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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.
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?
dP/dt=kP
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 :)
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 :)