Question

This is a calculus 2 question regarding population models.

1000(dp/dt)=p(100-p)

The initial population is 200 individuals. Find p(t)

Answer 1

looks like a calc 1 prob to me. can you not take the integral in that form?

Answer 2

1000dp/(p(100-p)) = dt
dp/(p(100-p)) ->
A/p + B/(100-p) = 1000/(p(100-p))
A = 10 , B = 10

(10/p + 10/(100-p))dp = dt
integral of both sides ..
10ln(p) - 10ln(100-p) = t + C
10ln(p/(100-p)) = t + C
ln(p/(100-p)) = t/10 + C (C is constant so i can divide it by 10 and still call it "C" .. if you want call it now C2)
p/(100-p) = e^(t/10 + C)
p/(100-p) is like -1 + 100/(100-p)
so
-1 + 100/(100-p) = e^(t/10) * e^C
100/(100-p) = Ce^(t/10) + 1
100 = 100Ce^(t/10) + 100 - p(Ce^(t/10) + 1)
100Ce^(t/10) = p(Ce^(t/10) + 1)
p = 100Ce^(t/10) / (Ce^(t/10) + 1)
and you can change it to :
p = 100 - 100/(Ce^(t/10) + 1)
where C = -2 since p(0) = 200