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First Order Differential Equations - Population With Harvesting = Need Help!

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squance:
Im stuck on the last part of a 3 part question on First Order DEs...

Consider the following differential equation :



as a model for a fish population, where t is in months. The constant h >0 represents the harvesting rate.

***At what tiem does the population die out if h = 100/3 and P(0) = 50????

The answer im supposed to get is 3.63 months but i can't seem to get this answer...

HELP!!  :'(

Collin Li:
Show us how you tried to get the answer.

squance:
Well...

I did this...

dP/dt = P(1 - P/100) - h
dP/dt = P - P^2/100 - (100/3)
dP/dt = 300P - 3P^2 - 10000 (multiply everything to eliminate denominator)
Integral (3P^2 - 300P) dP/dt = Integral (-10000)

Integrate using separation of variables

and I got P^3 - 300P^2/2 = -10000t + C

Then I sub in t = 0, P = 50 into the above equation to get C = -250000

So equation becomes P^3 - 300P^2/2 = -10000t -250000

Then I don't know what to sub in next or whatever but I know im doing it wrong because I can't get the answer





                                                                                                                                                                                                                                 

squance:
And I also did this another way:

dP/dt = P(1- (P/100) - h
dP/dt = P - p^2/100 - 100/3
dP/dt = -1/100(P^2 - 100P + 1/3)

integral 1/ (P^2 - 100P + 3) dP/dt = integral (-1/100)

And when intergrating I get this .

1/p - 100 logP + 3P = -(1/100)t + C

And sub in t = 0, P = 50 and I get -19.87

so 1/p - 100 log P + 3P = -(1/100)t -19.87

Im not sure how we are supposed to find the time if we are already given the time (ie. P(0) = 50 which is t = 0..

Collin Li:

--- Quote from: squance on September 24, 2008, 01:02:50 pm ---dP/dt = P - P^2/100 - (100/3)
dP/dt = 300P - 3P^2 - 10000 (multiply everything to eliminate denominator)
Integral (3P^2 - 300P) dP/dt = Integral (-10000)

--- End quote ---

These steps are incorrect. Firstly, if you are eliminating the denominator, you have to bring it to the LHS as well.

So you should have got:



Then, your third line is also incorrect. If I bring the terms to the LHS, they won't be a factor, they will be added terms. Separation does not work here. You have to divide by the whole quadratic on the RHS:



You should find it is an irreducible quadratic, and hence you can complete the square and integrate it into an function.

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