# Diff Eq: Equilibrium Solution sketching?

## Homework Statement

$\frac{dy}{dt}=r(1-\frac{y}{K})y$

## The Attempt at a Solution

$0=r(1-\frac{y}{K})y$
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?

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Mark44
Mentor
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.

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?

Mark44
Mentor
I don't believe there are any algebraic techniques to find the local max or min - you'll need to use calculus.

Thank you Mark44. [URL]http://smiliesftw.com/x/bowdown.gif[/URL]

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Mark44
Mentor
Thank you Mark44. [PLAIN]http://smiliesftw.com/x/bowdown.gif[/QUOTE] [Broken]
That's a really cool smily!

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