22_pdfsam_math 54 differential equation solutions odd

22_pdfsam_math 54 differential equation solutions odd -...

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Chapter 1 y n +1 = y n + hf ( x n ,y n )= y n +0 . 1 · ± x n y n ² ,n =0 , 1 , 2 ,... . To fnd an approximation For the solution at the point x 1 = x 0 . 1=0 . 1, we let n =0in the last recursive Formula to fnd y 1 = y 0 . 1 · ± x 0 y 0 ² =4+0 . 1 · (0) = 4 . To approximate the value oF the solution at the point x 2 = x 1 . . 2, we let n =1inthe last recursive Formula to obtain y 2 = y 1 . 1 · ± x 1 y 1 ² . 1 · ± 0 . 1 4 ² =4 1 400 =3 . 9975 3 . 998 . Continuing in this way we fnd x 3 = x 2 . . 3 3 = y 2 . 1 · ± x 2 y 2 ² . 9975 + 0 . 1 · ± 0 . 2 3 . 9975 ² 3 . 992 , x 4 . 4 4 3 . 985 , x 5 . 5 5 3 . 975 , where all oF the answers have been rounded o± to three decimal places. 3. Here f ( x, y y (2 y ), x 0 ,and y 0 = 3. We again use recursive Formulas From Euler’s method with h . 1. Setting n = 0, 1, 2, 3, and 4 and rounding o± results to three decimal places, we get x 1 = x 0 . . 1 1 = y 0 . 1 · [ y 0 (2 y 0 )] = 3 + 0 . 1 · [3(2 3)] = 2 . 700; x 2 . 1+0 . . 2 2 =2 . 700 + 0 . 1 · [2 . 700(2 2 . 700)] = 2 . 511; x 3 . 2+0 . . 3 3 . 511 + 0 . 1 · [2 . 511(2
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