# Differential Equations - Population Dynamics

by hawks32
Tags: differential, dynamics, equations, population
 P: 3 1. The problem statement The DE governing a fish pop. P(t) with harvesting proportional to the population is given by: P'(t)=(b-kP)P-hP where b>0 is birthrate, kP is deathrate, where k>0, and h is the harvesting rate. Model assumes that the death rate per individual is proportional to the pop. size. An equilibrium point for the DE is a value of P so that P'(t)=0. Find general solution of the DE, when.. a) h>b b) h=b c) h
 HW Helper P: 6,193 P'(t)=(b-kP)P-hP P'(t)=bP-kP2-hP P'(t)= (b-h)P-kP2 P'(t)= ((b-h)-kP)P P'(t)= dP/dt so put it in the form f(P) dP= f(t) dt then integrate both sides.
 P: 3 okay, i worked the integral of dp/dt = ((b-h)-kP)P out as... 1/(b-h) * ln(p/b-h-kP) + C = t is that correct?
PF Patron
P: 619

## Differential Equations - Population Dynamics

Umm you have a $$P^2$$ in there. You should try partial fractions.
 P: 3 I did use partial fractions. 1/((b-h)-kP)P dp Let a = b-h integral of 1/(a-kP)P = integral of A/a-kP + B/P Solved for A & B, A = k/a, B = 1/a So integral (k/a)/(a-kP) + (1/a)/p end up with -(1/a)ln(a-kP) + (1/a)lnP ==> 1/a ln(P/(a-kP)) + C sub back a = b-h 1/(b-h) * ln(p/(b-h-kP)) + C Did I do something wrong?

 Related Discussions Calculus & Beyond Homework 1 Calculus & Beyond Homework 2 Calculus & Beyond Homework 9 Calculus & Beyond Homework 1 Introductory Physics Homework 5