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30_pdfsam_math 54 differential equation solutions odd

# 30_pdfsam_math 54 differential equation solutions odd - 4...

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Chapter 1 T will take the place of the y . Also we see that h = 0 . 1 and f ( t, T ) = K ( M 4 T 4 ), where K = 40 4 and M = 70. The recursive formulas (2) and (3) on page 25 of the text become t n +1 = t n + 0 . 1 , T n +1 = T n + hf ( t n , T n ) = T n + 0 . 1 ( 40 4 ) ( 70 4 T 4 n ) , n = 0 , 1 , 2 , . . . . From the initial condition, T (0) = 100, we see that t 0 = 0 and T 0 = 100. Therefore, for n = 0, t 1 = t 0 + 0 . 1 = 0 + 0 . 1 = 0 . 1 , T 1 = T 0 + 0 . 1 ( 40 4 ) ( 70 4 T 4 0 ) = 100 + 0 . 1 ( 40 4 ) ( 70 4 100 4 ) 97 . 0316 , where we have rounded off to four decimal places. For n = 1, we have t 2 = t 1 + 0 . 1 = 0 . 1 + 0 . 1 = 0 . 2 , T 2 = T 1 + 0 . 1 ( 40 4 ) ( 70 4 T 4 1 ) = 97 . 0316 + 0 . 1 ( 40 4 ) ( 70 4 97 . 0316 4 ) 94 . 5068 . By continuing this way, we fill in Table 1-D. Table 1–D : Euler’s method approximations for the solution of T = K ( M 4
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Unformatted text preview: 4 ), T (0) = 100, with K = 40 − 4 , M = 70, and h = 0 . 1. n t n T n n t n T n n t n T n 100 7 0.7 85.9402 14 1.4 79.5681 1 0.1 97.0316 8 0.8 84.7472 15 1.5 78.9403 2 0.2 94.5068 9 0.9 83.6702 16 1.6 78.3613 3 0.3 92.3286 10 1.0 82.6936 17 1.7 77.8263 4 0.4 90.4279 11 1.1 81.8049 18 1.8 77.3311 5 0.5 88.7538 12 1.2 80.9934 19 1.9 76.8721 6 0.6 87.2678 13 1.3 80.2504 20 2.0 76.4459 From this table we see that T (1) = T ( t 10 ) ≈ T 10 = 82 . 694 and T (2) = T ( t 20 ) ≈ T 20 = 76 . 446 . 26...
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