System of ODE's and direction field

  • #1
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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!
 

Answers and Replies

  • #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:
 
  • #3
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Hey Tim,

Thanks for the response.

I meant to ask what is the relationship of the vector <x'(t),y'(t)> with dy/dx?
 
  • #4
tiny-tim
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I meant to ask what is the relationship of the vector <x'(t),y'(t)> with dy/dx?
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:
 
  • #5
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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?
 
  • #6
tiny-tim
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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?
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:
 
  • #7
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Ah ok. Thanks!
 

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