54_pdfsam_math 54 differential equation solutions odd

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

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Chapter 2 the Simpson’s rule becomes 1 Z 0 µ ( s ) sds 0 . 1 3 [ µ (0)(0) + 4 µ (0 . 1)(0 . 1) + 2 µ (0 . 2)(0 . 2) + 4 µ (0 . 3)(0 . 3) +2 µ (0 . 4)(0 . 4) + 4 µ (0 . 5)(0 . 5) + 2 µ (0 . 6)(0 . 6) + 4 µ (0 . 7)(0 . 7) +2 µ (0 . 8)(0 . 8) + 4 µ (0 . 9)(0 . 9) + µ (1 . 0)(1 . 0)] 1 . 064539 . Therefore, y (1) 1 µ (1) 1 Z 0 µ ( s ) sds + 2 µ (1) = 1 3 . 076723 · 1 . 064539 + 2 3 . 076723 =0 . 9960 . (c) We rewrite the diFerential equation in the form used in Euler’s method, dy dx = x p 1+sin 2 xy, y (0) = 2 , and conclude that f ( x, y )= x 1+sin 2 xy . Thus the recursive formulas (2) and (3) on page 25 of the text become x n +1 = x n + h, y n +1 = y n + h ± x n p 1+sin 2 x n y n ² ,n =0 , 1 ,... , x 0 =0 , y 0 =2 . W i th h =0 . 1 we need (1 0) / 0 . 1 steps to get an approximation at x =1. n =0: x 1 =0 . 1 ,y 1 =(2)+0 . 1[(0) p 1+sin
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