Solving Logistic Equation Homework Problem

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SUMMARY

The discussion focuses on solving the logistic equation for equilibrium points represented by the formula kP(1 - P/N) - C = 0. The user attempts to derive the quadratic equation -kP² + kNP - CN = 0 and applies the quadratic formula to find P. The discrepancy between the user's solution and the textbook's solution is clarified, highlighting the correct interpretation of the quadratic formula and its application in this context.

PREREQUISITES
  • Understanding of logistic equations and their applications
  • Familiarity with quadratic equations and the quadratic formula
  • Basic knowledge of algebraic manipulation
  • Concept of equilibrium points in mathematical modeling
NEXT STEPS
  • Study the derivation of logistic growth models in mathematical biology
  • Learn about the implications of equilibrium points in population dynamics
  • Explore advanced applications of the quadratic formula in real-world problems
  • Investigate the differences between continuous and discrete models in logistic equations
USEFUL FOR

Students studying mathematical biology, educators teaching logistic growth models, and anyone interested in understanding equilibrium points in population dynamics.

roam
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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:
 
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roam said:
[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]
 
scurty said:
[itex]2k = \sqrt{\text{(what?)}}[/itex]

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

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