By Deturck D., Wilf H.
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057196, and since this is in sufficiently close agreement with the previous result, we declare that the iteration has converged. 1). 5y, y(0) = 1 as the column Trap(x). For comparison, we show Midpoint(x) and Exact(x). 25 .. 13316 .. 13282 .. 13315 .. 00 .. 48203 .. 48032 .. 48169 .. 6 Comparison of the methods We are now in possession of three methods for the numerical solution of differential equations. 2) and the midpoint rule yn+1 = yn−1 + 2hyn . 3) In order to compare the performance of the three techniques it will be helpful to have a standard differential equation on which to test them.
3) is again exactly satisfied, as the reader should check. 3) holds once more. How long does this continue? 3) is not 0, but is instead −h3 /2. , that it is an integration rule of order two (“order” is an overworked word in differential equations). It follows by linearity that the rule is exact on any quadratic polynomial. By way of contrast, it is easy to verify that Euler’s method is exact for a linear function, but fails on x2 . 2) is of the form const ∗ h2 ∗ y (X), it is perhaps reasonable to expect the error term for the trapezoidal rule to look like const ∗ h3 ∗ y (X).
40 The Numerical Solution of Differential Equations This kind of situation will come up again and again as we look at more accurate methods, because to obtain greater precision without computing higher derivatives we will get the next approximate value of y from a recurrence formula that may involve not just one or two, but several of its predecessors. To get such a formula started we will have to find several starting values in addition to the one that is given in the statement of the initial-value problem.