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Diff Eq: Equilibrium Solution sketching?

  • Thread starter cdotter
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  • #1
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Homework Statement


[itex]\frac{dy}{dt}=r(1-\frac{y}{K})y[/itex]


Homework Equations





The Attempt at a Solution


[itex]0=r(1-\frac{y}{K})y[/itex]
y=0 and K.

Plotting dy/dt vs y, the intercepts would then be (0,0) and (0,K).

The book says "the vertex of the parabola is (K/2, rK/4)." Is this something from algebra that I'm forgetting? How do I know/find this?
 

Answers and Replies

  • #2
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If you plot the equation dy/dt = r(1 - y/K)y, with dy/dt considered to be a function of y, the graph is a parbola that opens downward. The vertex will be on a vertical line midway between the two y intercepts, namely at y = K/2.

This is the same stuff you learned a while back when you were studying the graphs of parabolas.
 
  • #3
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What about something like dy/dt = y(y-1)(y-2). I can easily find the points as y=0, 1, and 2. How do I find the vertices? Is there an easy way with algebra or do I just use maximum/minimum from calculus I?
 
  • #4
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I don't believe there are any algebraic techniques to find the local max or min - you'll need to use calculus.
 
  • #5
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Thank you Mark44. [URL]http://smiliesftw.com/x/bowdown.gif[/URL]
 
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  • #6
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Thank you Mark44. [PLAIN]http://smiliesftw.com/x/bowdown.gif[/QUOTE] [Broken]
That's a really cool smily!
 
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