25_pdfsam_math 54 differential equation solutions odd

# 25_pdfsam_math 54 differential equation solutions odd - y =...

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Exercises 1.4 ( x n ,y n )w i th( x n +1 ,y n +1 ), n =0 , 1 ,..., 9. Since f ( x, y )= x 2 yx 1 y 2 , the recursive formulas have the form x n +1 = x n +0 . 1 , y n +1 = y n +0 . 1 ± 1 x 2 n y n x n y 2 n ² ,n =0 , 1 ,..., 9 , x 0 =1 , y 0 = 1. Therefore, x 1 =1+ 0 . 1=1 . 1 ,y 1 = 1+0 . 1 ± 1 1 2 1 1 ( 1) 2 ² = 0 . 9; x 2 =1 . 1+0 . 1=1 . 2 ,y 2 = 0 . 9+0 . 1 ± 1 1 . 1 2 0 . 9 1 . 1 ( 0 . 9) 2 ² ≈− 0 . 81653719 ; x 3 =1 . 2+0 . 1=1 . 3 ,y 3 = 0 . 81653719 + 0 . 1 ± 1 1 . 2 2 0 . 81653719 1 . 2 ( 0 . 81653719) 2 ² ≈− 0 . 74572128 ; x 4 =1 . 3+0 . 1=1 . 4 ,y 4 = 0 . 74572128 + 0 . 1 ± 1 1 . 3 2 0 . 74572128 1 . 3 ( 0 . 74572128) 2 ² ≈− 0 . 68479653 ; etc . The results of these computations (rounded to Fve decimal places) are shown in Table 1-B. Table 1–B : Euler’s method approximations for the solutions of y 0
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Unformatted text preview: y = x − 2 − yx − 1 − y 2 , y (1) = − 1, on [1 , 2] with h = 0 . 1. n x n y n n x n y n 1.0 − 1 . 00000 6 1.6 − . 58511 1 1.1 − . 90000 7 1.7 − . 54371 2 1.2 − . 81654 8 1.8 − . 50669 3 1.3 − . 74572 9 1.9 − . 47335 4 1.4 − . 68480 10 2.0 − . 44314 5 1.5 − . 63176 The function y ( x ) = − 1 /x = x − 1 , obviously, satisFes the initial condition, y (1) = − 1. ±urther 21...
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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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