# Sketching a node, can anyone answer this query?

• December 20th 2012, 04:51 AM
darren86
Solved: Sketching a node, can anyone answer this query?
i've got the dynamical system:

xdot=-2x
ydot=-y

Now I have solved the system and found the general solution:

x(t)=Ae(^-2t)v + Be(^-t)u, where v=(1,0), u=(0,1), A,B=constants

Now, it can be shown that dy/dx=y/2x, hence dy/dx=0 only when y=0, and for positive x and y dy/dx>0

however, as x=Ae(^-2t), and y=Be(^-t), we have that if t is negative, x is larger than y, and if t is positive, y is larger than x, implying dy/dx = 0 at some point x>0,y>0 (assuming for example the 2 constants are positive and equal).

How is this possible?
• December 20th 2012, 06:02 AM
BobP
Re: Sketching a node, can anyone answer this query?
Isn't dy/dx = 0 only when y = 0 ?
• December 20th 2012, 06:30 AM
darren86
Re: Sketching a node, can anyone answer this query?
Yes thanks, typo, still the problem is the same, any ideas?
• December 20th 2012, 08:14 AM
BobP
Re: Sketching a node, can anyone answer this query?
If

$\frac{dy}{dx}=\frac{y}{2x},$ then $y^{2}=Cx.$

The parabolas open up to the right, there isn't a point at which $\frac{dy}{dx} = 0$ ?