Question

Consider a fishery in which harvesting is taking placeat a constant rateh. The population at the fisherywill be modeled by

Answer 1

Answer:

The general solution to this differential equation is $P(t) = \frac{Ke^{kt} + h}{k}$

Step-by-step explanation:

We are given the following differential equation:

$\frac{dP}{dt} = kP - h$

Solving by separation of variables:

$\frac{dP}{kP-h} = dt$

Integrating both sides:

$\int \frac{dP}{kP-h} = \int dt$

On the left side, by substitution, u = kP - h, du = kDp, Dp = du/k. Then

$\frac{1}{k} \ln{kP-h} = t + K$

In which K is the constant of integration.

$\ln{kP-h} = kt + K$

$e^{\ln{kP-h}} = e^{kt + K}$

$kP - h = Ke^{kt}$

$kP = Ke^{kt} + h$

$P(t) = \frac{Ke^{kt} + h}{k}$

The general solution to this differential equation is $P(t) = \frac{Ke^{kt} + h}{k}$