HW_08_Solutions - Y 25 = 0 . 03748 , Y 15 = 0 . 07965 , Y...

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A. Problem 1. Derivation y ( x + h ) = y ( x - h ) + Z x + h x - h f ( s, y ( s )) ds Simpson’s rule: Have three points, x-h, x, x+h = y ( x + h ) = y ( x - h ) + 1 3 h [ f ( x n - 1 , y n - 1 ) + 4 f ( x n , y n ) + f ( x n +1 , y n +1 )] y n +1 = y n - 1 + 1 3 h [ f ( x n - 1 , y n - 1 ) + 4 f ( x n , y n ) + f ( x n +1 , y n +1 )] This method is implicit because y n +1 (what we’ve solving for) is found in functions on the right-hand side. This method is multi-step because it requires two points ( n - 1 , n ) to begin. We need another method to get Frst approximation so we can start. B. Problem 4. Predictor-corrector using the Trapezoidal rule for the linear sys- tem(IVP) Y 0 1 = f 1 ( x, Y 1 , Y 2 ) = - 4 Y 1 + Y 2 , Y 1 (0) = 1 Y 0 2 = f 2 ( x, Y 1 , Y 2 ) = Y 1 - 4 Y 2 , Y 2 (0) = - 1 h = 0 . 25 x 2 = 0 . 25 ¯ Y 12 = Y 11 + hf 1 (( x, Y 11 , Y 21 ) = 1 + h ( - 4 Y 11 + Y 21 ) = - 0 . 25 ¯ Y 22 = Y 21 + hf 2 (( x, Y 11 , Y 21 ) = - 1 + h ( Y 11 - 4 Y 21 ) = 0 . 25 Y 12 = Y 11 + h 2 ( f 1 ( x, Y 11 , Y 21 ) + f 1 ( x, ¯ Y 12 , ¯ Y 22 ) = 0 . 53125 Y 22 = Y 21 + h 2 ( f 2 ( x, Y 11 , Y 21 ) + f 2 ( x, ¯ Y 12 , ¯ Y 22 ) = - 0 . 53125 1
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x 3 = 0 . 5 ¯ Y 13 = Y 12 + hf 1 (( x, Y 12 , Y 22 ) = - 0 . 13281 ¯ Y 23 = Y 22 + hf 2 (( x, Y 12 , Y 22 ) = 0 . 13281 Y 13 = Y 12 + h 2 ( f 1 ( x, Y 12 , Y 22 ) + f 1 ( x, ¯ Y 13 , ¯ Y 23 ) = 0 . 28223 Y 23 = Y 22 + h 2 ( f 2 ( x, Y 12 , Y 22 ) + f 2 ( x, ¯ Y 13 , ¯ Y 23 ) = - 0 . 28223 x 4 = 0 . 75 ¯ Y 14 = - 0 . 07056 , ¯ Y 24 = 0 . 07056 , Y 14 = 0 . 14993 , Y 24 = - 0 . 14993 x 5 = 1 . 0 ¯ Y 15 = - 0 . 03748 ,
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Unformatted text preview: Y 25 = 0 . 03748 , Y 15 = 0 . 07965 , Y 25 =-. 07965 C. Problem 5. Predictor-corrector using AB4-AM4 for the linear system(IVP) Y 1 = y, Y 2 = y Converting the 2nd order diferential equation into a linear system, Y 1 = f 1 ( x, Y 1 , Y 2 ) = Y 2 , Y 1 (0) = 1 8 Y 2 = f 2 ( x, Y 1 , Y 2 ) =-sin ( Y 1 ) , Y 2 (0) = 0 h = 10 x 2 = 10 Y 12 = Y 11 + h ( f 1 ( x, Y 11 , Y 21 ) = Y 11 + hY 21 = 0 . 39270 Y 22 = Y 21 + h ( f 2 ( x, Y 11 , Y 21 ) = Y 21 + h (-sin ( Y 11 )) =-. 12022 x 3 = 5 Y 13 = Y 12 + h ( f 1 ( x, Y 12 , Y 22 ) = Y 12 + hY 22 = 0 . 35493 Y 23 = Y 22 + h ( f 2 ( x, Y 12 , Y 22 ) = Y 22 + h (-sin ( Y 12 )) =-. 24045 2...
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This note was uploaded on 02/22/2010 for the course CHE 348 taught by Professor Chelikowsky during the Spring '08 term at University of Texas at Austin.

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HW_08_Solutions - Y 25 = 0 . 03748 , Y 15 = 0 . 07965 , Y...

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