Stability and Critical points-

[pat666]

Stability and Critical points---Please Help

Homework Statement

I need to know the critical points and stability of this ode
(2xy-5)dx+(x^2+y^2)dy=0

Homework Equations

The Attempt at a Solution

i've graphed this is Mathematica (attached) but I don't know if its stable or what the critical points are.

 

[HallsofIvy]

Your equation can be written as
\frac{dy}{dx}= \frac{5- 2xy}{x^2+ y^2}

Critical points occur where both numerator and denominator are 0. Of course, x^2+ y^2= 0 only at (0, 0) and that does not make 5- 2xy= 0. This equation has NO critical points.

 

[pat666]

Thanks again HallsofIvy - - lifesaver
what about stability?
could you help me out with this one please
dy/dx=(-2x-y^2)/4xy) I put the dy/dx on LHS as per your example.
-2x-y^2=o at (0,0) which does make the denominator 0 as well, so it does have critcal points but what are they?

THANKS!

Also any idea what either of these 2ODE's could be used to model?

 

[HallsofIvy]

As you must said, (0, 0) makes both numerator and denominator 0 so (0, 0) is the only critical point.

I personally find it simpler to think of problems like these as two separate equations. If
\frac{dy}{dx}= \frac{F(x, y)}{G(x,y)}
then, introducing the parameter t, we must have
\frac{dy}{dt}= F(x,y)
\frac{dx}{dt}= G(x,y)

If we think of t as "time", we can think of these equations as describing motion in the xy plane.

"Critcal points" (or, as I would say, "equilibrium" points) are where everything is "balanced"- neither x nor y changes: F(x,y)= 0, G(x,y)= 0.

Now, what happens if x and y are close to a critical point? dx/dt and dy/dt are not 0 so the point will "move" but will it move toward or away from the critical point? If such a point always moves closer to the critical point, that is a stable point. If it can move away from the critical point, that is an unstable point.

Here, you critical point is (0, 0) and your "equations of motion" are
\frac{dx}{dt}= 4xy
and
\frac{dy}{dt}= -(2x+ y^2)

If x is a small positive number and y is 0, that gives
\frac{dx}{dt}= 0
\frac{dy}{dt}= -2x
so the "motion" is really a circulation, not going toward or away from (0, 0).

If y is a small positive number and x is 0, we have
\frac{dx}{dt}= 0
\frac{dy}{dt}= -y^2
downward, so now the point is moving toward (0, 0).

But if y is a small negative number and is 0, because of that square, we have
\frac{dx}{dt}= 0
\frac{dy}{dt}= -y^2
still downward but now way from (0, 0).

Because it is possible to move away, this is an unstable point.

 

[pat666]

Can this be done by looking at the/a graph, I was asked to "comment" on stability and critical points and got the feeling that it didn't require extra math. I have learn't phase planes but I said that a phase plane could not be drawn for this one, but that could be because I don't know the correct mathematica code to do it for this particular one?
Also, hope I'm not asking to much but you seem to be the only person on PF who understands this stuff and is willing to help. Could you take a look at this thread and see what you think PLEEEEEASE: https://www.physicsforums.com/showthread.php?p=3215933#post3215933

Thankyou