169_pdfsam_math 54 differential equation solutions odd

169_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: Exercises 3.7 This gives u = 2 + 0.5 = 2.5 , 1 v = 0.1 + (0.0245 + 2 · 0.033306 + 2 · 0.036114 + 0.053410) ≈ 0.136125 . 6 We compute kj ’s again and find an approximate value of v (3). 2.5 5 + 1 (0.136125)3 + 2.5 + (0.136125)2 = 0.053419; 2 2 2.5 + 0.5/2 k2 = (0.5) (2.5 + 0.5/2) + 1 (0.136125 + 0.053419/2)3 2 5 (0.136125 + 0.053419/2)2 = 0.083702 ; + (2.5 + 0.5/2) + 2 2.5 + 0.5/2 k3 = (0.5) (2.5 + 0.5/2) + 1 (0.136125 + 0.083702/2)3 2 5 + (2.5 + 0.5/2) + (0.136125 + 0.083702/2)2 = 0.101558 ; 2 2.5 + 0.5 + 1 (0.136125 + 0.101558)3 k4 = (0.5) (2.5 + 0.5) 2 5 + (2.5 + 0.5) + (0.136125 + 0.101558)2 = 0.205709 . 2 k1 = (0.5) 2.5 Therefore, at u = 2.5 + 0.5 = 3.0, v = 0.136125 + 1 (0.053419 + 2 · 0.083702 + 2 · 0.101558 + 0.205709) ≈ 0.241066 . 6 This time the relative error is ε= 0.241066 − 0.236027 v (3; 2−1) − v (3; 20) = ≈ 0.020903 > 0.0001 . −1 ) v (3; 2 0.241066 Thus we set m = 2, N = 2m = 4, h = 1/N = 0.25, repeat computations with this new step, and find that v (3; 2−2) ≈ 0.241854 and ε= 0.241854 − 0.241066 v (3; 2−2) − v (3; 2−1) = ≈ 0.003258 > 0.0001 . −1 ) v (3; 2 0.241854 165 ...
View Full Document

Ask a homework question - tutors are online