Differential Equations Solutions 112

# Differential Equations Solutions 112 - 122 Chapter 20...

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122 Chapter 20. Solutions: Solution of Ordinary DiFerential Equations perhaps h = h/ 2. perhaps h = h/ 2 p where, since we need δ 2 5 p τ , we defne p = (log δ log τ ) / (5 log 2). CHALLENGE 20.10. We know that iF our old values are correct, y P n +1 y ( t n +1 )= 3 h 4 8 y (4) ( η ) . y C n +1 y ( t n +1 h 4 24 y (4) ( ν ) . Subtracting, we obtain y P n +1 y C n +1 = 3 h 4 8 y (4) ( η ) ( h 4 24 y (4) ( ν )) , where η, ν are in the interval containing y P n +1 , y C n +1 , and the true value. Since 3 / 8+ 1 / 24 = 10 / 24, the error in the corrector can be estimated as ± = | y P n +1 y C n +1 | / 10. Now, iF ±>τ , we might reduce h by a Factor oF 2 and retake the step. IF ±<<τ , we might double h in preparation For the next step (expecting that the local error might increase by a Factor oF 2 4 ). CHALLENGE 20.11. No answer provided. CHALLENGE 20.12. y ± ( t D y H ( y )( t )
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## This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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