# Finding critical points for phase planes

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1. Dec 3, 2015

### NiallBucks

1. The problem statement, all variables and given/known data
X'=x-y^2 Y'=x^2 -xy -2x

2. Relevant equations
Find the critical point of this system.

3. The attempt at a solution
I know one point is (0,0).
When I tried it I got sqrt(x)= 2 and sqrt(x)= 1 but the point (1,1) doesn't fit later parts of the question. I'm also aware that since a sqrt can have positive and negative values then (4, -2) can also equal zero. Am I missing something obvious?

2. Dec 3, 2015

### HallsofIvy

Staff Emeritus
Frankly, I don't know what you mean by "When I tried it I got sqrt(x)= 2..." When you tried what? And why write "sqrt(x)= 2" rather than "x= 4"? In any case, a critical point of the system will be where x'= x- y^2= 0 and y'= x^2- xy- 2x= 0. From the first equation, x= y^2. Replacing x with y^2 in the second equation gives y^4- y^3- 2y^2= y^2(y^2- y- 1)= y^2(y+ 1)(y- 2)= 0.

3. Dec 3, 2015

### NiallBucks

Sorry by 'try it' I meant solved for zero and I used sqrt(x) because I was trying to solve it in x instead of y. Your way makes much more sense and solved the problem! Thanks a million

4. Dec 3, 2015

### Staff: Mentor

There are two things wrong with this sentence.
1) By definition, the square root of a nonnegative number represents a single value. For example, $\sqrt{4} = + 2$. Period. You are mistaken to think that $\sqrt{4} = \pm 2$.
2) "then (4, -2) can also equal zero." -- No. (4, -2) is a point in the plane. It cannot equal a single number such as zero.

BTW, (4, -2) is also NOT a solution of the equations below.
Set both X' and Y' to 0 and solve the resulting equations simultaneously.
IOW, solve
$x - y^2 = 0$
$x^2 - xy - 2x = 0$
It helps to use one of the equations to substitute in the other.
There are three solutions, one of which is (0, 0), which you already found. (1, 1) is NOT a solution, nor is (4, -2)..