# Critical Points

1. Aug 11, 2007

### EugP

1. The problem statement, all variables and given/known data
Find all the critical points:

$$\frac{dx}{dt} = 1 - xy$$

$$\frac{dy}{dt} = x - y^3$$

2. Relevant equations
$$\frac{dx}{dt} = 0$$

$$\frac{dy}{dt} = 0$$

3. The attempt at a solution

Here's what I did:

$$1 - xy= 0$$ .......... $$x - y^3 = 0$$

$$1 - xy= 0$$ .......... $$x = y^3$$

$$1 - y^4= 0$$ .......... $$x = y^3$$

$$y= 1, -1$$ .......... $$x = y^3$$

$$y= 1, -1$$ .......... $$x = 1, -1$$

So the critical points are:

$$(1, 1) (1, -1) (-1, 1) (-1, -1)$$

But in the book the answer is only:

$$(1, 1) (-1, -1)$$

Can someone explain to me why?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 11, 2007

### HallsofIvy

It doesn't really matter, but are those caculated from a given function or were you give the "direction field"?

I find your way of writing this a bit peculiar! You know that dx/dt= 1- xy= 0 and dy/dt= x- y3= 0. That's your first line. The second line repeats the first equation and solves the second for x: x= y3.
In the third line you replace the x in the first equation by y3, from the second equation, and have 1- y4= 0. That has the two (real) solutions y= 1 and y= -1. If y= 1, then the first equation becomes 1- x= 0 so x= 1: a critical point is (1,1). If y= -1 then the first equation become 1+ x= 0 so x= -1: a critical point is (-1, -1).

Because you get x= 1 only when y= 1, not when y=-1. Similarly, x= -1 is correct only when y= -1. You are looking for critical points, not just values of x and y that you then combine willy-nilly.

3. Aug 11, 2007

### EugP

Oh now I understand. Thanks a lot!