Posted by Colin on September 13, 2002 at 10:54:00:
In Reply to: initial value problem posted by Colin on September 11, 2002 at 11:12:16:
: (Doing Diff. Eq is very humbling experience)
: Here's the latest problem:
: y' - ty = 2 - t^2, y(0)=y0
: We are asked to "find the critical value of y0 exactly. How does the solution corresponding to this value of y0 behave as t -> Infinity?"
: The solution of the IVP is
: y = (Sqrt(Pi)/2)(e^(t^2/2))Erf(t/Sqrt(2)) + t + y0(e^(t^2/2))
: In this case, the book describes the "critical initial value" as "value that separates solutions that behave in two quite different ways...for instance, solutions that become positively unbounded from those that become negatively unbounded...".
: Further on, it states "the possible points of discontinuity, or singularity, of the solution can be identified (without solving the problem) merely by finding the points of discontinuity of the coefficients". (I'm not sure if this is applicable here...)
: How do I find the "exact critical value of y0"???
: thank you,
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