Method of Forbenius

[where_was_i]

This is from my differential equations homework:

(x2-1)y’’ - (x-1)y’ - 3y = 0

I’m getting p0 = q0 = 0 when using the method of Frobenius. It always worries me when things come out simply. Can someone tell me if what I've got is right?

Comments

[joshua_green]

Well, y = 0 is certainly a solution.  What are your initial conditions?

[where_was_i]

It's just centered about x=1, no initial values given. It's only asking for the indicial (or auxiliary) equation, and the exponents.

[joshua_green]

What are p0 and q0?

[where_was_i]

The second order equation is rewritten as y''+p(x)y'+q(x)y=0. Then p0 = limx→0 (x-x0)p(x) and q0 = limx→0 (x-x0)2q(x). Then the indicial equation comes from assumming y=xr, where y''+p0y'+q0y=0, and solving for r. The 'exponents' are the one or two values of r that come from the indicial. x0=1.