Deriving Logistic Population Model (ODE question)

[SirPartypants]

Homework Statement

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

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

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

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:
\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1}{C-P} \, \mathrm{d}P
Which is...
\ln(P)-\ln(C-P)=\ln(P/C-P) (**)

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?

 

[Curious3141]

Homework Statement

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

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

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

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:
\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1}{C-P} \, \mathrm{d}P
Which is...
\ln(P)-\ln(C-P)=\ln(P/C-P) (**)

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, \ln(P)-\ln(C-P)=\ln(P/C-P) should be written \ln(P)-\ln(C-P)=\ln(P/(C-P)).

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 \ln k in one case and \ln k' 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 P = P_0 at t= 0. 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.

 

[Curious3141]

To obviate this sort of issue, I prefer to take the definite integral on both sides. The bounds will be (P_0, P(t)) on the LHS and (0,t) on the RHS.