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...
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...