The limit of a solution of the logistic equation as time tends to infinity $$ \frac{dP}{dt} = 3P(4 - P),\quad P(0) = 2.$$
What value does $P$ approach as $t$ gets large, ie. as $t \to\infty$.
How do I solve this?
Is the idea to this question to first rearrange the equation so that there is a constant on the Right Hand Side, then you can integrate both sides with respect to $dt$?
Thanks
 A: $$
\frac{dP}{dt} = 3P(4 - P),\quad P(0) = 2.
$$
Two posted answers so far have answered this by solving the differential equation.
If there is any reason for anyone to learn about differential equations, they should learn how to answer this question without solving this equation.
Notice that $P$ is intially between $2$ and $4$.  That means $P(4-P)$ is positive.  That means $P$ is increasing.  As long as $P$ is between $2$ and $4$, then $P$ is increasing.  But notice that if $P$ is more than $4$, then $P$ is decreasing.  And when $P$ is exactly $4$, then $dP/dt$ is zero, so $P$ is constant.
So:


*

*If $P$ is less than $4$ (but more than $0$) then $P$ gets bigger;

*If $P$ is more than $4$ then $P$ gets smaller;

*If $P$ should reach exactly $4$ then $P$ remains constant.


That tells you what $P$ approaches as $t\to\infty$.
A: The rigorous way to solve this would be to solve the differential equation, get an expression for $P(t)$ and then take the limit $\lim_{t \to \infty}P(t)$.
But if you start out by assuming the limit exists, then at the limit, $\frac{dP}{dt}=0$ (i.e. loosely speaking, "P stops changing").
So solve $3P(4-P) = 0$ giving $P=4$ as the limiting value (ignore the other root $P=0$ because $P(0) = 2$ and $P$ is always increasing).
Note that what I did is a "quick and dirty" solution. If you're asked this in homework, a test or an examination, you should start by solving the differential equation. It's a separable first order differential equation.
You start by separating the variables:
$$\frac{dP}{3P(4-P)} = dt$$
Then integrate both sides. I find it easier to set the initial conditions as the lower bound of a definite integral.
$$\int_2^{P(t)}\frac{dP}{3P(4-P)} = \int_0^t 1dt$$
The integrand on the left hand side can be resolved quickly with partial fractions, and the rest is algebra.
A: Rearrange first, as you say.
$$
\frac{dP}{P(4-P)} = 3\, dt\\
\because \frac1{P(4-P)} = \frac1{4}\left( \frac1{P} + \frac1{4-P} \right)\\
\therefore 3\int_0^t dt = \int_{P(0)}^{P(t)} \frac{dP}{P(4-P)} = \frac1{4}\int_{2}^{P} \left( \frac1{P} + \frac1{4-P} \right) dP \\
3t = \frac1{4}\left( \ln \left(\frac{P}{2}\right) - \ln \left(\frac{4-P}{2}\right)\right) = \frac1{4}\ln\frac{P}{4-P} \\
12t = \ln \frac{P}{4-P} \\
$$
Then rearrange this and we get
$$
P = \frac{4}{1+e^{-12t}}
$$
Now as $t\to \infty, e^{-t} \to 0$, so what can you say about P?
