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

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- May 2nd 2013, 10:19 PMmathlover10solution 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

- May 2nd 2013, 10:57 PMchiroRe: 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 ? - May 2nd 2013, 11:39 PMmathlover10Re: 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: 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?

- May 3rd 2013, 01:16 AMmathlover10Re: solution of differential equation for vector field
anyways is it more powerful if you work less rigorously since you can generalize?

- May 3rd 2013, 03:27 AMchiroRe: 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).