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I had to solve the equation

$$\frac{dN}{dt} = rN \left( 1 -\frac{N}{K} \right)$$

where $r$ and $K$ are constant and $N(0) = N_0$.

I solved it and got

$$N(t) = \frac{kAe^{rt}}{1 - Ae^{rt}}$$

where $A = \frac{N_0}{k - N_0}$. Now I am being told to sketch the graph of this solution $N(t)$ against $t$ for $t \geq 0$ at fixed values of parameters $r, K$ and for two fixed values on $N_0$, one in the range $0 < N_0 < K$ and the other in the range $N_0 > K$.

How do I do this? I haven't got a clue what the graph would look like.

EDIT: I was thinking, if I divide through by $e^{rt}$ then I get $\frac{kA}{e^{-rt} - A}$ which seems easier to sketch as it now only has $1$ function of $t$, but I'm still unsure how to sketch this.

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Is this a laser rate equation? – Ron Gordon Feb 8 '13 at 1:27
@rlgordonma No, Verhulst Model (of population growth) – Kaish Feb 8 '13 at 17:40
up vote 0 down vote accepted

It is immediate from the differential equation that the rate of change of $N$ is $0$ if either $N=0$ or $N=K$, but if $N$ is between $0$ and $K$, then the rate is positive, so $N$ is increasing. If $N$ is close to either $0$ or $K$, then $dN/dt$ is close to $0$ so $N$ is increasing slowly; if $N$ is far from both (say halfway between) then $N$ is increasing fast.

That much tells you a lot about the graph, and you can see it's true just by thinking about what the differential equation says. So it approaches $K$ more slowly as it approaches it. Therefore you've got a horizontal asymptote at $N=K$, and for the same reason a horizontal asymptote at $N=0$.

If you can show that $N\left(1-\frac KN\right)$ is biggest when $N$ is halfway between $0$ and $K$, that tells you where the curve will be steepest.

If you're a bit more ambitious, try showing you have symmetry about the point at which $N$ is halfway between $0$ and $K$, i.e. if $t_0$ is the time at which $N$ is halfway between, then at time $t_0+t$, $N$ will exceed its value at $t_0$ by the same amount by which it falls short of that value at time $t_0-t$.

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