Answer:
Decreases
Step-by-step explanation:
We need to determine the integral of the DE;
[tex]dP/dt=P(aP-b)[/tex]
[tex]dP=P(aP-b)dt[/tex]
[tex]1/(dP^2-bP)dP=dt[/tex]
We can solve this by integration by parts on the left side. We expand the fraction 1/P²:
[tex]1/(d-b/P)\cdot{P^2} dP[/tex]
let
[tex]u=d-b/P[/tex]
[tex]du/dP=b/P^2[/tex]
[tex]dP=[/tex][tex]\int\limits {P^2/b} \, du[/tex]
[tex]P=lnu/b[/tex]
Substitute u in:
[tex]P=ln(d-b/P)/b[/tex]
Therefore the equation is:
[tex]ln(d-b/P)/b=t[/tex]
We simplify:
[tex]d-b/P=e^b^t[/tex]
[tex]P=b/(d-e^b^t)[/tex]
As t increases to infinity P will decrease