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Unformatted text preview: MATH 128A, SUMMER 2009: HOMEWORK 6 Problems 5.3: 1(c), 5.4: 1(c) and 13(c), 5.5: 1(c), modiﬁed versions of 5.6: 1(c) and 4: Use the following methods to solve the problem of 5.2: 1(c) from last week (y = 1 + y/t, t ∈ [1, 2], y (1) = 2,). Put your answers (for each value of t) in a table, rounded to four signiﬁcant digits (x.xxx). (1) Actual solution: y = t(ln t + 2); (2) Taylor method of order 2, h = 0.25; (3) Modiﬁed Euler method, h = 0.25; (4) Adams-Bashforth two-step explicit method, h = 0.25, with exact starting values; (5) Adams-Moulton one-step implicit method (wi+1 = wi + h (f (ti+1 , wi+1 ) + f (ti , wi ))), h = 0.25, with 2 exact starting values; (6) Adams second-order predictor-corrector method (one-step Adams-Bashforth, i.e. Euler, feeding the above Adams-Moulton formula), h = 0.25, with exact starting values; (7) Runge-Kutta method of order 4 (see rk4.m), h = 0.25; (8) Runge-Kutta-Fehlberg method (see rk4.m), tol = 10−4 , hmin = 0.05, hmax = 0.25; RK problem: Assume that the diﬀerential equation y = f (t, y ) takes the form y = g (y ). (Equations of this form are called autonomous.) Find the best choice of coeﬃcients a, b1 , b2 , and c in the following Runge-Kutta method: k1 = hg (wi ) k2 = hg (wi + ak1 ) wi+1 = wi + b1 k1 + b2 k2 What is the order of the resulting method’s truncation error? Multistep Method Problem: Find a0 , a1 , b0 , and b1 so that the following explicit 2-step method has the highest-order truncation error possible: wi+1 = a1 wi + a0 wi−1 + h[b1 f (ti , wi ) + b0 f (ti−1 , wi−1 )] Hint : Let y (t) = c0 + c1 (t − ti + c2 (t − ti )2 + c3 (t − ti )3 + . . .. Match as many terms as possible in the power series y (ti + h) and a1 y (ti ) + a0 y (ti − h) + h[b1 y (ti ) + b0 y (ti − h)]. Note : In spite of the good truncation error, the resulting method is unsuitable for all problems. Date : Due Thursday 7/30.
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