18 a using y 114 e 3 x 1 we see that the local

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18. (a) Using y = 114 e 3( x 1) we see that the local truncation error is y ( c ) h 3 6 = 114 e 3( x 1) h 3 6 = 19 h 3 e 3( c 1) . (b) Since e 3( x 1) is a decreasing function for 1 x 1 . 5, e 3( c 1) e 3(1 1) = 1 for 1 c 1 . 5 and y ( c ) h 3 6 19(0 . 1) 3 (1) = 0 . 019 . (c) Using the improved Euler’s method with h = 0 . 1 we obtain y (1 . 5) 2 . 080108. With h = 0 . 05 we obtain y (1 . 5) 2 . 059166. (d) Since y (1 . 5) = 2 . 053216, the error for h = 0 . 1 is E 0 . 1 = 0 . 026892, while the error for h = 0 . 05 is E 0 . 05 = 0 . 005950. With global truncation error O ( h 2 ) we expect E 0 . 1 /E 0 . 05 4. We actually have E 0 . 1 /E 0 . 05 = 4 . 52. EXERCISES 6.2 Runge-Kutta Methods 3. 6. 96
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x n y n 0.00 0.5000 0.10 0.5213 0.20 0.5358 0.30 0.5443 0.40 0.5482 0.50 0.5493 x n y n 0.00 0.5000 0.10 0.5250 0.20 0.5498 0.30 0.5744 0.40 0.5987 0.50 0.6225 x n h 0.05 h 0.1 1.00 1.0000 1.0000 1.05 1.1112 1.10 1.2511 1.2511 1.15 1.4348 1.20 1.6934 1.6934 1.25 2.1047 1.30 2.9560 2.9425 1.35 7.8981 1.40 1.0608 10 15 903.0282 1.1 1.2 1.3 1.4 x 5 10 15 20 y x n y n 0.0 1.00000000 init.cond. 0.2 0.73280000 RK4 0.4 0.64608032 RK4 0.6 0.65851653 RK4 0.8 0.72319464 ABM 6.3 Multistep Methods 9. 12. 15. (a) (b) 18. (a) Using y (5) = 1026 e 3( x 1) we see that the local truncation error is y (5) ( c ) h 5 120 = 8 . 55 h 5 e 3( c 1) . (b) Since e 3( x 1) is a decreasing function for 1 x 1 . 5, e 3( c 1) e 3(1 1) = 1 for 1 c 1 . 5 and y (5) ( c ) h 5 120 8 . 55(0 . 1) 5 (1) = 0 . 0000855 .
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