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Differential Equations - Population Dynamics

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hawks32
#1
Jul26-09, 02:29 PM
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<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.
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rock.freak667
#2
Jul26-09, 03:09 PM
HW Helper
P: 6,202
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.
hawks32
#3
Jul26-09, 04:22 PM
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?

djeitnstine
#4
Jul26-09, 04:28 PM
PF Gold
djeitnstine's Avatar
P: 619
Differential Equations - Population Dynamics

Umm you have a [tex]P^2[/tex] in there. You should try partial fractions.
hawks32
#5
Jul26-09, 05:38 PM
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?


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