2.2.45 Problem 45

Internal problem ID [9167]
Book : Second order enumerated odes
Section : section 2
Problem number : 45
Date solved : Friday, April 25, 2025 at 05:58:49 PM
CAS classification : [_Bessel]

Solved as second order Bessel ode

Time used: 0.100 (sec)

Solve

$$\begin{array}{rl}{x}^2{y}^{\prime \prime }+x{y}^{\prime }+(x^2-5)y& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& x^2{y}^{\prime \prime }+x{y}^{\prime }+(x^2-5)y=0\end{array}$$

Bessel ode has the form

$$\begin{array}{}\text{(2)}& x^2{y}^{\prime \prime }+x{y}^{\prime }+(-n^2+x^2)y=0\end{array}$$

The generalized form of Bessel ode is given by Bowman (1958) as the following

$$\begin{array}{}\text{(3)}& x^2{y}^{\prime \prime }+(1-2\alpha )x{y}^{\prime }+({\beta }^2{\gamma }^2{x}^{2\gamma }-n^2{\gamma }^2+{\alpha }^2)y=0\end{array}$$

With the standard solution

$$\begin{array}{}\text{(4)}& y& =x^{\alpha }(c_1\mathrm{BesselJ}(n,\beta x^{\gamma })+c_2\mathrm{BesselY}(n,\beta x^{\gamma }))\end{array}$$

Comparing (3) to (1) and solving for $\alpha ,\beta ,n,\gamma$ gives

$$\begin{array}{rl}\alpha & =0\\ \beta & =1\\ n& =-\sqrt{5}\\ \gamma & =1\end{array}$$

Substituting all the above into (4) gives the solution as

$$\begin{array}{r}y=c_1\mathrm{BesselJ}(-\sqrt{5},x)+c_2\mathrm{BesselY}(-\sqrt{5},x)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\mathrm{BesselJ}(-\sqrt{5},x)+c_2\mathrm{BesselY}(-\sqrt{5},x)\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 19

ode:=x^2*diff(diff(y(x),x),x)+x*diff(y(x),x)+(x^2-5)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1\mathrm{BesselJ}(\sqrt{5},x)+c_2\mathrm{BesselY}(\sqrt{5},x)$$

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 
<- No Liouvillian solutions exists 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   <- Bessel successful 
<- special function solution successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x\left(\frac{d}{dx}y\left(x\right)\right)+(x^2-5)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{(x^2-5)y\left(x\right)}{x^2}-\frac{\frac{d}{dx}y\left(x\right)}{x}\\ \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{\frac{d}{dx}y\left(x\right)}{x}+\frac{(x^2-5)y\left(x\right)}{x^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{1}{x},P_3\left(x\right)=\frac{x^2-5}{x^2}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=1\\ & \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}=-5\\ & \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}\\ & & x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x\left(\frac{d}{dx}y\left(x\right)\right)+(x^2-5)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\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+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(r^2-5)x^r+a_1(r^2+2r-4)x^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k(k^2+2kr+r^2-5)+a_{k-2})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r^2-5=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{\sqrt{5},-\sqrt{5}\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(r^2+2r-4)=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(k^2+2kr+r^2-5)+a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+2}({(k+2)}^2+2(k+2)r+r^2-5)+a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{a_k}{k^2+2kr+r^2+4k+4r-1}\\ \bullet & & \text{Recursion relation for}r=\sqrt{5}\\ & & a_{k+2}=-\frac{a_k}{k^2+2k\sqrt{5}+4+4k+4\sqrt{5}}\\ \bullet & & \text{Solution for}r=\sqrt{5}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\sqrt{5}},a_{k+2}=-\frac{a_k}{k^2+2k\sqrt{5}+4+4k+4\sqrt{5}},a_1=0]\\ \bullet & & \text{Recursion relation for}r=-\sqrt{5}\\ & & a_{k+2}=-\frac{a_k}{k^2-2k\sqrt{5}+4+4k-4\sqrt{5}}\\ \bullet & & \text{Solution for}r=-\sqrt{5}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-\sqrt{5}},a_{k+2}=-\frac{a_k}{k^2-2k\sqrt{5}+4+4k-4\sqrt{5}},a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\sqrt{5}})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k-\sqrt{5}}),a_{k+2}=-\frac{a_k}{k^2+2k\sqrt{5}+4+4k+4\sqrt{5}},a_1=0,b_{k+2}=-\frac{b_k}{k^2-2k\sqrt{5}+4+4k-4\sqrt{5}},b_1=0]\end{array}$$

✓ Mathematica. Time used: 0.079 (sec). Leaf size: 26

ode=x^2*D[y[x],{x,2}]+x*D[y[x],x]+(x^2-5)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_1\mathrm{BesselJ}(\sqrt{5},x)+c_2\mathrm{BesselY}(\sqrt{5},x)$$

✓ Sympy. Time used: 0.199 (sec). Leaf size: 19

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*Derivative(y(x), (x, 2)) + x*Derivative(y(x), x) + (x**2 - 5)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=C_1{J}_{\sqrt{5}}\left(x\right)+C_2{Y}_{\sqrt{5}}\left(x\right)$$