This preview shows page 1. Sign up to view the full content.
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 RungeKutta 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 ...
View
Full
Document
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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details