Register to reply

(Regular) Singular Points

by Ted123
Tags: points, regular, singular
Share this thread:
Ted123
#1
Jun6-11, 08:24 AM
P: 447
1. The problem statement, all variables and given/known data

Locate the singular points of [tex]x^3(x-1)y'' - 2(x-1)y' + 3xy =0[/tex] and decide which, if any, are regular.

3. The attempt at a solution

In standard form the DE is [tex]y'' - \frac{2}{x^3} y' + \frac{3}{x^2(x-1)} y = 0.[/tex]

Are the singular points [tex]x=0,\pm 1\;?[/tex]

Regular singular points [tex]x_0[/tex] of [tex]y'' + p(x)y' + q(x)y =0[/tex] satisfy [tex](x-x_0)p(x) ,\; (x-x_0)^2q(x)[/tex] both finite as [tex]x \to x_0.[/tex]

Considering [tex]x \left( -\frac{2}{x^3} \right),\; (x-1) \left( -\frac{2}{x^3} \right) ,\; (x+1) \left( -\frac{2}{x^3} \right)[/tex] none of which are finite as [tex]x\to x_0[/tex] Does this mean there are no regular singular points?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
HallsofIvy
#2
Jun6-11, 08:34 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Quote Quote by Ted123 View Post
1. The problem statement, all variables and given/known data

Locate the singular points of [tex]x^3(x-1)y'' - 2(x-1)y' + 3xy =0[/tex] and decide which, if any, are regular.

3. The attempt at a solution

In standard form the DE is [tex]y'' - \frac{2}{x^3} y' + \frac{3}{x^2(x-1)} y = 0.[/tex]

Are the singular points [tex]x=0,\pm 1\;?[/tex]

Regular singular points [tex]x_0[/tex] of [tex]y'' + p(x)y' + q(x)y =0[/tex] satisfy [tex](x-x_0)p(x) ,\; (x-x_0)^2q(x)[/tex] both finite as [tex]x \to x_0.[/tex]

Considering [tex]x \left( -\frac{2}{x^3} \right),\; (x-1) \left( -\frac{2}{x^3} \right) ,\; (x+1) \left( -\frac{2}{x^3} \right)[/tex] none of which are finite as [tex]x\to x_0[/tex] Does this mean there are no regular singular points?
What's wrong with [itex](x- 1)\left(\frac{2}{x^3}\right)[/itex] as x goes to 1?
Ted123
#3
Jun6-11, 08:40 AM
P: 447
Quote Quote by HallsofIvy View Post
What's wrong with [itex](x- 1)\left(\frac{2}{x^3}\right)[/itex] as x goes to 1?
Doh - I've tended to 0 on every limit!


Register to reply

Related Discussions
Regular Singular Point Calculus & Beyond Homework 2
Regular singular points of 2nd order ODE Calculus & Beyond Homework 1
Regular singular points (definition) Differential Equations 3
Why derive Regular Singular Points? Differential Equations 1
Series solutions near a regular singular point Differential Equations 1