MethodMath
MR
May 16, 2026

How to model population growth using logistic differential equations?

I need to understand the logistic growth model:

dPdt=rP(1PK)\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)

where rr is the growth rate and KK is the carrying capacity.

I know this is a separable ODE, but I'd like help with:

  1. Step-by-step separation of variables and integration using partial fractions
  2. Finding the equilibrium solutions and their stability
  3. What happens when P0>KP_0 > K? Does the population decrease to KK?
  4. How to fit this model to real-world data

Can someone solve:

dPdt=0.5P(1P1000),P(0)=50\frac{dP}{dt} = 0.5P\left(1 - \frac{P}{1000}\right), \quad P(0) = 50

1 answers424 views
Loading comments...

1 Answer

PJ
Prof. James ChenMay 16, 2026 Accepted
Let's solve $\frac{dP}{dt} = 0.5P(1 - P/1000)$ with $P(0) = 50$. Step 1: Separate variables: $$\frac{dP}{P(1 - P/1000)} = 0.5 \, dt$$ Step 2: Partial fractions. Rewrite as: $$\frac{1}{P(1 - P/1000)} = \frac{1}{P} + \frac{1/1000}{1 - P/1000}$$ Integrating: $$\ln|P| - \ln|1 - P/1000| = 0.5t + C$$ $$\ln\left|\frac{P}{1 - P/1000}\right| = 0.5t + C$$ Step 3: Solve for $P$: $$\frac{P}{1 - P/1000} = Ae^{0.5t}$$ Using $P(0) = 50$: $$\frac{50}{1 - 0.05} = \frac{50}{0.95} = A \implies A \approx 52.63$$ $$P(t) = \frac{1000Ae^{0.5t}}{1000 + Ae^{0.5t}} = \frac{1000}{1 + 19e^{-0.5t}}$$ **Equilibrium analysis:** - $P = 0$ is unstable (population grows away from 0) - $P = K = 1000$ is stable (population approaches 1000 from either direction) If $P_0 > 1000$, the growth rate becomes negative (the $1 - P/K$ term is negative), so the population decreases toward $K$. The logistic equation always approaches the carrying capacity in the long run.
Loading comments...
Login or Register to post an answer