168_pdfsam_math 54 differential equation solutions odd

168_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 3 2 + 1/ 2 + 1 (0.1 + 0.088356/2)3 2 5 (0.1 + 0.088356/2)2 = 0.120795 ; + (2 + 1/2) + 2 2+1 k4 = hf (u + h, v + k3 ) = (1) (2 + 1) + 1 (0.1 + 0.120795)3 2 5 + (2 + 1) + (0.1 + 0.120795)2 = 0.348857 . 2 k3 = hf (u + h/2, v + k2 /2) = (1) (2 + 1/2) So, u = u + h = 2 + 1 = 3, 1 v = v + (0.049 + 2 · 0.088356 + 2 · 0.120795 + 0.348857) ≈ 0.236027 . 6 Because the relative error between two successive approximations, v (3; 20) = 0.236027 and v = 0.1 is ε = |(0.236027 − 0.1)/0.236027| ≈ 0.576320 > 0.0001, we go back to Step 2 and set m = 1, take N = 2m = 2 on Step 3, compute h = 1/N = 0.5, and use the 4th order Runge-Kutta subroutine on page 138 of the text to find v (3; 0.5). This takes two steps and yields k1 = (0.5) 2 2 5 + 1 (0.1)3 + 2 + (0.1)2 = 0.0245; 2 2 2 + 0.5/2 + 1 (0.1 + 0.0245/2)3 k2 = (0.5) (2 + 0.5/2) 2 5 + (2 + 0.5/2) + (0.1 + 0.0245/2)2 = 0.033306 ; 2 2 + 0.5/2 + 1 (0.1 + 0.033306/2)3 2 5 (0.1 + 0.033306/2)2 = 0.036114 ; + (2 + 0.5/2) + 2 2 + 0.5 + 1 (0.1 + 0.036114)3 2 5 + (2 + 0.5) + (0.1 + 0.036114)2 = 0.053410 . 2 k3 = (0.5) (2 + 0.5/2) k4 = (0.5) (2 + 0.5) 164 ...
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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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