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.