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$$ \frac{dP}{dt} = 3P(4 - P),\quad P(0) = 2.$$

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

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    $\begingroup$ I think one should learn the easy way to find $\lim\limits_{t\to\infty} P$ without actually solving the differential equation. Thus my posted answer does not separate and integrate. One should also learn how to separate and integrate, but it would be wrong to think that's the simplest way to answer the question posed here. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 16 '14 at 2:37
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$$ \frac{dP}{dt} = 3P(4 - P),\quad P(0) = 2. $$

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 $t\to\infty$.

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  • $\begingroup$ If you read my post carefully, then you should see I started out with the "quick and dirty way", which is the same as yours. But yours is inaccurate. You cannot say "When $P$ is more than $4$" because $P$ never actually reaches $4$! And $P$ is always increasing. $\endgroup$ – Deepak Oct 16 '14 at 2:35
  • $\begingroup$ @Deepak : Would you feel better if I had said if $P$ is ever more than $4$? It was a hypothetical to be used in a context in which one has not actually fully solved the equation. $\endgroup$ – Michael Hardy Oct 16 '14 at 2:39
  • $\begingroup$ . . . . ok, I've changed it to "if". ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 16 '14 at 2:40
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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.

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  • $\begingroup$ Definitely the hard way!! $\endgroup$ – Michael Hardy Oct 16 '14 at 2:28
  • $\begingroup$ @MichaelHardy I started out by giving what I consider the "easy way" before going on to the rigorous "hard way". Is there an even easier way than my "easy way"? $\endgroup$ – Deepak Oct 16 '14 at 2:32
  • $\begingroup$ You show that if $P$ reaches $4$ then $P$ remains constant, but you are less than explicit in saying what happens when $P\ne 4$. At any rate I disagree with the proposition that the ONLY rigorous way to do this is to solve the equation. Certainly you would solve the equation if you want to know how fast $P$ approaches $4$, but the argument that stops short of solving the equation can be stated rigorously. $\endgroup$ – Michael Hardy Oct 16 '14 at 2:34
  • $\begingroup$ @MichaelHardy I pointed out a major semantic error in your own answer above. And the only rigorous way to prove that convergence does, in fact, happen as expected is to solve the equation and take the limit. $\endgroup$ – Deepak Oct 16 '14 at 2:37
  • $\begingroup$ I've changed "when" to "if" in my answer, so I hope you'll be happier with that. $\endgroup$ – Michael Hardy Oct 16 '14 at 2:41
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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?

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  • $\begingroup$ Definitely the hard way!! $\endgroup$ – Michael Hardy Oct 16 '14 at 2:26

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