116 if a0 0 then all the coecients must be zero from

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Unformatted text preview: + c1 )x−1 , so and we obtain the general solution y = c1 1 log |x| + c2 . x x In this case, only one of the basis elements in the general solution is a generalized power series. For equations which are not of Cauchy-Euler form the Frobenius method is more involved. Let us consider the example 2x d2 y dy + + y = 0, 2 dx dx (1.15) which can be rewritten as d2 y dy + P (x) + Q(x)y = 0, dx2 dx where 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=0 Then ∞ dy (n + r)an xn+r−1 , = dx n=0 and ∞ d2 y = (n + r)(n + r − 1)an xn+r−2 dx2 n=0 ∞ d2 y 2x 2 = 2(n + r)(n + r − 1)an xn+r−1 . dx 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.

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