152_pdfsam_math 54 differential equation solutions odd

# 152_pdfsam_math 54 differential equation solutions odd - h...

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Chapter 3 Starting with the inputs x = x 0 =0 , y = y 0 = 2, and h =0 . 005 we obtain F =(0 2+2) 2 =0 , G =[0 2+2+0 . 005(1 + 0)] 2 =0 . 000025 . Thus, in Step 4 we compute x =0+0 . 005 = 0 . 005 , y = 2+0 . 005(0 + 0 . 000025)(1 / 2) ≈− 2 . Thus the approximate value of the solution at x =0 . 005 is 2. We continue with Steps 3 and 4 of the improved Euler’s method subroutine until we arrive at x =1 . 270 and y ≈− 0 . 04658269. The next iteration, with x =1 . 275, yields y 0 . 006295411. This tells us that y = 0 is occurs somewhere between x =1 . 270 and x =1 . 275. Therefore, rounding oF to two decimal places yields x =1 . 27. 17. In this initial value problem, f ( x, y )= 20 y , x 0 =0,and y 0 = 1. By applying formula (4) on page 125 of the text, we can ±nd a general formula for y n in terms of h . Indeed, y n = y n 1 + h ( 20 y n 1 )=(1 20 h ) y n 1 = ··· =(1 20 h ) n y 0 =(1 20 h ) n =[ c ( h )] n , where c ( h )=1 20 h
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Unformatted text preview: h , we have h = 0 . 1 ⇒ c (0 . 1) = − 1 ⇒ x n = 0 . 1 n, y n = ( − 1) n , n = 1 , . . . , 10; h = 0 . 025 ⇒ c (0 . 025) = 0 . 5 ⇒ x n = 0 . 025 n, y n = (0 . 5) n , n = 1 , . . . , 40; h = 0 . 2 ⇒ c (0 . 2) = − 3 ⇒ x n = 0 . 2 n, y n = ( − 3) n , n = 1 , . . . , 5 . These values are shown in Table 3-E. Thus, for h = 0 . 1 we have alternating y n = ± 1; for h = 0 . 2, y n ’s have an increasing magnitude and alternating sign; h = 0 . 025 is a good step size. ²rom this example we conclude that, in Euler’s method, one should be very careful in choosing a step size. Wrong choice can even lead to a diverging process. 148...
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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 Berkeley.

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