x^2+4)\left(\frac{d}{dx}y\left(x\right)\right)+(7x^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{(7x^2+1)y\left(x\right)}{4x^2}+\frac{(x^2-4)\left(\frac{d}{dx}y\left(x\right)\right)}{2x}\\ \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{(x^2-4)\left(\frac{d}{dx}y\left(x\right)\right)}{2x}+\frac{(7x^2+1)y\left(x\right)}{4x^2}=0\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=-\frac{x^2-4}{2x},P_3\left(x\right)=\frac{7x^2+1}{4x^2}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=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}{4}\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & 4x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-2x(x^2-4)\left(\frac{d}{dx}y\left(x\right)\right)+(7x^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}^2\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x^2\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}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0{(1+2r)}^2{x}^r+a_1{(3+2r)}^2{x}^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k{(2k+2r+1)}^2-a_{k-2}(2k-11+2r))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{(3+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}\\ & & a_k{(2k+2r+1)}^2+(-2k+11-2r)a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+2}{(2k+5+2r)}^2+a_k(-2k-2r+7)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{a_k(2k+2r-7)}{{(2k+5+2r)}^2}\\ \bullet & & \text{Recursion relation for}r=-\frac{1}{2}\text{; series terminates at}k=4\\ & & a_{k+2}=\frac{a_k(2k-8)}{{(2k+4)}^2}\\ \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(2k-8)}{{(2k+4)}^2},a_1=0]\end{array}$$

✓ Mathematica. Time used: 0.4 (sec). Leaf size: 70

ode=4*x^2*D[y[x],{x,2}]+2*x*(4-x^2)*D[y[x],x]+(1+7*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{\sqrt{e}(x^4-16x^2+32)(c_2{\int }_1^x\frac{e^{\frac{K[1{]}^2}{4}-1}}{K[1]{(K[1{]}^4-16K[1{]}^2+32)}^2}dK[1]+c_1)}{\sqrt{x}}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*x**2*Derivative(y(x), (x, 2)) + 2*x*(4 - x**2)*Derivative(y(x), x) + (7*x**2 + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False