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Math Help - Phase Portrait using Isoclines?

  1. #1
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    Apr 2009
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    Phase Portrait using Isoclines?

    I want to sketch the phase portrait (solution curves) of this DE

    y'=y-x^2

    Could I use isoclines? i.e. something like this

    0=y-x^2
    =>y=x^2

    So i would sketch a parabola with slope 0 hairs on it? Then do the same thing for slope 1.

    1=y-x^2
    =>y=\frac{1}{x^2}

    So i would sketch 1/x^2 with slope 1 hairs on the curve?
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  2. #2
    Super Member
    Joined
    Aug 2008
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    You can just plot small vectors at a rectangular grid of numbers in the x-y plane say from -5 to 5, the slope of the vectors of course is the value y-x^2 at each point so for example at the point (1,1), I'd draw a small line with a slope of 0. At the point (0,2), the line would have a slope of 2 and so forth. Once you understand the meaning of that, then try StreamPlot in Mathematica which plots the vector field of {x'(t), y'(t)} so if I let x=t then I'd need to plot the field {1,y-t^2} like so:

    StreamPlot[{1, y - t^2}, {t, -5, 5}, {y, -5, 5}] and that gives me the phase portrait below which I've also drawn a solution (in red) to the DE for y(0)=1.

    Also check out "Differential Equations" by Blanchard, Devaney, and Hall. It's the best reference for these types of problems.
    Attached Thumbnails Attached Thumbnails Phase Portrait using Isoclines?-phase-portrait.jpg  
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