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

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Collin Li:

--- Quote from: squance on September 24, 2008, 01:15:14 pm ---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

--- End quote ---

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.


--- Quote from: coblin on September 24, 2008, 07:16:40 pm ---You almost got there, but you had a few numerical errors on the way...





--- End quote ---

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....

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