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A textbook of Computer Based Numerical and Statiscal Techniques part 38. By joining statistical analysis with computer-based numerical methods, this book bridges the gap between theory and practice with software-based examples, flow charts, and applications. Designed for engineering students as well as practicing engineers and scientists, the book has numerous examples with in-text solutions. | 356 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES k4 hf x0 h y0 k3 0.2 0 0.2 1 - 0.0198 0.039208 1 .3 y 0.1 2 y0 k1 2k2 2k3 k4 6 1 1 - 0 2 -0.02 2 -0.0198 -0.039208 1.0000 - 0.198013 0.9801986 0.9802 The exact value of y 0.2 is 0.9802. Example 3. Solve the equation y x y with y0 1 by Runge-Kutta rule from x 0 to x 0.4 with h 0.1. Sol. Here f x y x y h 0.1 given y0 1 when x0 0. We have k1 hf x0 y0 0.1 0 1 0.1 f h k1 k2 hf I x0 y0 2 I 0.1 0.05 1.05 0.11 f h k2 k3 hf I x0 y0 2 I 0.1 0.05 1.055 0.1105 k4 hf x0 h y0 k3 0.1 0.1 1.1105 0.12105 1 y1 y x 0.1 y0 k1 2k2 2k3 k4 1 1 0.1 0.22 0.2210 0.12105 111034 6. Similarly for finding y2 y x 0.2 we get k1 hf x1y1 0.1 0.1 1.11034 0.121034 f h k1 k2 hf I x1 2 y y1 0.1 0.15 1.11034 0.660517 0.13208 f h k2 k3 hf I x1 2 y y- I 0.1 0.15 1.11034 0.06604 0.13208 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATION 357 k4 - hf x1 h yi k3 0.1 0.20 1.11034 0.13263 - 0.14263 1 y2 - y x-0.2 - y1 k1 2k2 2k3 k4 1 -1.11034 0.121034 2 0.13208 0.13263 0.14429 -1.2428 Similarly for finding y3 y x 0.3 we get k1 hf x2 y2 0.1 0.2 1.2428 0.14428 I h 1 k2 - hf I X2 y2 - 0.1 0.25 132428 0.07214 - 0.15649 I h k2 k3 - hf I X2 2 y2 I - 0.1 0.25 1.2428 0.07824 - 0.15710 k4 - hf x2 h y2 k3 - 0.1 0.30 1.2428 0.15710 - 0.16999 .-. y3 y x 0.3 y2 k1 2K2 2K3 k4 0.13997 Similarly for finding y4 y x 0.4 we get k1 0.1 0.3 1.3997 0.16997 k2 0.1 0.35 1.3997 0.08949 0.18347 k3 0.1 0.35 1.3997 0.9170 0.18414 k4 0.1 0.4 1.3997 0.18414 0.19838 k1 - 0.16997 k2 - 0.18347 k3 - 0.18414 k1 - 0.19838 1 y4 -1.3997 - 0.16997 2 0.18347 0.18414 0.19838 y4 -1.5836. Ans. Example 4. Given y - x with y 0 2 find y 0.1 and y 0.2 correct to 4 decimal places. dx Sol. We have x0 0 y0 2 h 0.1 Then we get k1 hf x0 y0 0.1 2 - 0 0.2 358 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES I h 1 k2 hf I X0 y0 2 2 02 0.1 0 02 2 0.205 Therefore 0.1 n 0.205 2 ----- 2 0.20525 k4 hf X0 h y0 k3 0.1 2 0.20525 - 0 0.1 0.210525 2 I h 2 k3 hf I X0 y0 y V0 1 k1 2k2 2k3 k4 6 2 0.2051708 2.2051708 y
