Solving a Logistic model equation with harvesting I have the following Logistic model equation (left out the values for the constants for simplicity), which I'm unable to solve for $P(t)$.
$\dfrac {dP} {dt} = kP  \left (1- \dfrac P {P_\infty} \right)-H$
If the harvesting constant, $H$ was not present, then the ODE could be solved by Bernoulli's equation. The problem is that I am not sure how to start solving this specific type of differential equation.
Can someone point me in the right direction?
Thank you for your time!
 A: The equation is in separated variables:
$$
\frac{dP}{k\,P  \left (1- \dfrac P {P_\infty} \right)-H}=dt.
$$
A: There are in fact three cases to consider for the logistic model with harvesting.  Multiplying out the expression on the right side of the differential equation produces  $$ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} \ P^2 \ + \ k P \ - \ H \ \ , $$
where I am replacing $ \ P_{\infty} \ $ with $ \ M \ \ , $ since the former label can be misleading in the harvesting model, as we shall see.  We may solve the quadratic polynomial for its zeroes, which will give us the equilibrium solutions of the differential equation (if they exist).  For $ \ P^2 \ - \ MP \ + \ \frac{HM}{k} \ = \  0 \ \ , $ we obtain
$$ P \ \ = \ \ \frac{M \ \pm \ \sqrt{M^2 \ - \ \frac{4HM}{k}}}{2} \ \ . $$
How we factor the right side of the differential equation then depends upon the "size" of the harvesting term.
$ \mathbf{{M^2 - \frac{4HM}{k}}  \ > \ 0 \ \Rightarrow \ H < \frac{kM}{4} \ \ - \ } $ ("subcritical harvesting"):  we have two equilibrium solutions smaller than $ \ M \ $ , the "carrying capacity" in the standard logistic model, that we could call $ \ M_{+} \ $ and $ \ M_{-} \ $ ; the differential equation becomes $ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} · (P - M_{+}) · (P - M_{-}) \ \ , $ which is separable and can be solved by using partial-fraction decomposition; we in turn have three cases for the initial condition, $ \ P_0 < M_{-} \ \ , \ \ M_{-} < P_0 < M_{+} \ \ , \  $ and $  \ P_0 > M_{+} \ \ ; $ we find that $ \ M_{+} \ $ is the "attractor" (stable) equilibrium solution and $ \ M_{-} \ $ is the "repeller" (unstable) equilibrium
$ \mathbf{ H > \frac{kM}{4} \ \ - \ } $ ("over-harvesting"):  the polynomial $ \ P^2 \ - \ MP \ + \ \frac{HM}{k} \ $ cannot be factored in real numbers, so there is no   equilibrium solution; the equation is separable, but we must use "completion-of-squares" to write $ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} · \left(P - \frac{M}{2} \right)^2  \ - \ \left(\frac{4H \ - \ kM}{4} \right) \ \ ; $ we will have an "arctanh-type"  solution, with the population falling to zero ("and beyond") regardless of the initial condition
$ \mathbf{ H = \frac{kM}{4} \ \ - \ } $ ("critical harvesting"):   the differential equation is again separable, and can be written as $ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} · \left(P - \frac{M}{2} \right)^2   \ \ ; $ there is a single critical solution which is a peculiar equilibrium solution sometimes called "semi-stable": $ \frac{M}{2} \ $ is an attractor for $ \ P_0 > \frac{M}{2} \ $ , but is a repeller for $ \ P_0 < \frac{M}{2} \ \ ; $ this can be thought of as the "merger" of the two equilibria in the subcritical case
Note that in all three of these models, a population at a sufficiently low level (or any over-harvested population) will "crash" to zero, since $ \ \frac{dP}{dt} < 0 \ $ at $ \ P = 0 \ $ ; the population will become abruptly extinct, rather than disappearing gradually.
