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Karl Valten
Dec 1st, 2007, 05:54:31 PM
Ok, I'm stuck on this problem for Calculus II homework and my TA is not helping out at all. Any math geniuses have any inkling to this:


Using the logistic function P'(t)= kP(N-P)-C, solve the differential equation in the case k = .03, N = 60, C = 33.75

Substituting everything in I get:

dP/dt = (.03)P(60-P)-33.75


Am I on the right track that the final equation should have the general form of:

P(t) = Atan(Bt+C)+D ?


Any help would be appreciated.

Jeseth Cloak
Dec 2nd, 2007, 03:33:18 AM
Since it doesn't seem like you're having much luck, I recommend trying your question at http://answers.yahoo.com instead. I'd help out if I could, but I only know about mathematics as it relates to tax laws...

x_x

Byl Laprovik
Dec 2nd, 2007, 03:35:50 AM
I had Cal II in high school, and never touched the stuff again. Everything past basic derivatives is lost to me. Sorry.

Dasquian Belargic
Dec 2nd, 2007, 08:24:22 AM
I think my brain just exploded a little, having read that.

Ilias Nytrau
Dec 2nd, 2007, 08:33:37 AM
I never took calc. AT ALL.

My brain didn't explode, because it automatically looks at something, tries to figure the thing out, then shrugs it off when it either:

A) figures it out

or

B) gives up.

It went B. But that DOES look somewhat interesting, if not tiring on my almost done graveshift brain. Good luck, though. :lol

Nathanial K'cansce
Dec 2nd, 2007, 02:01:50 PM
If I had any clue where my calc books are (yes, plural) I'd offer some more help. Is the (N-P) in the exponent? Which then means you do have one of those funky forms as an answer. (I'm just not sure which one it is).

If not, I shall smite thee cause you just multiple out and it's an easy integral.

Karl Valten
Dec 2nd, 2007, 02:10:56 PM
If I had any clue where my calc books are (yes, plural) I'd offer some more help. Is the (N-P) in the exponent? Which then means you do have one of those funky forms as an answer. (I'm just not sure which one it is).

If not, I shall smite thee cause you just multiple out and it's an easy integral.

Normally yes, but after putting the differential in standard form, I end up with an inseparable function in the denominator of one integral. So yes, its one of them funky forms, hence the tangent form instead of the natural log form.

I just don't know if I'm right (or which answer is right) because I've never seen this layout before.