# Use differential equation logistic model to solve model

How should I go about with solving the mathematical eqn for the following scenario using differential equations?

In a guppy farm, it is estimated that an enclosure is able to support a maximum of 15,000 guppies. The owner started with 2000 guppies two years ago and realizes that their population has doubled presently. The owner is able to start harvesting from the enclosure when the number of guppies reaches 6000. Derive a math model for the guppy population, G(t).

When will the farm reach the harvesting stage for guppies?

I am using the logistic eqn so
$$\frac{dx}{dt} = rx(t)*(1- \frac{x(t)}{15})$$

I am assuming that the growth rate is appx 2000 every 2 years [i.e 1000 every year] so means the growth rate
$$r = \frac{1}{15}$$
where everything is in thousands.

Are my values correct and if so, how do I go ahead with solving that logistic eqn?

Thanks!

Any help will be appreciated!

• Did you try searching for “logistic equation”? There are plenty of questions about how to solve it already, for exampe here: math.stackexchange.com/questions/78560/… Jun 1 at 15:04
• @HansLundmark hi I see, but are the values I have obtained and the following equations correct? :) Jun 1 at 15:10

Using $$N$$ for the max population, set $$y=Nx^{-1}-1$$. Then $$y'=-Nx^{-2}x'=-rNy.$$ The main point to take away is that $$y(t)$$ will be an exponential function. Thus we have \begin{align} y(0)&=\frac{15}2-1=\frac{13}2 \\ y(T)&=q^Ty(0)=\frac{15}4-1=\frac{11}4\implies q^T=\frac{11}{26} \end{align} Now you get to solve $$y(t)=q^ty(0)=\frac{15}{6}-1=\frac32\implies q^t=\frac3{13}$$