146_pdfsam_math 54 differential equation solutions odd

146_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 3 y2 = 1 + 4h + 8h2 y1 = 1 + 4h + 8h2 y3 = 1 + 4h + 8h2 y2 = 1 + 4h + 8h2 Continuing this way we see that yn = 1 1 + 4h + 8h2 3 n 1 3 1 3 1 + 4h + 8h2 = 1 + 4h + 8h2 2 1 2 1 + 4h + 8h2 , 3 1 3 1 + 4h + 8h2 . = 3 . (3.16) (This can be proved by induction using equation (3.15) above.) We are looking for an approximation to our solution at the point x = 1/2. Therefore, we have h= 1/2 − x0 1/2 − 0 1 = = n n 2n ⇒ n= 1 . 2h Substituting this value for n into equation (3.16) yields yn = 1 1 + 4h + 8h2 3 1/(2h) . 7. For this problem, f (x, y ) = x − y 2 . We need to approximate the solution on the interval [1, 1.5] using a step size of h = 0.1. Thus the number of steps needed is N = 5. The inputs to the subroutine on page 129 are x0 = 1, y0 = 0, c = 1.5, and N = 5. For Step 3 of the subroutine we have F = f (x, y ) = x − y 2 , G = f (x + h, y + hF ) = (x + h) − (y + hF )2 = (x + h) − y + h(x − y 2) Starting with x = x0 = 1 and y = y0 = 0 we get h = 0.1 (as specified) and F = 1 − 0 2 = 1, G = (1 + 0.1) − 0 + 0.1(1 − 02 ) Hence in Step 4 we compute x = 1 + 0.1 = 1.1 , y = 0 + 0.05(1 + 1.09) = 0.1045 . 142 2 2 . = 1.1 − (0.1)2 = 1.09 . ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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