Why is x = 0 an ordinary point for y′′ − xy = 0?

[phil ess]

Homework Statement

Find two linearly independent power series solutions for the differential equation y′′ − xy = 0 about the ordinary point x0 = 0. Your answer should include a general formula for the coefficients.

The Attempt at a Solution

Im having trouble seeing how x0 = 0 is an ordinary point (i assume ordinary point means regular singular point?).

For it to be an ordinary point (x-x0)p(x) and (x-x0)q(x) have to be analytic at 0 right?

(x-x0)p(x) = x(0/x) = 0

(x-x0)q(x) = x(-1/x²) = -1/x which is not analytic at 0?

So how can x0=0 be an ordinary point?
Please help!

 

[lanedance]

so after a quick google to refresh my memory

if the DE is in the form
$$y'' + p(x) y' + q(x) y = 0$$

if p(x) & q(x) are analytic at x₀, its an ordinary point, if either diverges, its a singular point

if its a singular point, but both (x-x₀)p(x) & (x-x₀)²q(x) are analytic at x₀ its a regular singular point

http://mathworld.wolfram.com/RegularSingularPoint.html

 

[HallsofIvy]

Homework Statement

Find two linearly independent power series solutions for the differential equation y′′ − xy = 0 about the ordinary point x0 = 0. Your answer should include a general formula for the coefficients.

The Attempt at a Solution

Im having trouble seeing how x0 = 0 is an ordinary point (i assume ordinary point means regular singular point?).

Well, that's where you are wrong- an "ordinary point" is not a singular point at all! And that means you don't use Frobenius' method, just a regular series solution.

For it to be an ordinary point (x-x0)p(x) and (x-x0)q(x) have to be analytic at 0 right?

No. For an ordinary point p(x) and q(x) must be analytic.

(x-x0)p(x) = x(0/x) = 0

(x-x0)q(x) = x(-1/x²) = -1/x which is not analytic at 0?

So how can x0=0 be an ordinary point?
Please help!