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(Regular) Singular Points

by Ted123
Tags: points, regular, singular
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Ted123
#1
Jun6-11, 08:24 AM
P: 449
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?
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HallsofIvy
#2
Jun6-11, 08:34 AM
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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: 449
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!


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