122Chapter 20.Solutions: Solution of Ordinary DiFerential Equations•perhapsh=h/2.•perhapsh=h/2pwhere, since we needδ2−5p≈τ, we defnep= (logδ−logτ)/(5 log 2).CHALLENGE 20.10.We know that iF our old values are correct,yPn+1−y(tn+1)=3h48y(4)(η).yCn+1−y(tn+1−h424y(4)(ν).Subtracting, we obtainyPn+1−yCn+1=3h48y(4)(η)−(−h424y(4)(ν)),whereη, νare in the interval containingyPn+1,yCn+1, and the true value. Since 3/8+1/24 = 10/24, the error in the corrector can be estimated as±=|yPn+1−yCn+1|/10.Now, iF±>τ, we might reducehby a Factor oF 2 and retake the step. IF±<<τ,we might doublehin preparation For the next step (expecting that the local errormight increase by a Factor oF 24).CHALLENGE 20.11.No answer provided.CHALLENGE 20.12.y±(tD∇yH(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.