Find equilibrium points given 2 differential equation

[fireychariot]

Homework Statement

\dot{x} = -pxy + qx, \dot{y} = rxy - sy

where p,q,r and s are positive constants (p does not equal r)

Question is : Determine all the equilibrium points for the system of differential equations given above, expressing your answers in terms of p,q,r and s

The Attempt at a Solution

I do know one point is (0,0) however am stuck in finding out the others. I factorise them to get

x(-py + q) = 0
y(xr - s) = 0

Any hints will be much appreciated

 

[HallsofIvy]

Homework Statement

\dot{x} = -pxy + qx, \dot{y} = rxy - sy

where p,q,r and s are positive constants (p does not equal r)

Question is : Determine all the equilibrium points for the system of differential equations given above, expressing your answers in terms of p,q,r and s

The Attempt at a Solution

I do know one point is (0,0) however am stuck in finding out the others. I factorise them to get

x(-py + q) = 0
y(xr - s) = 0

Okay, and then what? If you are taking a differential equations course, we should be able to assume that you can do basic algebra. You should know "if ab= 0 then either a= 0 or b= 0."

Any hints will be much appreciated

 

[fireychariot]

-py+q=0 so do I say y = p/q how do I find the x coordinate from that? Is that when x =0 or do I substitute it into my second equation?

 

[fireychariot]

-py+q=0 so do I say y = p/q how do I find the x coordinate from that? Is that when x =0 or do I substitute it into my second equation?

I mean y = q/p sorry. What's confusing is the fact that it is variables as the constants instead of numbers so any more hints would be grateful.

 

[HallsofIvy]

?? Constants are numbers.

Your two equations are
x(-py + q) = 0
y(xr - s) = 0

clearly, (0, 0) is a root. In fact if y= 0, -py+q is not 0 (unless q happens to be 0 which I am assuming is not true) so we must have x= 0 also.

If y is NOT 0 then we must have xr- s= 0 so that x= s/r. Putting that into the first equation gives s/r(-py+ q)= 0 so that y= q/p. (s/r, q/p) is also an equilibrium point.

If x is not 0 we can do the same thing but get the same point again.