# Differential Equations - Population Dynamics

1. Jul 26, 2009

### hawks32

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<b

3. The attempt at a solution
I'm having problems figuring out how to set up parts a) and c). I'm horrible at DE, so if any one could help point me in the right direction, it would be greatly appreciated.

2. Jul 26, 2009

### rock.freak667

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.

3. Jul 26, 2009

### hawks32

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?

4. Jul 26, 2009

### djeitnstine

Umm you have a $$P^2$$ in there. You should try partial fractions.

5. Jul 26, 2009

### 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

Did I do something wrong?