hw4solns - h = . 1, y ( ) = 2 y 1 : k 1 = 0 k 2 =-. 049979...

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Section 2.6 3. y 0 = - y 2 cos x , h = . 1, y (0) = 1. We have x 0 = 0, x 1 = . 1, x 2 = . 2. y 1 : k 1 = f (0 , 1) = - (1) 2 cos 0 = - 1 k 2 = f ± 0 + . 05 , 1 + ( . 1)( - 1) 2 ² = - ( . 95) 2 cos( . 05) = - . 901372 k 3 = f ± 0 + . 05 , 1 + ( . 1) 2 ( - . 902499) ² = - . 9107543 k 4 = f (0 + . 1 , 1 + ( . 1)( - . 911786)) = - . 8220166 y 1 = 1 + ( . 1) 6 ( - 1 + 2( - . 901372) + 2( - . 910754) + ( - . 8220166)) = . 9092288 y 2 : k 1 = - . 822567 k 2 = - . 745136 k 3 = - . 751797 k 4 = - . 68177 y 2 = . 8342587 4. y 0 = sin x y 3 ,
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Unformatted text preview: h = . 1, y ( ) = 2 y 1 : k 1 = 0 k 2 =-. 049979 k 3 =-. 0503557 k 4 =-. 101356889 y 1 = . 994966 y 2 : k 1 =-. 101356 k 2 =-. 15405988 k 3 =-. 1552968 k 4 =-. 211447 y 2 = . 9794409...
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This note was uploaded on 03/11/2008 for the course EC 201 taught by Professor Brown during the Winter '07 term at Cal Poly Pomona.

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hw4solns - h = . 1, y ( ) = 2 y 1 : k 1 = 0 k 2 =-. 049979...

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