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# E y f t y0 y0 0 this problem which is of little

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Unformatted text preview: is of little interest, can be solved by quadrature to yield Zt y(t) = y0 + f ( )d : 0 We can easily construct high-order approximate methods for this problem by using numerical integration. Thus, for example, the simple left-rectangular rule would lead to Euler's method. The midpoint rule with a step size of h would give us y(h) = y0 + hf (h=2) + O(h3): Thus, by shifting the evaluation point to the center of the interval we obtained a higherorder approximation. Neglecting the local error term and generalizing the method to the interval tn 1 < t tn yields ; yn = yn 1 + hf (tn 1 + h=2): ; ; 2 Runge 21] sought to extend this idea to true di erential equations having the form of (3.1.1). Thus, we might consider yn = yn 1 + hf (tn ; h=2 yn ; 1=2 ; ) as an extension of the simple midpoint rule to (3.1.1). The question of how to de ne the numerical solution yn 1=2 at the center of the interval remains unanswered. A simple possibility that immediately comes to mind is to evaluate it by Euler's method. This gives yn 1=2 = yn 1 + h f (tn 1 yn 1) 2 ; ; ; ; ;...
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