# Solving $\frac{dP}{dt}=aP-bP^2$ (population dynamics)

This equation represents the dynamics of a population. I am being asked to explain whether, according to this model, there is a possibility for unbounded growth or guaranteed decay.

Usually, in order to interpret systems like this, I would first find a solution to the differential equation. The problem is, because I cannot express $$\frac{dP}{dt}=aP-bP^2$$ in the form $$\frac{dP}{dt}+f(t)P=g(t)$$, I cannot solve using an integrating factor. Can this equation in fact be solved? Do I even need to solve it in order to answer the question of whether there will be unbounded growth or guaranteed decay?

• Just divide both sides by $aP-bP^2$ and use partial fractions to integrate the left side. – Alex R. May 16 '19 at 5:42
• The equation is separable – Dylan May 17 '19 at 8:53
• This equation is also of the Bernoulli type, setting $Q=P^{-1}$ results in the linear equation $Q'=b-aQ$ which you apparently know how to solve. – Lutz Lehmann May 17 '19 at 19:21

$$\frac {dP} {dt}=aP-bP^2$$ $$\frac {dP} {P(a-bP)}=dt$$ $$(\frac {1} {P} + \frac {b} {a-bP} )dP=adt$$

I think now you can just integrate both sides. There is no need of that form

It would be nice if you mention that $$P$$ represents the population, $$a,b$$ are constants?

$$\frac{dP}{dt} = aP - bP^2$$ is a differential equation, a system which varies with time, hence we say these systems as "$$\textit{Dynamical Systems}$$".

$$\frac{dP}{dt} > 0$$ means the population is increasing with time and $$\frac{dP}{dt} < 0$$ means the population is decreasing.

Observe that for unbounded growth ($$P \rightarrow \infty$$) we need $$\frac{dP}{dt} >0$$ for all times, hence $$\frac{dP}{dt} = aP -bP^2 >0$$ implies $$P>0$$ and $$P< \frac{a}{b}$$. So if $$0 < P < \frac{a}{b}$$, the n the population grows. Also for decay ($$P \rightarrow 0$$) we need $$\frac{dP}{dt} <0$$ or $$aP -bP^2 <0$$ implies $$P>\frac{a}{b}$$, so if population exceed $$\frac{a}{b}$$ then there is decrease in population.

Then we ask ourselves what happens when $$P =\frac{a}{b}$$? when $$P =\frac{a}{b}$$, we have $$\frac{dP}{dt} = 0$$ or the population remains stagnant or fixed and doesnot change with time. It is the steady state solution. Observe that $$P=0$$ is also a steady state solution. Usually, these are known as fixed points of the dynamical system.

If you observe to classify the stability of the fixed point then $$P =\frac{a}{b}$$ is a stable fixed point. Hence if you are in the regime of $$0 then as the population increases it reaches $$P = \frac{a}{b}$$, which is fixed point hence the population remains fixed

If $$p > \frac{a}{b}$$, then the population decreases and at a certain time reaches $$P =\frac{a}{b}$$, and becomes steady again. Hence in both cases we reach steady population. I don't think the over growth or decay is possible here.

Possibility of unbounded growth or decay :

Observe that if $$b=0$$, then we have we have $$\frac{dP}{dt} =aP$$ and if $$a>0$$, the population grows exponentially but now there is no fixed point $$P =\frac{a}{b}$$ to make it steady and hence here the population grows without bounds.

Now if $$a<0$$, then the population decreases indefinitely and tends to zero implying decay of population.

You can solve the system explicitly too as suggested above in comments and answers!

The essence is you can really visualize the dynamics of the system simply by analyzing the system, without actually solving it (this is the case once your dimension of the system increases like here

You can find much here - notes

• So helpful, thank you. – Harman May 17 '19 at 16:42