162_pdfsam_math 54 differential equation solutions odd

162_pdfsam_math 54 differential equation solutions odd - 1...

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Chapter 3 y 1 + 1 6 [0 + 2(0 . 03125) + 2(0 . 0273438) + 0 . 0556641] 1 . 02881 . Thus the approximate value of the solution at 0 . 25 is 1 . 02881. By repeating Steps 3 and 4 of the algorithm we fill in the following Table 3-I. Table 3–I : 4th order Runge-Kutta subroutine approximations for y = x + 1 y at x = 1 with h = 0 . 25 . x 0 0.25 0.50 0.75 1.0 y 1 1.02881 1.10654 1.22238 1.36789 Thus, our approximation at x = 1 is approximately 1 . 36789. Comparing this with Problem 5, we see we have obtained accuracy to four decimal places as we did with the Taylor method of order four, but without having to compute any partial derivatives. 11. In this problem, f ( x, y ) = 2 x 4 y 2 . To find the root of the solution within two decimal places of accuracy, we choose a step size h = 0 . 005 in 4th order Runge-Kutta subroutine. It will require (2
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Unformatted text preview: 1) / . 005 = 200 steps to approximate the solution on [1 , 2]. With the initial input x = x = 1, y = y = − . 414, we get k 1 = hf ( x, y ) = 0 . 005[2(1) − 4 − ( − . 414) 2 ] = 0 . 009143; k 2 = hf ( x + h/ 2 , y + k 1 / 2) = 0 . 005[2(1 + 0 . 005 / 2) − 4 − ( − . 414 + 0 . 009143 / 2) 2 ] = 0 . 009062; k 3 = hf ( x + h/ 2 , y + k 2 / 2) = 0 . 005[2(1 + 0 . 005 / 2) − 4 − ( − . 414 + 0 . 009062 / 2) 2 ] = 0 . 009062; k 4 = hf ( x + h, y + k 3 ) = 0 . 005[2(1 + 0 . 005) − 4 − ( − . 414 + 0 . 009062) 2 ] = 0 . 008983; ⇓ x = 1 + 0 . 005 = 1 . 005 , y = − . 414 + 1 6 (0 . 009143 + 2 · . 009062 + 2 · . 009062 + 0 . 008983) ≈ − . 404937; . . . 158...
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