330_pdfsam_math 54 differential equation solutions odd

# 330_pdfsam_math 54 differential equation solutions odd - =...

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Chapter 5 100 200 300 400 500 –80 –60 –40 –20 0 v x Figure 5–A : Poincar´ e section for Problem 3. and, therefore, x n = x (2 πn ) e 0 . 1 πn sin(1 . 997498 πn ) , v n = v (2 πn ) e 0 . 1 πn (0 . 05 sin(1 . 997498 πn )+0 . 998749 cos(1 . 997498 πn )) + 10 . The values of x n and v n for n =0 , 1 ,..., 20 are listed in Table 5-F, and points ( x n ,v n )are shown in Figure 5-A. When n →∞ ,thepo ints( x n ,v n ) become unbounded because of e 0 . 1 πn term. 5. We want to construct the Poincar´ emapu s ing t =2 πn for x ( t ) given in equation (5) on page 295 of the text with A
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Unformatted text preview: = 0, = 1 / 3, and b = 0 . 22. Since tan = 2 1 b = 4 . 040404 , we take = tan 1 ( 4 . 040404) = 1 . 328172 and get x n = x (2 n ) = e . 22 n sin(0 . 629321 n ) (1 . 092050) sin(1 . 328172) , v n = x (2 n ) = . 11 e . 22 n sin(0 . 629321 n ) + (1 . 258642) e . 22 n cos(0 . 629321 n ) +(1 . 092050) cos(1 . 328172) . 326...
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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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