295_pdfsam_math 54 differential equation solutions odd

295_pdfsam_math 54 - Exercises 5.3 k2,1 = hf2(tn x1;n x2;n = −hx1;n 1 rx2 n 1 h k1,1 k2,1 k2,1 x2;n k1,2 = hf1 tn x1;n = h x2;n 2 2 2 2 k2,2 =

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Unformatted text preview: Exercises 5.3 k2,1 = hf2 (tn , x1;n , x2;n ) = −hx1;n 1 + rx2 n , 1; h k1,1 k2,1 k2,1 , x2;n + k1,2 = hf1 tn + , x1;n + = h x2;n + , 2 2 2 2 k2,2 = hf2 h k1,1 k2,1 , x2;n + tn + , x1;n + 2 2 2 k1,1 = −h x1;n + 2 = h x2;n + k2,2 , 2 k1,2 2 1 + r x1;n + k1,2 2 2 k1,1 1 + r x1;n + 2 2 , h k1,2 k2,2 , x2;n + k1,3 = hf1 tn + , x1;n + 2 2 2 h k1,2 k2,2 , x2;n + k2,3 = hf2 tn + , x1;n + 2 2 2 = −h x1;n + , k1,4 = hf1 (tn + h, x1;n + k1,3 , x2;n + k2,3 ) = h (x2;n + k2,3 ), k2,4 = hf2 (tn + h, x1;n + k1,3 , x2;n + k2,3 ) = −h (x1;n + k1,3 ) 1 + r (x1;n + k1,3 )2 . Using these values, we find tn+1 = tn + h = tn + 0.1 , 1 x1;n+1 = x1;n + (k1,1 + 2k1,2 + 2k1,3 + k1,4 ) , 6 1 x2;n+1 = x2;n + (k2,1 + 2k2,2 + 2k2,3 + k2,4 ) . 6 In Table 5-D we give the approximate period for r = 1 and 2 with a = 1, 2 and 3, from this we see that the period varies as r is varied or as a is varied. Table 5–D: Approximate period of the solution to Problem 23. r 1 2 a=1 4.8 4.0 a=2 3.3 2.4 a=3 2.3 1.7 25. With x1 = y , x2 = y , and x3 = y , the initial value problem can be expressed as the system x1 = x2 , x2 = x3 , x3 = t − x3 − x2 1 , x1 (0) = 1, x2 (0) = 1, x3 (0) = 1. 291 ...
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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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