37_pdfsam_math 54 differential equation solutions odd

# 37_pdfsam_math 54 differential equation solutions odd - 2 x...

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Exercises 2.2 Substituting this value into equation (2.3) from part (b) yields y (0 . 5) [1 + 3(0 . 544999003)] 1 / 3 1 . 38121 . Repeating these calculations for n =3 , 4, and 5 yields Table 2-A. Table 2–A : Successive approximations for y (0 . 5) using Simpson’s rule. Number of Intervals y ( 0 . 5 ) 6 1.38120606 8 1.38120520 10 1.38120497 Since these values do not change by more than 5 in the fourth place, we can conclude that the Frst three places are accurate and that we have obtained an approximate solution y (0 . 5) 1 . 381 . 29. (a) Separating variables and integrating yields dy y 1 / 3 = dx Z dy y 1 / 3 = Z dx 1 2 / 3 y 2 / 3 = x + C 1 y = ± 2 3 x + 2 3 C 1 ² 3 / 2 =
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Unformatted text preview: 2 x 3 + C 3 / 2 . (b) Using the initial condition, y (0) = 0, we Fnd that 0 = y (0) = 2(0) 3 + C 3 / 2 = C 3 / 2 C = 0 , and so y = (2 x/ 3 + 0) 3 / 2 = (2 x/ 3) 3 / 2 , x 0, is a solution to the initial value problem. (c) The function y ( x ) 0, clearly, satisFes both, the dierential equation dy/dx = y 1 / 3 and the initial condition y (0) = 0. (d) In notation of Theorem 1 on page 12, f ( x, y ) = y 1 / 3 and so f y = d dy ( y 1 / 3 ) = 1 3 y 2 / 3 = 1 3 y 2 / 3 . 33...
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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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