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**Unformatted text preview: **Exercises 5.4
x3 (1; 1) = 0.03125 . Repeating the algorithm with h = 2−1 , 2−2 , 2−3 we obtain the approximations in Table 5-E. Table 5–E: Approximations of the Solution to Problem 25.
n 0 1 2 3 h 1.0 0.5 0.25 0.125 y (1) ≈ x1 (1; 2−n ) 1.29167 1.26039 1.25960 1.25958 x2(1; 2−n ) 0.28125 0.34509 0.34696 0.34704 x3(1; 2−n ) 0.03125 −0.06642 −0.06957 −0.06971 We stopped at n = 3 since x1 (1; 2−3 ) − x1 (1; 2−2 ) 1.25958 − 1.25960 = = 0.00002 < 0.01 , −3 ) x1 (1; 2 1.25958 x2 (1; 2−3 ) − x2 (1; 2−2 ) 0.34704 − 0.34696 = = 0.00023 < 0.01 , x2 (1; 2−3 ) 0.34704 −0.06971 + 0.06957 x3 (1; 2−3 ) − x3 (1; 2−2 ) = = 0.00201 < 0.01 . −3 ) x3 (1; 2 −0.06971 Hence y (1) ≈ x1 1; 2−3 = 1.25958 , with tolerance 0.01 . 27. See the answer in the text. 29. See the answer in the text. EXERCISES 5.4: Introduction to the Phase Plane, page 274 and 1. Substitution of x(t) = e3t , y (t) = et into the system yields d 3t dx = e = 3e3t = 3 et dt dt
3 = 3y 3 , 293 ...

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