Chapter 6 - 6 Numerical Solutions of Ordinary Differential...

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h = 0.1 h = 0.05 x y x(n) y(n) 1.00 5.0000 1.00 5.0000 1.10 3.9900 1.05 4.4475 1.20 3.2546 1.10 3.9763 1.30 2.7236 1.15 3.5751 1.40 2.3451 1.20 3.2342 1.50 2.0801 1.25 2.9452 1.30 2.7009 1.35 2.4952 1.40 2.3226 1.45 2.1786 1.50 2.0592 h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 0.00 2.0000 0.00 2.0000 0.10 1.6600 0.05 1.8150 0.20 1.4172 0.10 1.6571 0.30 1.2541 0.15 1.5237 0.40 1.1564 0.20 1.4124 0.50 1.1122 0.25 1.3212 0.30 1.2482 0.35 1.1916 0.40 1.1499 0.45 1.1217 0.50 1.1056 h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 0.00 0.0000 0.00 0.0000 0.10 0.1005 0.05 0.0501 0.20 0.2030 0.10 0.1004 0.30 0.3098 0.15 0.1512 0.40 0.4234 0.20 0.2028 0.50 0.5470 0.25 0.2554 0.30 0.3095 0.35 0.3652 0.40 0.4230 0.45 0.4832 0.50 0.5465 6 Numerical Solutions of Ordinary Differential Equations Exercises 6.1 All tables in this chapter were constructed in a spreadsheet program which does not support subscripts. Consequently, x n and y n will be indicated as x ( n ) and y ( n ), respectively. 1. 2. 3. 266
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h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 0.00 1.0000 0.00 1.0000 0.10 1.1110 0.05 1.0526 0.20 1.2515 0.10 1.1113 0.30 1.4361 0.15 1.1775 0.40 1.6880 0.20 1.2526 0.50 2.0488 0.25 1.3388 0.30 1.4387 0.35 1.5556 0.40 1.6939 0.45 1.8598 0.50 2.0619 h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 0.00 0.0000 0.00 0.0000 0.10 0.0952 0.05 0.0488 0.20 0.1822 0.10 0.0953 0.30 0.2622 0.15 0.1397 0.40 0.3363 0.20 0.1823 0.50 0.4053 0.25 0.2231 0.30 0.2623 0.35 0.3001 0.40 0.3364 0.45 0.3715 0.50 0.4054 h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 0.00 0.0000 0.00 0.0000 0.10 0.0050 0.05 0.0013 0.20 0.0200 0.10 0.0050 0.30 0.0451 0.15 0.0113 0.40 0.0805 0.20 0.0200 0.50 0.1266 0.25 0.0313 0.30 0.0451 0.35 0.0615 0.40 0.0805 0.45 0.1022 0.50 0.1266 h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 0.00 0.5000 0.00 0.5000 0.10 0.5215 0.05 0.5116 0.20 0.5362 0.10 0.5214 0.30 0.5449 0.15 0.5294 0.40 0.5490 0.20 0.5359 0.50 0.5503 0.25 0.5408 0.30 0.5444 0.35 0.5469 0.40 0.5484 0.45 0.5492 0.50 0.5495 Exercises 6.1 4. 5. 6. 7. 267
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h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 0.00 1.0000 0.00 1.0000 0.10 1.1079 0.05 1.0519 0.20 1.2337 0.10 1.1079 0.30 1.3806 0.15 1.1684 0.40 1.5529 0.20 1.2337 0.50 1.7557 0.25 1.3043 0.30 1.3807 0.35 1.4634 0.40 1.5530 0.45 1.6503 0.50 1.7560 h = 0.1 h = 0.05 x(n) y(n) x(n) y(n) 1.00 1.0000 1.00 1.0000 1.10 1.0095 1.05 1.0024 1.20 1.0404 1.10 1.0100 1.30 1.0967 1.15 1.0228 1.40 1.1866 1.20 1.0414 1.50 1.3260 1.25 1.0663 1.30 1.0984 1.35 1.1389 1.40 1.1895 1.45 1.2526 1.50 1.3315 h = 0.1 h = 0.05 x(n) y(n} x(n) y(n} 0.00 0.5000 0.00 0.5000 0.10 0.5250 0.05 0.5125 0.20 0.5498 0.10 0.5250 0.30 0.5744 0.15 0.5374 0.40 0.5986 0.20 0.5498 0.50 0.6224 0.25 0.5622 0.30 0.5744 0.35 0.5866 0.40 0.5987 0.45 0.6106 0.50 0.6224 h = 0.1 h = 0.05 x(n) y(n} exact x(n) y(n} exact 0.00 2.0000 2.0000 0.00 2.0000 2.0000 0.10 2.1220 2.1230 0.05 2.0553 2.0554 0.20 2.3049 2.3085 0.10 2.1228 2.1230 0.30 2.5858 2.5958 0.15 2.2056 2.2061 0.40 3.0378 3.0650 0.20 2.3075 2.3085 0.50 3.8254 3.9082 0.25 2.4342 2.4358 0.30 2.5931 2.5958 0.35 2.7953 2.7997 0.40 3.0574 3.0650 0.45 3.4057 3.4189 0.50 3.8840 3.9082 Exercises 6.1 8. 9. 10. 11. 268
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1.1 1.2 1.3 1.4 x 5 10 15 20 y IMPROVED h=0.1 EULER EULER x(n) y(n) y(n) 1.00 1.0000 1.0000 1.10 1.2000 1.2469 1.20 1.4938 1.6668 1.30 1.9711 2.6427 1.40 2.9060 8.7988 Exercises 6.1 12. (a) (b) 13. (a) Using the Euler method we obtain y (0 . 1) y 1 = 1 . 2. (b) Using y = 4 e 2 x we see that the local truncation error is y ( c ) h 2 2 = 4 e 2 c (0 . 1) 2 2 = 0 . 02 e 2 c . Since e 2 x is an increasing function, e 2 c e 2(0 . 1) = e 0 . 2 for 0 c 0 . 1. Thus an upper bound for the local truncation error is 0 . 02 e 0 . 2 = 0 . 0244. (c) Since y (0 . 1) = e 0 . 2 = 1 . 2214, the actual error is y (0 . 1) y 1 = 0 . 0214, which is less than 0 . 0244. (d) Using the Euler method with h = 0 . 05 we obtain y (0 . 1) y 2 = 1 . 21. (e) The error in (d) is 1 . 2214 1 . 21 = 0 . 0114. With global truncation error O ( h ), when the step size is halved we expect the error for h = 0 . 05 to be one-half the error when h = 0 . 1. Comparing 0 . 0114 with 0 . 214 we see that this is the case. 14. (a) Using the improved Euler method we obtain y (0 . 1) y 1 = 1 . 22. (b) Using y = 8 e 2 x we see that the local truncation error is y ( c ) h 3 6 = 8 e 2 c (0 . 1) 3 6 = 0 . 001333 e 2 c . Since e 2 x is an increasing function, e 2 c e 2(0 . 1) = e 0 . 2 for 0 c 0 . 1. Thus an upper bound for the local truncation error is 0 . 001333 e 0 . 2 = 0 . 001628. (c) Since y (0 . 1) = e 0 . 2 = 1 . 221403, the actual error is y (0 . 1) y 1 = 0 . 001403 which is less than 0 . 001628. (d) Using the improved Euler method with h = 0 . 05 we obtain y (0 . 1) y 2 = 1 . 221025. (e) The error in (d) is 1 . 221403 1 . 221025 = 0 . 000378. With global truncation error O ( h 2 ), when the step size is halved we expect the error for h = 0 . 05 to be one-fourth the error for h = 0 . 1. Comparing 0 . 000378 with 0 . 001403 we see that this is the case.
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