Recent content by hawks32

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...

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?

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...