# Question #74c42

Mar 20, 2017

See below

#### Explanation:

The DE follows from the verbal description:

$\frac{\mathrm{dn}}{\mathrm{dt}} = k n$

This is separable as follows:

$\frac{1}{n} \frac{\mathrm{dn}}{\mathrm{dt}} = k$

Integrate both sides wrt t:

$\int \frac{1}{n} \frac{\mathrm{dn}}{\mathrm{dt}} \setminus \mathrm{dt} = \int k \setminus \mathrm{dt}$

By chain rule:

$\int \frac{1}{n} \setminus \mathrm{dn} = \int k \setminus \mathrm{dt}$

$\ln n = k t + C$

$n = {e}^{k t + C} = {e}^{k t} \cdot {e}^{C}$

$\implies n = A {e}^{k t}$

For the next part, we have:

$\frac{\mathrm{dn}}{\mathrm{dt}} = k n - p$

For variety we write this as:

$\frac{\mathrm{dn}}{\mathrm{dt}} - k n = - p$

An integrating factor ${e}^{\int - k \mathrm{dt}}$ can make this exact as:

${e}^{- k t} \left(\frac{\mathrm{dn}}{\mathrm{dt}} - k n\right) = - {e}^{- k t} p$

Which is:

$\frac{d}{\mathrm{dt}} \left(n {e}^{- k t}\right) = - {e}^{- k t} p$

So:

$n {e}^{- k t} = \int - p {e}^{- k t} \setminus \mathrm{dt} = \frac{p}{k} {e}^{- k t} + C$

$n = \frac{p}{k} + C {e}^{k t}$

Using the IV's and constants:

$500 = \frac{100}{2} + C {e}^{0} \implies C = 450$

$n = 50 + 450 {e}^{k t}$

There are several practical limitations to the use of this model.

First, there is presumably only so much living space and food in the pond. Therefore the amount of fish that the pond can actually sustain is a limiting factor too.

It should also be the case that the anglers catch more fish as the pond population increases.