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After solving the problem $$\begin{cases}dP = r(P-T)P\\P(0)=P_0\end{cases}$$

where r, T, and P(initial) are constant, I am asked to use the solution to find the limit of P(t) as t goes to infinity where P(initial) < T. Also, what does this mean for the population?

For the answer for the integral, I got

$$\begin{cases}P=P_0e^{rTt+c} + T\end{cases}$$

but feel like the e exponent is incorrect.

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  • $\begingroup$ That is not the solution. Since $P^2$ would yield ~$P_0^2\mathrm{e}^{2(rT+c)}$ which is not balanced - I suggest going back over your calculation. $\endgroup$ – Chinny84 Jan 30 '17 at 21:34
  • $\begingroup$ plus - I know the form of the solution - this is a very well known equation/solution $\endgroup$ – Chinny84 Jan 30 '17 at 21:34
  • $\begingroup$ Initially after integration, I get $$\frac{ln|P-T|-ln|P|}{T} = rt + C$$ How do I go about the simplification? $${ln|P-T|-ln|P|} = Trt + C$$ is there a simplification of constants here? $$ln|1-\frac{T}{P}| = Trt + C$$ I am confused with how the constants translate after inverting with e. @Chinny84 $\endgroup$ – Bryan Chen Jan 30 '17 at 21:47
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You had the correct solution as per your comments. $$ \ln\left|1-\frac{T}{P}\right| = rTt + C $$ we can re-arrange by the following $$ 1-\frac{T}{P} = \mathrm{e}^{rTt + C} = A\mathrm{e}^{rTt} $$ which is simply $$ P(t) = \frac{T}{1-A\mathrm{e}^{rTt}} $$ now placing the initial conditions, namely $P(t=0) = P_0$ we can see $$ P(t=0) = P_0 = \frac{T}{1-A} $$ or $$ 1-A = \frac{T}{P_0} \implies A = 1- \frac{T}{P_0} $$ inserting into the solution we have $$ P(t) = \frac{T}{1 - \left(1- \frac{T}{P_0}\right)\mathrm{e}^{rTt}} $$ So you are know in a position to figure out what happens when $t\to\infty$ when we vary $P_0$ wrt $T$.

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