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Topic: [ap-calculus] Re: is this separable?
Replies: 1   Last Post: Mar 3, 2005 9:30 AM

 Richard J Maher Posts: 80 Registered: 12/6/04
[ap-calculus] Re: is this separable?
Posted: Mar 3, 2005 9:30 AM

Hello Gay,

On Wed, 2 Mar 2005, Gay L Lawrimore wrote:

1. Do you think AB will have to solve one like this and
2. I still would like to know how to do this.

dP/dt = 2P(3-P) Show that P(t) = 3/(1+2e ^(-6t)) is a solution.
In previous parts they say sketch a solution which passes through
(0,4) and suppose P(t) = 1, but not in this part.

I'm not sure about 1. but as far as 2. goes, write

dP/(P(P-3)) = -2 dt

Now, 1/(P(P-3)) can be written as

(-1/3)/P + (1/3)/(P-3)

{you can use algebra, partial fractions, or just fooling around}
so that the left hand side then integrates to

ln(abs((P-3)/P)))

and the right hand side integrates to

-6t+C*

where C* is your constant of integration. If you exponentiate both
sides you then get - after combining a couple of steps -

(P-3)/P = Ae^(-6t)

where A is an arbitrary constant. Do some algebra and you get

P = 3/(1-Ae(-6t))

I'm not sure what the (0,4) has to do with anything - I don't have
the text you mentioned - but I do know that if we require P(0)=1, then
A = -2 and we get the solution you mentioned.

Hope this helps.

Dick Maher

Richard J. Maher
Mathematics and Statistics
Loyola University Chicago
6525 N. Sheridan Rd.
Chicago, Illinois 60626
1-773-508-3565
rjm@math.luc.edu

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