# Re: actually something slightly different...

[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]

Posted by Brad Paul on October 29, 2002 at 08:21:23:

In Reply to: actually something slightly different... posted by colin on October 28, 2002 at 21:58:05:

: : Am I to understand that your question is:

: : What values of y0 will give finite y as t goes to infinity?

: what I'm looking for is the value of the constant coefficient y0 which is a "critical point" in the family of solutions y(t). The point which separates solutions that behave in two quite different ways- in this case becoming either negatively or positively unbounded.

: I have plotted the function using Mathematica, and can see that this critical value is around approx - 0.8865.

: The book wants us to solve for the "exact value of y0"... and it shows the answer as y0 = - Sqrt(Pi)/2.

: But I am unable to solve the equation y(t) which I've provided for y0. Do you know how to do it?

So we are looking for a critical point at t->infinity. Because I'm a
physicist I'm going to do the math like a physicist.

As t-> infinity, Erf[t]->1 therefore we have

y->Sqrt(Pi)/2 (1) Et^2/2-t+y0Et^2/2
as t->infinity

rewrite

y->(Sqrt(Pi)/2 +y0)Et^2/2-t as t->infinity

t, when compared to Et^2/2 at t->infinity is small. Or no
matter what linear coefficient is on Et^2/2 (i.e. a), t
will always be smaller than a Et^2/2 out at infinity. So
let's drop the small and insignificant t.

y->(Sqrt(Pi)/2 +y0)Et^2/2

What value of y0 is the division line between this
coefficient being positive or negative?

y0=-Sqrt(Pi)/2

Name:
E-Mail:

Subject: