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

[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]

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)

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.

[Collin Li]

And I also did this another way:

...

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

This is also wrong:

(in general), which is what you appear to have done.

[squance]

Well...

I've tried what you said...

I completed the square for the 300P-3p^2 -10000

which came out to be -3(p-50)^2 - 2500/3

then I placed it back into the fraction so it becomes...

300/ (-3(p-50)^2-2500/3)

300/-3((P-50)^2 + 2500/9)

which becomes integral(-100/(P-50)^2 + 2500/9)

and when integrate that  i get...

-6 arctan(3(p-50)/50) = t + c

But when I sub t = 0 and P = 50 into equation, it gives me C = 0 and I think that is wrong

But i dunno...im on the verge of giving up on this question and moving on to the next..

[Collin Li]

You almost got there, but you had a few numerical errors on the way...







Therefore:





Using the fact that ,





So, solving for when :

[Collin Li]

Heh, the reason why I didn't do this at the start is because I didn't actually see that was given, which makes the decision-making process more difficult.

The size of determines whether the quadratic is irreducible or not (fork into partial fractions, or recognising an ). Of course, you could just do partial fractions with complex linear roots... (eww)

[squance]

Thanks.

But im still not sure with some of the steps of working you had there.

You almost got there, but you had a few numerical errors on the way...

This line im not sure about...
Did you take the -1/100 out and put it on the right hand side and then multiplied it by the root of the a^2 factor thingy (that is, the 50/sqrt 3).

Im not sure why you used the 50/sqrt 3 thing...i thought we only used that when we start integrating....

[Collin Li]

Yeah, that's exactly what I did.

I used because I foresaw that I would be using this formula:



I put that there because, because on the denominator we have

[squance]

I think I got a little confused because the formula on my formula sheet says

[Collin Li]

Same thing though yeah? is just a constant so that formula has just divided the over to the RHS.

Yeah, I used the other one, because that's what you get in VCE Specialist Maths (the way I'm used to teaching). In practice I would probably use the formula you just presented though, just to avoid the confusion of manipulating the to appear on the numerator.

[squance]

Ok. Thanks for clearing things up.
I know the basic concepts of differential equations but I just have issues with algebraically manipulating things...which usually costs me some marks on tests and asssignements. But hopefully I'll soon be able to do these kinds of questions with ease.

Thanks again. Your awesome