294_pdfsam_math 54 differential equation solutions odd

# 294_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 5 k2,3 k1,2 = h 60 − x1;0 + 2 k2,2 − (19.42) x2;0 + 2 2 /77.7 = 0.38124 , k1,4 = h (x2;0 + k2,3 ) = 0.5 ((0) + 0.38124) = 0.19062 , k2,4 = h 60 − (x1;0 + k1,3 ) − (19.42) (x2;0 + k2,3 )2 /77.7 = 0.36732 . Using the recursive formulas, we ﬁnd t1 = t0 + h = 0 + 0.5 = 0.5 1 x1 (0.5) ≈ x1;1 = x1;0 + (k1,1 + 2k1,2 + 2k1,3 + k1,4 ) = 0.09573 , 6 1 x2 (0.5) ≈ x2;1 = x2;0 + (k2,1 + 2k2,2 + 2k2,3 + k2,4 ) = 0.37980 . 6 Next, we repeat the procedure with n = 1, 2, . . . , 9. The results of these computations (the values of x1;n only) are presented in Table 5-C. Table 5–C: Approximations of the solution to Problem 21. n 0 1 2 3 4 5 tn 0 0.5 1.0 1.5 2.0 2.5 x1;n ≈ H (tn) 0 0.09573 0.37389 0.81045 1.37361 2.03111 n 6 7 8 9 10 tn 3.0 3.5 4.0 4.5 5.0 x1;n ≈ H (tn) 2.75497 3.52322 4.31970 5.13307 5.95554 23. Let x1 = y and x2 = y to give the initial value problem x1 = f1 (t, x1 , x2 ) = x2 , x2 = f2 (t, x1 , x2 ) = −x1 (1 + rx2 ) , 1 x1 (0) = a, x2 (0) = 0. Now, using the deﬁnitions of tn , xi;n , ki,1 , ki,2 , ki,3 , and ki,4 on page 258 of the text, we have k1,1 = hf1 (tn , x1;n , x2;n ) = hx2;n , 290 ...
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