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I am new to the forum :) and was hoping that you guys could help me out with the problem i am facing when solving the logistic population model: taking different steps during the integration process seems to produce different results. I am very sorry for the long post :(

BACKGROUND

Here is the original DE model:

[itex]\frac{dN}{dt}[/itex]=rN(1-[itex]\frac{N}{K}[/itex]).

N is population number (endogenous variable), r is the reproduction rate (constant), K is the carrying capacity (contant). The (1-[itex]\frac{N}{K}[/itex]) part ensures that the population growth reduces as the population number approaches the carrying capacity limit.

To solve this DE, we integrate by separating the variables into:

[itex]\int[/itex] [itex]\frac{dN}{N(1-N/K)}[/itex]= [itex]\int[/itex] r dt

To solve the left side of the equation, we do a partial fraction technique which transforms the equation into:

[itex]\int[/itex] [itex]\frac{dN}{N}[/itex] + [itex]\int[/itex] [itex]\frac{dN}{K(1-N/K)}[/itex] = [itex]\int[/itex] r dt

PROBLEM

My problem is with the second integral equation [itex]\int[/itex] [itex]\frac{dN}{K(1-N/K)}[/itex]. It seems that there are two ways to proceed with this equation.

1. The first is to multiply the constant K to the variables within the parentheses: K(1-N/K) which produce (K-N). Thus, we have the following integration equation:

[itex]\int[/itex] [itex]\frac{dN}{K-N}[/itex]

To solve this we use substitution: let u=K-N, so du=-dN, hence dN=-du. Substituting this to the second integration equation produces the following equation:

[itex]\int[/itex] [itex]\frac{-du}{u}[/itex] = - ln u = - ln (K-N)

2. The second approach is to directly substitute u=1-N/K, du=-dN/K, dN=-K du. Substituting this produces:

[itex]\int[/itex] [itex]\frac{-du}{u}[/itex] = - ln u = - ln (1-N/K)

So, the first approach produces ln (K-N), while the second approach produces ln (1-N/K). Hm, I cant help to worry, that these two equation are not equal, are they?

Further combining this with the rest of the equation further produce different solutions.

1. With the first approach:

[itex]\int[/itex] [itex]\frac{dN}{N}[/itex] + [itex]\int[/itex] [itex]\frac{dN}{K(1-N/K)}[/itex] = [itex]\int[/itex] r dt

ln N - ln (K-N) = rt + C

ln [itex]\frac{N}{K-N}[/itex] = rt + C

[itex]\frac{N}{K-N}[/itex] = e[itex]^{rt+C}[/itex]

N(1+ e[itex]^{rt+C}[/itex])=Ke[itex]^{rt+C}[/itex]

N=[itex]\frac{KCe^{rt}}{1+Ce^{rt}}[/itex]

or, dividing by Ce[itex]^{rt}[/itex] some would also write that:

N=[itex]\frac{K}{1+Ce^{-rt}}[/itex]

2. With the second approach:

[itex]\int[/itex] [itex]\frac{dN}{N}[/itex] + [itex]\int[/itex] [itex]\frac{dN}{K(1-N/K)}[/itex] = [itex]\int[/itex] r dt

ln N - ln (1-N/K) = rt + C

ln ([itex]\frac{N}{1-N/K}[/itex]) = rt + C

[itex]\frac{N}{1-N/K}[/itex] = e[itex]^{rt+C}[/itex]

N = (1-N/K) e[itex]^{rt+C}[/itex]

N(1+[itex]e^{rt+C}\frac{N}{K}[/itex])= e[itex]^{rt+C}[/itex]

N=[itex]\frac{Ce^{rt}} {1+\frac{NCe^{rt}}{K}}[/itex]

multiplying by Ke[itex]^{-rt}[/itex] produces

N=[itex]\frac {KC} {Ke^{-rt}+C}[/itex]

QUESTION

In conclusion, taking different approach when solving the initial integral equation [itex]\int[/itex] [itex]\frac{dN}{K(1-N/K)}[/itex] produces different result, which lead to different final equation.

Directly multiplying K into (1-N/K) produces -ln (K-N) as a result of its integration, thus lead to the final solution of: N=[itex]\frac{K}{1+Ce^{-rt}}[/itex].

Directly substituting u=1-N/K, produces -ln (1-N/K) as a result of its integration, thus lead to the final solution of: N=[itex]\frac {KC} {Ke^{-rt}+C}[/itex].

I have checked the literatures, and it seems that different papers use different approaches and lead to different results. The wikipedia article on logistic model for example uses the second equation as the solution of the model, while others uses the first equation (i.e.: http://www.math.northwestern.edu/~mlerma/courses/math214-2-03f/notes/c2-logist.pdf).

I have checked the wolfram alpha math apps to integrate the [itex]\int[/itex] [itex]\frac{dN}{K(1-N/K)}[/itex]. It generates ln (K-N) as the solution, however also notes that this solution can also be written in the form of (1-N/K), as the two equation is equal in some restricted N and K. This confuses me even more. Does this mean that ln (K-N) equal ln (1-N/K)? The differential of both of them is offcourse equal, which is [itex]\frac{1}{K-N}[/itex]. But both equation produce different result, dont they?

In sum, my question is:

a. Why do different approaches of integration (directly multiplying vs directly substituting) produce different result? Is it normal that integration such simple equation produces different solution? How do we know which one is correct? ln(N-K) or ln(1-N/K)?

b. Or, to put it in another way, could it be possible that ln(N-K) equals to ln (1-N/K)? If they are, then the puzzled can be solved, but i dont see how they can be equal. Or more generally, is it possible that N=[itex]\frac{K}{1+Ce^{-rt}}[/itex] equal to N=[itex]\frac {KC} {Ke^{-rt}+C}[/itex].

Please help,,as i am starting to have nightmares and all sort,,haunted by this paradox :surprised

Again, I am very sorry for the long post :(

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# Is ln(N-K) equal to ln(1-N/K);different approach different result,why?

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