{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

168_pdfsam_math 54 differential equation solutions odd

168_pdfsam_math 54 differential equation solutions odd -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online