Kuttas third-order method is defined by y n+1 = y n + 1/6(k 1 + 4k 2
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Table 21.3
Table 21.5
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ALGORITHM RUNGE-KUTTA (f, xo, Yo, h, N). This algorithm computes the solution of the initial value problem y' = f(x, y), y(Xx) = Yo at equidistant points X1 = xo + h, x2 = xo + 2h, , XN = xo + Nh; (9) here f is such that this problem has a unique solution on the interval [Xo, Xx] (see Sec. 1.7). INPUT: Function f, initial values xo, Yo, step size h, number of steps N OUTPUT: Approximation yn+1 to the solution y(Xn+1) at xn+1 = xo + (n + 1)h, where n = 0, 1, · , N – 1 For n = 0, 1, .., N – 1 do: ki = hf(xn, Yn) k2 = hf(xn + h, yn + k1) k3 = hf(xn + h, yn + żk2) k4 = hf(Xn + h, yn + k3) Xn+1 = Xn + h Yn+1 = Yn + & (k1 + 2k2 + 2k3 + k4) OUTPUT xn+1» Yn+1 End Stop End RUNGE-KUTTA Error y = et – x – 1 Improved Euler (Table 21.3) Runge-Kutta (Table 21.5) Euler (Table 21.1) 0.000003 0.000007 0.000011 0.2 0.021403 0.091825 0.0014 0.021 0.052 0.094 0.152 0.4 0.0034 0.0063 0.0102 0.6 0.222119 0.425541 0.8 0.718282 0.230 0.0156 0.000020 0.000031 1.0
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