System of ODE's and direction field

[IniquiTrance]

Given the system of diffe-eq's:

$$x'(t)=x(t)+9y(t)$$

$$y'(t)=-2x(t)-5y(t)$$

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:

$$(x+9y)\hat{i}+(-2x-5y)\hat{j}$$

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

$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$$

Are they supposed to be equivalent?

Thanks!

 

[tiny-tim]

Hi IniquiTrance!

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

that's the beauty of using parameters!

 

[IniquiTrance]

Hey Tim,

Thanks for the response.

I meant to ask what is the relationship of the vector <x'(t),y'(t)> with dy/dx?

 

[tiny-tim]

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.

 

[IniquiTrance]

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?

 

[tiny-tim]

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

 

[IniquiTrance]

Ah ok. Thanks!