Deriving Logistic Population Model (ODE question)

  • #1

Homework Statement


Solve the logistic population model:
[itex] dP/dt=rP(1-P/C); P(0)=P_{0}[/itex]

2. The attempt at a solution
First, I separated variables to get:
[itex]\int \! \frac{1}{P(1-P/C)} \, \mathrm{d}P = \int \! r \, \mathrm{d}t[/itex]

Then, I took the left hand side and split into partial fractions:
(1) - [itex]\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1/C}{1-P/C} \, \mathrm{d}P[/itex]
If I integrate, I get the following:
[itex]\ln(P)-\ln(1-P/C)=\ln(\frac{P}{1-P/C})[/itex] (*)

However, my problem is this. If I take (1) and multiply the second integral by C/C (which should be fine, its 1), I get the following:
[itex]\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1}{C-P} \, \mathrm{d}P[/itex]
Which is...
[itex]\ln(P)-\ln(C-P)=\ln(P/C-P)[/itex] (**)

However, (*) and (**) are not the same. I'm assuming there's something wrong with one of the ways that I integrated considering I came out with two different functions. Are they both correct and all that will change when I do the full problem out is the constant?
 

Answers and Replies

  • #2
Curious3141
Homework Helper
2,843
87

Homework Statement


Solve the logistic population model:
[itex] dP/dt=rP(1-P/C); P(0)=P_{0}[/itex]

2. The attempt at a solution
First, I separated variables to get:
[itex]\int \! \frac{1}{P(1-P/C)} \, \mathrm{d}P = \int \! r \, \mathrm{d}t[/itex]

Then, I took the left hand side and split into partial fractions:
(1) - [itex]\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1/C}{1-P/C} \, \mathrm{d}P[/itex]
If I integrate, I get the following:
[itex]\ln(P)-\ln(1-P/C)=\ln(\frac{P}{1-P/C})[/itex] (*)

However, my problem is this. If I take (1) and multiply the second integral by C/C (which should be fine, its 1), I get the following:
[itex]\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1}{C-P} \, \mathrm{d}P[/itex]
Which is...
[itex]\ln(P)-\ln(C-P)=\ln(P/C-P)[/itex] (**)

However, (*) and (**) are not the same. I'm assuming there's something wrong with one of the ways that I integrated considering I came out with two different functions. Are they both correct and all that will change when I do the full problem out is the constant?

BTW, [itex]\ln(P)-\ln(C-P)=\ln(P/C-P)[/itex] should be written [itex]\ln(P)-\ln(C-P)=\ln(P/(C-P))[/itex].

Both answers are in fact equivalent. Remember that you haven't taken into account the arbitrary constant of integration. Try putting that into the LHS as [itex]\ln k[/itex] in one case and [itex]\ln k'[/itex] in the other. You'll be able to bring everything under the natural log, after which you can exponentiate both sides. Now impose the initial value condition [itex]P = P_0[/itex] at [itex]t= 0[/itex]. You'll find that no matter which form you use, the k and k' will be just right so that your final form is the same.
 
  • #3
Curious3141
Homework Helper
2,843
87
To obviate this sort of issue, I prefer to take the definite integral on both sides. The bounds will be [itex](P_0, P(t))[/itex] on the LHS and [itex](0,t)[/itex] on the RHS.
 

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