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System of ODE's and direction field

  1. May 18, 2009 #1
    Given the system of diffe-eq's:

    [tex]x'(t)=x(t)+9y(t)[/tex]

    [tex]y'(t)=-2x(t)-5y(t)[/tex]

    I solved these ok. My question is, when graphing the solution curves on a direction field, I set up the direction field using the vector:

    [tex](x+9y)\hat{i}+(-2x-5y)\hat{j}[/tex]

    My question is, what is the relationship between this vector, and:

    [tex]\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}[/tex]

    Are they supposed to be equivalent?

    Thanks!
     
  2. jcsd
  3. May 19, 2009 #2

    tiny-tim

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    Hi IniquiTrance! :smile:

    Yes, (dy/dt)/(dx/dt) and dy/dx are the same …

    that's the beauty of using parameters! :wink:
     
  4. May 19, 2009 #3
    Hey Tim,

    Thanks for the response.

    I meant to ask what is the relationship of the vector <x'(t),y'(t)> with dy/dx?
     
  5. May 19, 2009 #4

    tiny-tim

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    Well, <x'(t),y'(t)> is a vector, but dy/dx is just a number …

    I don't see what you're getting at. :confused:
     
  6. May 19, 2009 #5
    Hmm, it seems that dy/dx is the slope of the vector <x'(t),y'(t)>, is that the extent of the relationship between the two?
     
  7. May 20, 2009 #6

    tiny-tim

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    If you go into three or more dimensions, there's some linear-algebra thing that I can't remember,

    but yes, in two dimensions that's just about all there is :smile:
     
  8. May 20, 2009 #7
    Ah ok. Thanks!
     
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