291_pdfsam_math 54 differential equation solutions odd

291_pdfsam_math 54 - Exercises 5.3 k2,2 = h 2 x1;0 k1,3 k2,3 k1,4 k2,4 −1 −1 1 −3 1 =− 2 2 2 −1/2 −1/2 3 =3 1 −4 1 =− 2 2 4 −1/2

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Unformatted text preview: Exercises 5.3 k2,2 = h 2 x1;0 + k1,3 k2,3 k1,4 k2,4 −1 −1 1 −3 1+ =− , 2 2 2 −1/2 −1/2 3 =3 1+ −4 1+ =− , 2 2 4 −1/2 −1/2 3 =2 1+ −3 1+ =− , 2 2 4 −3 −3 1 = h [3 (x1;0 + k1,3 ) − 4 (x2;0 + k2,3 )] = 3 1 + −4 1+ =− , 4 4 4 −3 −3 1 = h [2 (x1;0 + k1,3 ) − 3 (x2;0 + k2,3 )] = 2 1 + −3 1+ =− . 4 4 4 k1,1 2 k1,2 = h 3 x1;0 + 2 k1,2 = h 2 x1;0 + 2 k2,1 2 k2,2 − 4 x2;0 + 2 k2,2 − 3 x2;0 + 2 − 3 x2;0 + =2 1+ Using the recursive formulas, we find t1 = t0 + h = 0 + 1 = 1 and x1;1 = x1;0 + x2;1 3 1 (−1) + 2(−1/2) + 2(−3/4) + (−1/4) (k1,1 + 2k1,2 + 2k1,3 + k1,4 ) = 1 + =, 6 6 8 1 (−1) + 2(−1/2) + 2(−3/4) + (−1/4) 3 = x2;0 + (k2,1 + 2k2,2 + 2k2,3 + k2,4 ) = 1 + = 6 6 8 as approximations to x1 (1) and x2 (1) with step h = 1. We repeat the algorithm with h = 2−m , m = 1, 2, . . . . The results of these computations are listed in Table 5-A. Table 5–A: Approximations of the solution to Problem 17. m 0 1 2 h = 2−m 1.0 0.5 0.25 x1 (1; h) 0.375 0.36817 0.36789 x2 (1; h) 0.375 0.36817 0.36789 We stopped at m = 2, since x1 (1; 2−1 ) − x1 (1; 2−2 ) = x2 (1; 2−1 ) − x2 (1; 2−2 ) = 0.36817 − 0.36789 = 0.00028 < 0.001 . Hence u(1) = v (1) ≈ 0.36789 . 287 ...
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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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