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Unformatted text preview: + c1 )x−1 , so and we obtain the general solution
y = c1 1
x In this case, only one of the basis elements in the general solution is a generalized
For equations which are not of Cauchy-Euler form the Frobenius method is
more involved. Let us consider the example
2x d2 y
+ y = 0,
dx (1.15) which can be rewritten as
+ P (x)
+ Q(x)y = 0,
18 P (x) = 1
2x Q(x) = 1
2x One easily checks that x = 0 is a regular singular point. We begin the Frobenius
method by assuming that the solution has the form
∞ y = xr ∞ an xn =
n=0 an xn+r .
(n + r)an xn+r−1 ,
and ∞ d2 y
(n + r)(n + r − 1)an xn+r−2
n=0 ∞ d2 y
2x 2 =
2(n + r)(n + r − 1)an xn+r−1 .
n=0 Substitution into the diﬀerential equation yields
∞ 2(n + r)(n + r − 1)an xn+r−1 + n=0 ∞ ∞ (n + r)an xn+r−1 + n=0 an xn+r = 0,
n=0 which simpliﬁes to
∞ xr (2n + 2r − 1)(n + r)an xn−1 + n=0 ∞ an xn = 0.
n=0 We can div...
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This document was uploaded on 01/12/2014.
- Winter '14