Trouble creating differential equation from a flow line

  • Thread starter highc1157
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  • #1
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Main Question or Discussion Point

Hi,

I am having trouble creating a differential equation from my flow line that models population of deer. The flow line has two nodes, a sink (attractor), and a source (repeller). I need to come up with a D.E from the flow line.

my "skeleton equation looks like this"

dp/dt = p(1-(p/4))((p/6)-1)(1-(y/8))

but this doesnt work, i've rearranged so many ways and doesnt match the flow line
 

Answers and Replies

  • #2
1,796
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How about:

[tex]\frac{dy}{dt}=y(1-y)(2-y)[/tex]

y=2 is a source and y=1 is a sink. May not match exactly what you want but it does include the type of equilibrium points you want.
 

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