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The unique solution in the
antifunnel,
, has the value . (This is what a standard fourth-order Runge-Kutta
approximation will give if you solve backwards from **t=-9,u=3** or from
; these points are on the top and the bottom of the
antifunnel, and thus should give upper and lower bounds for . An
error analysis done as in [4, chapter 3], and helped by the program *
Numerical Methods* from * MacMath*, shows that all written decimal
digits of are correct.) Thus a * numerical* solution of
, backward in time, starting with
, remains bounded for a
considerable length of time, as our computer experiments showed---but
eventually it will blow up, since the exact solution is
unstable as .

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**Up:** The Riccati Transformation
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