The problem is...

$$ \frac{dp}{dt} = 10p(1-p),$$ $p(0)=0.1$.

Solve and show that $p(t) \to 1$ as $t\to \infty.$

I know this is probably really simple, I was trying to go down the line of finding a general solution and then imposing the boundary condition. But I can't even see how to find the general solution... Then for the second bit taking the limit of the general solution $p(t)$?


This is the so-called logistic equation, which occurs often in population dynamics and many other contexts. There's a trick which works for this particular equation and is much simpler than separation of variables (in my opinion): change variables to $y(t)=1/p(t)$. Then the nonlinear equation for $p$ turns into an inhomogeneous linear equation for $y$, which can be solved immediately by the usual "homogeneous + particular solution" method (the homogeneous solution is an exponential, and the particular solution is a constant). Since this is tagged as homework, I'll let you have a go at the details yourself.


HINT: The method we can use here is called Separation of Variables. Take all the $p$'s to one side and $t$'s to the other, then integrate both sides like this: $$ \int \frac{dp}{10p(1-p)} = \int dt $$ and now integrate both sides. The right hand side is simply $ t+ C$ where C is some constant. To integrate the left hand side, use partial fractions. Set $$ \frac{1}{10p(1-p) } = \frac{A}{p} + \frac{B}{1-p} $$ and solve for $A $ and $B$, then both terms are easily integrated in terms of natural logs. After that, you will be able to solve for $p$ in terms of $t$ and some constant. Feel free to ask for more help if you need it!

  • $\begingroup$ Hey, I had tried this but I ended up with log(p(1-p))=10t+C when you say sole in terms of t, do you mean put the initial value into this equation? $\endgroup$ – BlueFishi Nov 3 '11 at 12:41
  • $\begingroup$ @BlueFishi: that's not write, you haven't integrate $\int \frac{dp}{10p(1-p)}$ correctly. $\endgroup$ – Ilya Nov 3 '11 at 12:45
  • 1
    $\begingroup$ @BlueFishi:following the ideas suggested above you should find the solution $P(t)=\frac{1}{1-9e^{-10t}}$ which indicates the behavior at infinity $\endgroup$ – Gardel Nov 3 '11 at 12:56
  • $\begingroup$ @Gardel: How did you get to the solution P(t)=1/(1-9e^10t) So far I've got, (p/(1-p))=e^(10(t+c)) If this is even right I can't see any way of changing it into the form p(t)=? $\endgroup$ – BlueFishi Nov 3 '11 at 13:16
  • $\begingroup$ @Gardel: for your proposed solution $P(t)$, we have $P(0) = 1/(1-9) = -1/8 \neq 0.1$. Check your signs! $\endgroup$ – Gerben Nov 3 '11 at 13:17

From your comment, it looks you have been able to integrate correctly, following Ragib's Hint and Gourtaur comment. But now your problem is (to finish the solution) to express $p(t)$. This rest part is a simple algebra. Let me express $p(t)$ in terms of $t$:

$\frac{p}{1-p}=e^{10t+10c}=e^{10t}.e^{10c}=k.e^{10t}$ (where $k=e^{10c}$ is a new constant)

$\Rightarrow \frac{p}{(1-p)+p}=\frac{ke^{10t}}{1+ke^{10t}}$ (I applied $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b+a}=\frac{c}{d+c}$. You can just multiply both sides by $(1-p)$, or cross-multiply and solve for $p$)

$\Rightarrow p=p(t)=\frac{1}{1+k'e^{-10t}}$ (dividing numerator and denominator of the fraction on RHS by $ke^{10t}$ and writing $k'=\frac{1}{k}$)

Now, from using the condition $p(0)=0.1=\frac{1}{10}$, we get $\frac{1}{10}=\frac{1}{1+k'}\Rightarrow k'=9$

Hence, you get $p(t)$ and when $t\to\infty$, $e^{-10t}$ tends to what?...


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