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NSOutWest Mar8-12 08:23 PM

Differential Equations - Logistic Model
 
I have the equation dP/dt = kP(1 - P/A). It is supposed to describe a logistical situatuon involving the carrying capacity of the system.

k is a constant, and A is the carrying capacity of the system. t is time and P is population as a function of time. P(0) = P0. I solved c (the integration constant) to be:

c = -ln|(P0)/(A - P0)|

I'm trying to solve the equation in terms of t.

In my calculus book, a similar equation is given with an explanation.

dP/dt = 0.1P(1 - P/300)

With an initial condition of P(0) = 50, c is found to be ln(1/5). A = 300 and k = 0.1.

I follow along well up to this point.

After solving for c, the book lists the rearranged equation as:

P(t) = 300/(1 + 5e-0.1t)

I don't understand how they went from one equation to the other, the closest I could come with the general equation was:

P(t) = (Aekt + Q)/(1 + ekt) where Q = (P0)/(A - P0)

Which would coincide with an equation of:

(300e0.1t + .2)/(1 + e0.1t)

Which, when graphed, is not equivalent to the equation given by the book.

Can anyone go over how to solve the general equation? I think I'm missing some crucial point.

Thank you!

tjackson3 Mar8-12 09:11 PM

Re: Differential Equations - Logistic Model
 
This equation is separable - meaning that you can get all of the terms involving P on one side (including dP), all of the terms involving t on the other (including dt), and integrate both sides.

So we write the equation as

[tex]\left(\frac{1}{P} + \frac{1/A}{1-P/A}\right)\ dP = k\ dt[/tex]

Integrate both sides to get ln|P| - ln|1-P/A| = kt + c, take the exponential of both sides, and you'll get your answer

NSOutWest Mar8-12 09:17 PM

Re: Differential Equations - Logistic Model
 
That's what I've done. I'm trying to solve for P, and I can't figure out where my method diverges from the book's explanation.

I was hoping someone could show me the steps to solving for P, so I can figure out where my error lies.

tjackson3 Mar8-12 09:24 PM

Re: Differential Equations - Logistic Model
 
Well, exp(ln|P|-ln|1-P/A|) = exp(kt + c)

The right hand side simplifies to exp(kt+c) = exp(kt)*exp(c) = C exp(kt), where C is a constant that we'll determine later.

The left hand side simplifies to exp(ln|P|-ln|1-P/A|) = exp(ln|P|) exp(-ln|1-P/A|) = P/(1-P/A)

So P/(1-P/A) =C exp(kt)
P = (A-P) C exp(kt)/A
P = AC exp(kt) - (PC/A)exp(kt)
P + (PC/A) exp(kt) = AC exp(kt)
P(1 + C/A exp(kt)) = AC exp(kt)
P = AC exp(kt)/(1+C/A exp(kt))

Multiplying the numerator and denominator of the RHS by exp(-kt)

P = AC/(exp(-kt) + C/A)


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