Chem Differential Eq HW Solutions Fall 2011 191

Chem Differential - a 1 = a 3 = a 5 = ·· = a 2 k 1 = ·· = 0 a is arbitrary a 2 =-2 2 a =-a a 4 =-2 4 a 2 = 1 2 a a 6 =-2 6 a 4 =-1 3 a a 8

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Section A.5 The Power Series Method, Part II A191 Solutions to Exercises A.5 1. For the differential equation y ± +2 xy =0 , p ( x ) = 1 is its own power series expansion about a = 0. So a = 0 is an ordinary point. To solve, let Let y = ± m =0 a m x m y ± = ± m =1 ma m x m - 1 . Then y ± +2 xy = ± m =1 ma m x m - 1 +2 x ± m =0 a m x m = ± m =1 ma m x m - 1 + ± m =0 2 a m x m +1 = ± m =0 ( m +1) a m +1 x m + ± m =1 2 a m - 1 x m = a 1 + ± m =1 [( m +1) a m +1 +2 a m - 1 ] x m . So y ± +2 xy = 0 implies that a 1 + ± m =1 [( m +1) a m +1 +2 a m - 1 ] x m =0 ; a 1 =0 ( m +1) a m +1 +2 a m - 1 =0 a m +1 = - 2 m +1 a m - 1 . From the recurrence relation, a
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Unformatted text preview: a 1 = a 3 = a 5 = ··· = a 2 k +1 = ··· = 0; a is arbitrary; a 2 =-2 2 a =-a , a 4 =-2 4 a 2 = 1 2! a , a 6 =-2 6 a 4 =-1 3! a , a 8 =-2 8 a 6 = 1 4! a , . . . a 2 k = (-1) k k ! a . So y = a ∞ ± k =0 (-1) k k ! x 2 k = a ∞ ± k =0 (-x 2 ) k k ! = a e-x 2 ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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