Logistic Equation

  • Thread starter roam
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  • #1
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Homework Statement



I'm having some difficulty understanding the last step of the following solved problem from my textbook:

We will compute the equilibrium points of [itex]kP \left( 1- \frac{P}{N} \right) - C[/itex].

[itex]kP \left( 1- \frac{P}{N} \right) - C =0[/itex]

[itex]-kP^2+kNP - CN =0[/itex]

The quadratic equations has solutions

[itex]P= \frac{N}{2} \pm \sqrt{\frac{N^2}{4}-\frac{CN}{k}}[/itex]


The Attempt at a Solution



So if we start off by

[itex]-kP^2+kNP - CN =0[/itex]

and using the quadratic equation:

[itex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/itex]

I get:

[itex]P= \frac{-kN \pm \sqrt{(kN)^2-4kC}}{2k}[/itex]

[itex]= \frac{N}{2} \pm \frac{\sqrt{k^2N^2-4kC}}{2k}[/itex]

So, why is my answer different from the one in the textbook? And how can I end up with the solution in the book (that I've posted above)? :confused:
 

Answers and Replies

  • #2
392
17
[itex]P= \frac{-kN \pm \sqrt{(kN)^2-4kC}}{2k}[/itex]

[itex]= \frac{N}{2} \pm \frac{\sqrt{k^2N^2-4kC}}{2k}[/itex]

[itex]2k = \sqrt{\text{(what?)}}[/itex]
 
  • #3
1,271
12
[itex]2k = \sqrt{\text{(what?)}}[/itex]

Oh, I see. I never thought of it. Thank you very much. :smile:
 

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