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 highorder approximate methods for this problem by using numerical integration. Thus, for example, the simple leftrectangular 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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 Spring '14
 JosephE.Flaherty
 Numerical Analysis, yn, Tn, Numerical ordinary differential equations

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