- #1
TheBestMilk
- 13
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Homework Statement
Given x' = y - 2
y' = [itex]\frac{1}{4}[/itex]x-[itex]\frac{1}{2}[/itex]
Draw the phase plane.
The Attempt at a Solution
First I found what I believe are the nullclines by taking x'=0 and y'=0.
x' = 0 = y -2
so y = 2.
y' = 0 = [itex]\frac{1}{4}[/itex]x-[itex]\frac{1}{2}[/itex]
so x = 2.
This gives us a stationary point at (2,2).
I understand that I need to plug in points in each quadrant to see what the slopes do, but what I'm confused about is how to find the apparent lines that the points don't cross.
I plotted this in pplane and found two linear diagonals which seemed to be lines of attraction.
How would I find those lines algebraically? I thought maybe x'=y' but that would only give me y = [itex]\frac{1}{4}[/itex]x+[itex]\frac{3}{2}[/itex], which is possible, but there is another line with a negative slope as well.
Any thoughts? Thanks!