Differental Equation homework (w/ calc 2 intergrals)

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    Calc 2 Homework
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Homework Statement


Carrying capacity N of a population

Assume: small population, dp/dt is proportional to P

if 0<P<N then p(t) increases
if P>N then P(t) decreases
equilibrium means dp/dt = 0


Homework Equations


dP/dT = KP(N-P)

Solve for P(t)


The Attempt at a Solution


∫dP/P(N-P) = ∫kdt ?

My teacher said we'd be using ln(...) and 1/[P(N-P)] = A/P + B/(N-P)


Please help! Thanks. :)
 
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mandymandy said:

Homework Statement


Carrying capacity N of a population

Assume: small population, dp/dt is proportional to P

if 0<P<N then p(t) increases
if P>N then P(t) decreases
equilibrium means dp/dt = 0


Homework Equations


dP/dT = KP(N-P)

Solve for P(t)


The Attempt at a Solution


∫dP/P(N-P) = ∫kdt ?

My teacher said we'd be using ln(...) and 1/[P(N-P)] = A/P + B/(N-P)


Please help! Thanks. :)

You are on the right track. Perhaps you need to review partial fractions to work the integral$$
\int \frac{1}{P(N-P)}\, dP$$Your teacher gave you a pretty good hint there.
 
I can solve it from here I don't know why I was so confused. :) thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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