- #1

- 190

- 0

[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!