24_pdfsam_math 54 differential equation solutions odd

24_pdfsam_math 54 - Chapter 1 Plugging these values into the recursive equations above yields t2 = 0.25 0.25 = 0.5 Continuing in this way gives t3

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Chapter 1 Plugging these values into the recursive equations above yields t 2 =0 . 25 + 0 . 25 = 0 . 5a n d x 2 . 25 + 0 . 25 · [1 + 0 . 25 · sin(0 . 0625)] = 0 . 503904 . Continuing in this way gives t 3 . 75 and x 3 . 503904 + 0 . 25 · [1 + 0 . 5 · sin(0 . 251952)] = 0 . 785066 , t 4 =1 . 00 and φ (1) x 4 . 13920 . For N =8,wehave h / 8=0 . 125 . Thus, the recursive formulas become t n +1 = t n +0 . 125 and x n +1 = x n . 125 · [1 + t n sin( t n x n )] . Using these formulas and starting with t 0 =0and x 0 = 0, we can ±ll in Table 1-A. From this we see that φ (1) x 8 . 19157, which is rounded to ±ve decimal places. Table 1–A : Euler’s method approximations for the solution of x 0 =1+ t sin( tx ), x (0) = 0, at t = 1 with 8 steps (
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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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