nagle_differential_equations_ISM_Part7

nagle_differential_equations_ISM_Part7 - Chapter 1 x 2 = x...

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Unformatted text preview: Chapter 1 x 2 = x 1 + h = 0 . 2 , y 2 = y 1 + 0 . 1 ( x 1 + y 1 ) = 1 . 1 + 0 . 1(0 . 1 + 1 . 1) = 1 . 22 , x 3 = x 2 + h = 0 . 3 , y 3 = y 2 + 0 . 1 ( x 2 + y 2 ) = 1 . 22 + 0 . 1(0 . 2 + 1 . 22) = 1 . 362 , x 4 = x 3 + h = 0 . 4 , y 4 = y 3 + 0 . 1 ( x 3 + y 3 ) = 1 . 362 + 0 . 1(0 . 3 + 1 . 362) = 1 . 528 , x 5 = x 4 + h = 0 . 5 , y 5 = y 4 + 0 . 1 ( x 4 + y 4 ) = 1 . 5282 + 0 . 1(0 . 4 + 1 . 5282) = 1 . 721 , where all of the answers have been rounded off to three decimal places. 6. In this problem, x = 1, y = 0, and f ( x,y ) = x- y 2 . So, we let n = 0 , 1 , 2 , 3, and 4, in the recursive formulas and find x 1 = x + h = 1 . 1 , y 1 = y + 0 . 1 ( x- y 2 ) = 0 + 0 . 1(1- 2 ) = 0 . 1 , x 2 = x 1 + h = 1 . 2 , y 2 = y 1 + 0 . 1 ( x 1- y 2 1 ) = 0 . 1 + 0 . 1(1 . 1- . 1 2 ) = 0 . 209 , x 3 = x 2 + h = 1 . 3 , y 3 = y 2 + 0 . 1 ( x 2- y 2 2 ) = 0 . 209 + 0 . 1(1 . 2- . 209 2 ) = 0 . 325 , x 4 = x 3 + h = 1 . 4 , y 4 = y 3 + 0 . 1 ( x 3- y 2 3 ) = 0 . 325 + 0 . 1(1 . 3- . 325 2 ) = 0 . 444 , x 5 = x 4 + h = 1 . 5 , y 5 = y 4 + 0 . 1 ( x 4- y 2 4 ) = 0 . 444 + 0 . 1(1 . 4- . 444 2 ) = 0 . 564 , where all of the answers have been rounded off to three decimal places.where all of the answers have been rounded off to three decimal places....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part7 - Chapter 1 x 2 = x...

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