Differential EQ Homework: Model Population Growth with dP/dt = kP(1-P/C)

[Miagi]

Homework Statement

A population increases exponentially in its stages, but cannot continue forever. C = carrying capacity.
Model rate of population change by dP/dt = kP(1-P/C) for P=population size

Homework Equations

1] a population is model dP/dt = 1.2P(1-P/4200)

2] For what value of P is the population increasing and decreasing.

3] What are the equilibrium solutions? what do equilibrium solutions mean for the population.

The Attempt at a Solution

I don't know how to approach this, I try solving the differential EQ but can't.
i tried the phase line test, it tells me that it increases when 0<P<4200?

 

[someperson05]

For 2, consider values of P above your value of 4200. What can you say about the sign of the derivative? Considering what it means in physical terms for P > 4200 may help you to understand what is going on.
(Traditionally you don't need to worry about population less than zero, but if the problem calls for it, that isn't too difficult as well. Anti-population doesn't mean much.)

For 3, think about what it means for something to be at "equilibrium". What does the value of a differential equation have to equal for "equilibrium"?

 

[Miagi]

I did the phase line test. P>4200, the population is decreasing. So if the population gets over the capacity, it will die off because that area/town/city's limit is 4200 capacity.

So the equilibrium is 4200? sorry I kind of don't understand equilibrium. Does that mean its where you set the Diff EQ equal to 0?

 

[HallsofIvy]

"Equilibrium" here means "unchanging"- in other words dP/dt= 0.

Look at the right side: kP(1- P/C). For what values of P is that 0? 1- P/C is positive if 1- P/C> 0, 1> P/C, C> P. Assuming that k is positive then P(1- P/C) will be positive as long as P>0 and P< C. If P> C, 1- PC is negative, and so P(1- P/C) is negative.

So as long as P< C, its derivative is positive so P increases. As soon as P is greater than C, its derivative becomes negative and P decreases. What do you think will happen "in the long run"?