2.1.275 Problem 278

Internal problem ID [9445]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 278
Date solved : Friday, April 25, 2025 at 06:08:29 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{u}^{\prime \prime }+\frac{4u^{\prime }}{x}-a^{2}u& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.362 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& u^{\prime \prime }+\frac{4u^{\prime }}{x}-a^{2}u& =0\text{(2)}& A{u}^{\prime \prime }+B{u}^{\prime }+Cu& =0\end{array}$$

Comparing (1) and (2) shows that

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

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =u{e}^{\int \frac{B}{2A}dx}\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{a^2{x}^2+2}{x^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =a^2{x}^2+2\\ t& =x^2\end{array}$$

Therefore eq. (4) becomes

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

Equation (7) is now solved. After finding $z(x)$ then $u$ is found using the inverse transformation

$$\begin{array}{rl}u& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

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 2.275: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-2\\ & =0\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=x^2$. 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 $0$ 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

$$r=\frac{2}{x^2}+a^2$$

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 $b=2$. Hence

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

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=0$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{0}{2}& & =0\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ 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

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

Let $a$ be the coefficient of $x^v=x^0$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx a+\frac{1}{a{x}^2}-\frac{1}{2a^3{x}^4}+\frac{1}{2a^5{x}^6}-\frac{5}{8a^7{x}^8}+\frac{7}{8a^9{x}^{10}}-\frac{21}{16a^{11}{x}^{12}}+\frac{33}{16a^{13}{x}^{14}}+\dots \end{array}$$

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

$$a=a$$

From Eq. (9) the sum up to $v=0$ gives

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

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^{-1}=\frac{1}{x}$ 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

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

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

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

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $\frac{1}{x}$ in $r$ will be the coefficient in $R$ of the term in $x$ of degree of $t$ minus one, divided by the leading coefficient in $t$. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{a^2{x}^2+2}{x^2}\\ & =Q+\frac{R}{x^2}\\ & =\left(a^2\right)+\left(\frac{2}{x^2}\right)\\ & =\frac{2}{x^2}+a^2\end{array}$$

Since the degree of $t$ is $2$, then we see that the coefficient of the term $x$ in the remainder $R$ is $0$. Dividing this by leading coefficient in $t$ which is $1$ gives $0$. Now $b$ can be found.

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

Hence

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

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

$$r=\frac{a^2{x}^2+2}{x^2}$$

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

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

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 ${\alpha }_{\mathrm{\infty }}^{-}=0$ then

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

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

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

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=1$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{x}-a)\left(1\right)+(\left(\frac{1}{x^2}\right)+{(-\frac{1}{x}-a)}^2-\left(\frac{a^2{x}^2+2}{x^2}\right))& =0\\ \frac{2a{a}_0-2}{x}=0\end{array}$$

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

$$\{a_0=\frac{1}{a}\}$$

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

$$\begin{array}{rl}p(x)& =x+\frac{1}{a}\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x+\frac{1}{a}){\mathrm{e}}^{\int (-\frac{1}{x}-a)dx}\\ & =(x+\frac{1}{a}){\mathrm{e}}^{-ax-\ln \left(x\right)}\\ & =\frac{(ax+1){\mathrm{e}}^{-ax}}{ax}\end{array}$$

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

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

Which simplifies to

$$u_1=\frac{(ax+1){\mathrm{e}}^{-ax}}{x^{3}a}$$

The second solution $u_2$ to the original ode is found using reduction of order

$$u_2=u_1\int \frac{e^{\int -\frac{B}{A}dx}}{u_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{u}_2& =u_1\int \frac{e^{\int -\frac{\frac{4}{x}}{1}dx}}{{\left(u_1\right)}^2}dx\\ & =u_1\int \frac{e^{-4\ln \left(x\right)}}{{\left(u_1\right)}^2}dx\\ & =u_1\left(\frac{(ax-1){\mathrm{e}}^{2ax}}{2a(ax+1)}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}u& =c_1{u}_1+c_2{u}_2\\ & =c_1\left(\frac{(ax+1){\mathrm{e}}^{-ax}}{x^{3}a}\right)+c_2\left(\frac{(ax+1){\mathrm{e}}^{-ax}}{x^{3}a}\left(\frac{(ax-1){\mathrm{e}}^{2ax}}{2a(ax+1)}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.005 (sec). Leaf size: 32

ode:=diff(diff(u(x),x),x)+4/x*diff(u(x),x)-a^2*u(x) = 0; 
dsolve(ode,u(x), singsol=all);
 

$$u=\frac{c_1{\mathrm{e}}^{ax}(ax-1)+c_2{\mathrm{e}}^{-ax}(ax+1)}{x^3}$$

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}\\ & & \frac{d}{dx}\frac{d}{dx}u\left(x\right)+\frac{4\left(\frac{d}{dx}u\left(x\right)\right)}{x}-a^{2}u\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}u\left(x\right)\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{4}{x},P_3\left(x\right)=-a^2]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=4\\ & \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=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & -a^{2}u\left(x\right)x+\left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)x+4\frac{d}{dx}u\left(x\right)=0\\ \bullet & & \text{Assume series solution for}u\left(x\right)\\ & & u\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 u\left(x\right)\text{to series expansion}\\ & & x\cdot u\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 u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=1}}{a}_{k-1}{x}^{k+r}\\ & \circ & \text{Convert}\frac{d}{dx}u\left(x\right)\text{to series expansion}\\ & & \frac{d}{dx}u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & \frac{d}{dx}u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+r+1)x^{k+r}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+r+1)(k+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(3+r)x^{-1+r}+a_1(1+r)(4+r)x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+1}(k+r+1)(k+4+r)-a^2{a}_{k-1})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(3+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-3,0\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(1+r)(4+r)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+r+1)(k+4+r)-a^2{a}_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & a_{k+2}(k+2+r)(k+5+r)-a^2{a}_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{a^2{a}_k}{(k+2+r)(k+5+r)}\\ \bullet & & \text{Recursion relation for}r=-3\\ & & a_{k+2}=\frac{a^2{a}_k}{(k-1)(k+2)}\\ \bullet & & \text{Solution for}r=-3\\ & & [u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-3},a_{k+2}=\frac{a^2{a}_k}{(k-1)(k+2)},-2a_1=0]\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=\frac{a^2{a}_k}{(k+2)(k+5)}\\ \bullet & & \text{Solution for}r=0\\ & & [u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=\frac{a^2{a}_k}{(k+2)(k+5)},4a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [u\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k-3})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{c}_k{x}^k),b_{k+2}=\frac{a^2{b}_k}{(k-1)(k+2)},-2b_1=0,c_{k+2}=\frac{a^2{c}_k}{(k+2)(k+5)},4c_1=0]\end{array}$$

✓ Mathematica. Time used: 0.073 (sec). Leaf size: 68

ode=D[u[x],{x,2}]+4/x*D[u[x],x]-a^2*u[x]==0; 
ic={}; 
DSolve[{ode,ic},u[x],x,IncludeSingularSolutions->True]
 

$$u(x)\to \frac{\sqrt{\frac{2}{\pi }}((ia{c}_{2}x+c_1)\sinh (ax)-(a{c}_{1}x+i{c}_2)\cosh (ax))}{a{x}^{5/2}\sqrt{-iax}}$$

✓ Sympy. Time used: 0.233 (sec). Leaf size: 37

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

$$u\left(x\right)=\frac{C_1{J}_{\frac{3}{2}}\left(x\sqrt{-a^2}\right)+C_2{Y}_{\frac{3}{2}}\left(x\sqrt{-a^2}\right)}{x^{\frac{3}{2}}}$$