335_pdfsam_math 54 differential equation solutions odd

335_pdfsam_math 54 - Review Problems(b Substitution t = 2n into xp(t and xp(t yields xn = x(2n = xh(2n vn = v(2n = xh(2n 1 1 0.22 1 2(0.22)2 sin

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Review Problems (b) Substitution t =2 πn into x p ( t )and x 0 p ( t ) yields x n = x (2 πn )= x h (2 πn )+ 1 p 1+2(0 . 22) 2 sin ± 22 πn + ψ ² , v n = v (2 πn )= x 0 h (2 πn )+ 1 0 . 22 + 2 p 1+2(0 . 22) 2 cos ± 22 πn + ψ ² . As n →∞ , x h (2 πn ) 0and x 0 h (2 πn ) 0. Therefore, for n large, x n 1 p 1+2(0 . 22) 2 sin ± 22 πn + ψ ² = a sin ± 2 2 πn + ψ ² , v n 1 0 . 22 + 2 p 1+2(0 . 22) 2 cos ± 22 πn + ψ ² = c + 2 a cos ± 2 2 πn + ψ ² . (c) From part (b) we conclude that, for n large x 2 n a 2 sin 2 ± 2 2 πn + ψ ² and ( v n c ) 2 2 a 2 cos 2 ± 2 2 πn + ψ ² . Dividing the latter by 2 and summing yields x 2 n + ( v n c ) 2 2 a 2 h sin 2 ± 2 2 πn + ψ ² +cos 2 ± 2 2 πn + ψ ²i = a 2 , and the error (coming from the transient part) tends to zero as
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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