frank0520
  • frank0520
can someone help me find/ draw the phase portraits of these two differential equations:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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frank0520
  • frank0520
frank0520
  • frank0520
anonymous
  • anonymous
For an autonomous equation like this one, the first thing to do is determine where the derivative \(\dfrac{dN}{dt}\) disappears - these are your equilibrium points. This clearly happens when either of the following equations hold: \[\begin{cases} rN=0\\[1ex] 1-\dfrac{N}{K}=0\\[1ex] \dfrac{N}{A}-1=0 \end{cases}\]

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anonymous
  • anonymous
Easy enough: the equilibrium points occur for \(N=0,K,A\). Do you have any facts about \(K\) and \(A\)? I assume they're constants, but do you know anything about their relative values? Are they both positive/negative? Is one larger than the other? etc
frank0520
  • frank0520
@SithsAndGiggles heres a picture of the assignment if it make any difference. I have completed everything except #3
anonymous
  • anonymous
So is it \(\dfrac{N}{A}\) or \(\dfrac{A}{N}\)? Your images conflict, but I assume it's the first judging by this wiki page: https://en.wikipedia.org/wiki/Allee_effect#Mathematical_models
anonymous
  • anonymous
Anyway, yes, both \(A\) and \(K\) are constants, and we assume \(00\\ y=3~~\implies~~y'>0\]You get the following phase portrait (in one dimension).|dw:1444188255191:dw|
frank0520
  • frank0520
Its the first one, but the questions is to show that this equation has the same phase portrait as the equation from wiki
anonymous
  • anonymous
Here are some sample solutions for the example.|dw:1444188408781:dw|
frank0520
  • frank0520
|dw:1444189228506:dw| would it look like this
anonymous
  • anonymous
You have three equilibrium points, and so there are four intervals you should be considering: (1) \(y>K\), (2) \(A
anonymous
  • anonymous
Sorry, replace \(y\) with \(N\) in that last comment. So let's say \(N<0\), in particular \(N=-1\). Then \[\frac{dN}{dt}=r(-1)\left( -\frac{1}{A}-1\right)\left(1+\frac{1}{K}\right)\] It looks like \(r\) is a positive constant, so keep that in mind. Since \(A\) is a positive number, the term \(-\dfrac{1}{A}-1\) is also negative. This multiplied by \(-1\) gives a positive number. Multiplied by \(r\), it's still positive. And since \(K>0\), \(1+\dfrac{1}{K}\) is too, and positive times positive is positive. Now let's say \(0
anonymous
  • anonymous
You do the same sort of thing with the other intervals. Pick a convenient test point and check the sign of the derivative. For \(AK\) I would use \(N=2K\).

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