Method of Forbenius

Apr 12, 2007 14:27

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?
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