Numerical Methods in Engineering and Science by B. S. Grewal

Author:B. S. Grewal [B. S. Grewal]
Language: eng
Format: epub, mobi
ISBN: 978-1-68392-128-8
Publisher: Mercury Learning and Information
Published: 2019-02-09T05:00:00+00:00

Using y0 = 0 and y4 = 2, (i) gives the system of equation,

(i = 1) 16y2 – 31y1 = 0.25; (ii)

(i = 2) 16y3 – 31y2 + 16y1 = 0.5 (iii)

(i = 3) 32 – 31y3 + 16y2 = 0.75, i.e., – 31y3 + 16y2 = – 31.25 (iv)

Solving the equations (ii), (iii), and (iv), we get

y1 = 0.5443, y2 = 1.0701, y3 = 1.5604

Hence y (0.25) = 0.5443, y(0.5) = 1.0701, y(0.75) = 1.5604

EXAMPLE 10.36

Determine values of y at the pivotal points of the interval (0, 1) if y satisfies the boundary value problem yiv + 81y = 81x2, y(0) = y(1) = y″(0) = y″ (1) = 0. (Take n = 3).

Solution:

Here h = 1/3 and the pivotal points are x0 = 0, x1 = 1/3, x2 = 2/3, x3 = 1. The corresponding y-values are y0(= 0), y1, y2, y3(= 0).

Replacing yiv by its central difference approximation, the differential equation becomes

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