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## Homework Statement

The rate of change of an alligator population P is proportional to the square of P. The swamp contained a dozen alligators in 1988, two dozen in 1998. When will there be four dozen alligators in the swamp? What happens thereafter?

## Homework Equations

dP/dt=kP

## The Attempt at a Solution

Let P=the number of alligators in dozens

Let t=time in years, where 1988=0

dP/dt=kP^2

∫dP/P^2=∫kdt

-1/P=kt+C

P=-1/(kt+C)

Given that P(0)=1, C=-1

For P(10)=2, k=1/20 (2=-1/(10k-1))

So now could I say that P(t)=-1/((t/20)-1), let P(t)=4 and solve for t?

My concern isn't really to get the answer but that I am not understanding the procedure for this. I feel like I'm just stumbling through.

Any help is really appreciated greatly. Thanks :)