When will the alligator population in the swamp reach four dozen?

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SUMMARY

The discussion focuses on solving a differential equation related to the alligator population in a swamp, specifically when it will reach four dozen alligators. The population growth is modeled by the equation dP/dt = kP², where P represents the number of alligators in dozens. The user successfully derived the equation P(t) = -1/((t/20) - 1) and is advised to set P(t) = 4 to find the corresponding time t. This approach confirms the correct understanding of the procedure for solving the problem.

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forest125
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Working on some DE homework and need a tad bit of guidance. I got fairly far, I'm just wondering what the next step is.

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 :)
 
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I think you are doing just fine. Yes, set P(t)=4 and solve for t.
 

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