solution of differential equation for vector field

show that the field lines y=y(x) of a vector function F(x,y) are solutions of the differential equation....so since the vector function is like the line whose slope is the derivative of y(x), then can I reason that $\displaystyle \frac{dy}{dx}=m (xo,yo)=\frac{\Delta y}{\Delta x}=\frac{Fy}{Fx}$

Re: solution of differential equation for vector field

Hey mathlover10.

What does Fy mean here? You said F(x,y) is a vector function (I'm assuming F(x,y) = <x,y(x)>) but what does Fy mean F(x,y) * y or dF/dy ?

Re: solution of differential equation for vector field

thanks very much! a curve y=y(x) is a field line of the vector function F(x,y) if at each point (xo,yo) on the curve, F(xo,yo) is tangent to the curve y(x). F(x,y)=IFx(x,y)+jFy(x,y). The vector F(xo,yo) starts at (xo,yo) and has a finite magnitude. show that the field lines are solutions of the differential equation: $\displaystyle \frac{dy}{dx}=\frac{Fy(x,y)}{Fx(x,y)}$ Fy is just the y component of the vector F. but the vector F is not exactly the tangent line to y(x) since it has a smaller range and domain from [xo,x] [yo,y]. since there is a unique tangent line to y(x) at xo,yo, then F(x,y) can be thought of as a line segment with direction of that tangent line and the slopes of the tangent line and F(x,y) should be equal and that is equal to Fy/Fx?

Re: solution of differential equation for vector field

anyways is it more powerful if you work less rigorously since you can generalize?

Re: solution of differential equation for vector field

If you consider it as a parameterization x(t) and y(t), then you can use the result dy/dx = (dy/dt) / (dx/dt).