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Math Help - Sketching a node, can anyone answer this query?

  1. #1
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    Solved: Sketching a node, can anyone answer this query?

    i've got the dynamical system:

    xdot=-2x
    ydot=-y

    Now I have solved the system and found the general solution:

    x(t)=Ae(^-2t)v + Be(^-t)u, where v=(1,0), u=(0,1), A,B=constants

    Now, it can be shown that dy/dx=y/2x, hence dy/dx=0 only when y=0, and for positive x and y dy/dx>0

    however, as x=Ae(^-2t), and y=Be(^-t), we have that if t is negative, x is larger than y, and if t is positive, y is larger than x, implying dy/dx = 0 at some point x>0,y>0 (assuming for example the 2 constants are positive and equal).

    How is this possible?
    Last edited by darren86; December 20th 2012 at 11:40 AM.
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  2. #2
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    Re: Sketching a node, can anyone answer this query?

    Isn't dy/dx = 0 only when y = 0 ?
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  3. #3
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    Re: Sketching a node, can anyone answer this query?

    Yes thanks, typo, still the problem is the same, any ideas?
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  4. #4
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    Re: Sketching a node, can anyone answer this query?

    If

    \frac{dy}{dx}=\frac{y}{2x}, then y^{2}=Cx.

    The parabolas open up to the right, there isn't a point at which \frac{dy}{dx} = 0 ?
    Last edited by BobP; December 20th 2012 at 12:02 PM.
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