Find equilibrium points given 2 differential equations

[reality99]

Homework Statement

Given a system of differential equations and asked to find the equilibrium points and classify them.

Homework Equations

Equation 1 ... dx/dt=y(13-x^2-y^2 )
Equation 2 ... dy/dt=12-x(13-x^2-y^2 )

The Attempt at a Solution

I know the solution comes when the derivative is 0, so I set both equations to 0 and tried to solve from there. For equation 1, I get one of the solutions as y=0, x=N (assuming x+y=N). I also get x=+/- sqrt(13-y^2), y=sqrt(13/2). Is this correct for the first equation? I'm having trouble understand how to interpret these...

For equation 2 I am more confused since setting it to 0 ends up having a cubic term equal to a constant. My professor said it's similar to the first one and gave me something about if xyA=0 for the first equation then y(12-xA)=0 for equation 2... I don't really understand what this means/how to apply it. I can get the second equation to (x-1)(x-3)(x+4)+(x(y^2))=0 so if y=0 (which I believe is how the solution is supposed to work out, I just don't know why) then x=-4,1,3... Any help or explanation is greatly appreciated!

 

[Dick]

The first equation tells you 0=y*(13-x^2-y^2). That means either y=0 or (13-x^2-y^2)=0. If 13-x^2-y^2=0 can you have an equilibrium point? Look at the second equation.

 

[reality99]

So the system is inconsistent?

 

[Dick]

So the system is inconsistent?

No. You have just ruled out the possibility that (13-x^2-y^2)=0 at an equilibrium point. There was another possibility suggested by the first equation, wasn't there?

 

[reality99]

No. You have just ruled out the possibility that (13-x^2-y^2)=0 at an equilibrium point. There was another possibility suggested by the first equation, wasn't there?

This gets more confusing the more I think about it... From the condition (13-x^2-y^2)=0, I got x=+/- sqrt(13-y^2), then plugging x back into (13-x^2-y^2)=0 means that y=0 and x=sqrt(13)?? Or am I going in circles?

 

[Dick]

This gets more confusing the more I think about it... From the condition (13-x^2-y^2)=0, I got x=+/- sqrt(13-y^2), then plugging x back into (13-x^2-y^2)=0 means that y=0 and x=sqrt(13)??

(13-x^2-y^2)=0 doesn't give you an equilibrium point. The second equation told you that. So forget about it. The first equation also works if y=0. Pursue that possibility.

 

[reality99]

Ok, I don't know what I was doing before but I think I understand it now. Maybe.

[strike]If y=0, then x=12 or +/-sqrt(13)---> (12,0) and (sqrt(13),0)

another possibility if y=/=0 would be (x, +/-sqrt(13-x^2)) right?[/strike]

EDIT: (because I made a stupid mistake above)

y=0, x=1,3,-4

when y=0 the second equation is 12-x(13-y^2-x^2)=0 which is 12-x(13-x^2)=x^3-13x+12=(x-1)(x-3)(x+4) [I wrote it down wrong the first time I did it above, (12-x)(13-x^2)=/=12-x(13-x^2)...duh]

 

[Dick]

That's it.

 

[reality99]

Thanks so much for the help! I really appreciate it!