2.1.325 Problem 332

Internal problem ID [9497]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 332
Date solved : Sunday, March 30, 2025 at 02:36:14 PM
CAS classification : [[_Emden, _Fowler]]

Solved as second order ode using Kovacic algorithm

Time used: 0.334 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =x^4\\ \text{(3)}& B& =0C& =\lambda \end{array}$$

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(-\frac{\lambda }{x^4})z(x)\end{array}$$

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 2.325: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-0\\ & =4\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=x^4$. There is a pole at $x=0$ of order $4$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $4$ then the necessary conditions for case one are met. Therefore

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

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

Looking at higher order poles of order $2v$≥$4$ (must be even order for case one).Then for each pole $c$, $[\sqrt{r}{]}_c$ is the sum of terms $\frac{1}{(x-c{)}^i}$ for $2\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ expanded around each pole $c$. Hence

$$\begin{array}{}\text{(1B)}& [\sqrt{r}{]}_c& =\sum _2^v\frac{a_i}{(x-c{)}^i}\end{array}$$

Let $a$ be the coefficient of the term $\frac{1}{(x-c{)}^v}$ in the above where $v$ is the pole order divided by 2. Let $b$ be the coefficient of $\frac{1}{(x-c{)}^{v+1}}$ in $r$ minus the coefficient of $\frac{1}{(x-c{)}^{v+1}}$ in $[\sqrt{r}{]}_c$. Then

$$\begin{array}{rl}{\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)\\ {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)\end{array}$$

The partial fraction decomposition of $r$ is

$$r=-\frac{\lambda }{x^4}$$

There is pole in $r$ at $x=0$ of order $4$, hence $v=2$. Expanding $\sqrt{r}$ as Laurent series about this pole $c=0$ gives

$$\begin{array}{}\text{(2B)}& [\sqrt{r}{]}_c\approx \frac{i\sqrt{\lambda }}{x^2}+\dots \end{array}$$

Using eq. (1B), taking the sum up to $v=2$ the above becomes

$$\begin{array}{}\text{(3B)}& [\sqrt{r}{]}_c=\frac{i\sqrt{\lambda }}{x^2}\end{array}$$

The above shows that the coefficient of $\frac{1}{(x-0{)}^2}$ is

$$a=i\sqrt{\lambda }$$

Now we need to find $b$. let $b$ be the coefficient of the term $\frac{1}{(x-c{)}^{v+1}}$ in $r$ minus the coefficient of the same term but in the sum $[\sqrt{r}{]}_c$ found in eq. (3B). Here $c$ is current pole which is $c=0$. This term becomes $\frac{1}{x^3}$. The coefficient of this term in the sum $[\sqrt{r}{]}_c$ is seen to be $0$ and the coefficient of this term $r$ is found from the partial fraction decomposition from above to be $0$. Therefore

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

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =\frac{i\sqrt{\lambda }}{x^2}\\ {\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)& & =\frac{1}{2}(\frac{0}{i\sqrt{\lambda }}+2)& & =1\\ {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)& & =\frac{1}{2}(-\frac{0}{i\sqrt{\lambda }}+2)& & =1\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $4>2$ then

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

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

$$r=-\frac{\lambda }{x^4}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$4$	$\frac{i\sqrt{\lambda }}{x^2}$	$1$	$1$

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

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

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

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

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

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

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

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

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

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

$$y_1=z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}$$

Since $B=0$ then the above reduces to

$$\begin{array}{rl}{y}_1& =z_1\\ & =x{\mathrm{e}}^{\frac{i\sqrt{\lambda }}{x}}\end{array}$$

Which simplifies to

$$y_1=x{\mathrm{e}}^{\frac{i\sqrt{\lambda }}{x}}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Since $B=0$ then the above becomes

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dx\\ & =x{\mathrm{e}}^{\frac{i\sqrt{\lambda }}{x}}\int \frac{1}{x^2{\mathrm{e}}^{\frac{2i\sqrt{\lambda }}{x}}}dx\\ & =x{\mathrm{e}}^{\frac{i\sqrt{\lambda }}{x}}(-\frac{i{\mathrm{e}}^{-\frac{2i\sqrt{\lambda }}{x}}}{2\sqrt{\lambda }})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(x{\mathrm{e}}^{\frac{i\sqrt{\lambda }}{x}}\right)+c_2\left(x{\mathrm{e}}^{\frac{i\sqrt{\lambda }}{x}}(-\frac{i{\mathrm{e}}^{-\frac{2i\sqrt{\lambda }}{x}}}{2\sqrt{\lambda }})\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.006 (sec). Leaf size: 31

ode:=x^4*diff(diff(y(x),x),x)+lambda*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=x(c_1\sinh \left(\frac{\sqrt{-\lambda }}{x}\right)+c_2\cosh \left(\frac{\sqrt{-\lambda }}{x}\right))$$

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 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful
 

✓ Mathematica. Time used: 0.142 (sec). Leaf size: 56

ode=x^4*D[y[x],{x,2}]+\[Lambda]*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_{1}x{e}^{-1+\frac{i\sqrt{\lambda }}{x}}-\frac{i{c}_{2}x{e}^{1-\frac{i\sqrt{\lambda }}{x}}}{2\sqrt{\lambda }}$$

✓ Sympy. Time used: 0.108 (sec). Leaf size: 58

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

$$y\left(x\right)=\sqrt{x}(\frac{C_1\sqrt{\frac{\sqrt{{\lambda }_{}}}{x}}{J}_{-\frac{1}{2}}\left(\frac{\sqrt{{\lambda }_{}}}{x}\right)}{\sqrt{-\frac{\sqrt{{\lambda }_{}}}{x}}}+C_2{Y}_{-\frac{1}{2}}(-\frac{\sqrt{{\lambda }_{}}}{x}))$$