Unsustainable Populations Differential Equations

Question

I need some help solving this differential equation: $\frac{\text{d}P}{\text{d}t}=kP\left(\frac{A-P}{A}\right)\left(\frac{P-m}{P}\right)$, where $P$ is the population, $t$ is time in years, $A$ is the population capacity, $m$ is the minimum population level required for sustainability, and $k$ is a constant. I think it is possible to solve this with partial fractions, but I'm unsure how.

Could you please provide full working as well.
Thanks for your help in advance.

Answer 1

Simplify the right hand side: $$ \begin{aligned} \frac{dP}{dt}&=kP\left( \frac{A-P}{A}\right)\left(\frac{P-m}{P} \right)\\ &=\frac{k}{A}(A-P)(P-m) \end{aligned} $$ which is a ODE of variables separable type: $$ \int_{P_0}^{P_t}\frac{1}{(A-P)(P-m)}dP=\int_{t_0}^t\frac{k}{A} dt $$ Now you use partial fractions on the left hand side to reduce the integrand to a sum of terms with logarithmic integrals:

$$ \begin{aligned} \frac{1}{(A-P)(P-m)}&=\frac{U}{A-P}+\frac{V}{P-m}\\ &=\frac{U(P-m)+V(A-P)}{(A-P)(P-m)} \end{aligned} $$ and since the coefficient of $P$ in the numerator must be $0$ we are forced to choose $U=V$ when the numerator becomes $U(-m+A)=1$ so $U=\frac{1}{A-m}$.

So the integral on the left becomes: $$ \int_{P_0}^{P_t}\frac{1}{(A-P)(P-m)}dP=\frac{1}{A-m}\left(\int_{P_0}^{P_t}\frac{1}{(A-P)}dP+\int_{P_0}^{P_t}\frac{1}{(P-m)}dP \right) $$

Answer 2

you have $$\frac{\text{d}P}{\text{d}t}=kP\left(\frac{A-P}{A}\right)\left(\frac{P-m}{P}\right)$$ or $$\frac{\text{d}P}{(A-P)(P-m)}=\frac kA\,dt$$ or $$\int_{P_0}^P\frac{\text{d}p}{(A-p)(p-m)}=\int_{t_0}^{t}\frac kA\,dx$$ or $$\frac{-\log (P-A)+\log (P_0-A)+\log (P-m)-\log (P_0-m)}{A-m}=\frac kA (t-t_0)$$ or $$\log\frac{P-m}{P-A}=\log{\frac{P_0-m}{P_0-A}}+k\frac {A-m}A (t-t_0)$$