Final answer:
To find the critical points of the given system of equations, set both equations equal to zero and solve for x and y. The critical points are (0, 1) and (1/2, 1/2). The type and stability of each critical point can be determined by finding the eigenvalues of the Jacobian matrix at each point, which indicate that both critical points are saddle points.
Step-by-step explanation:
Given the system of equations:
x' = x^2 - 3xy + 2xy = x + y - 1
To find the critical points, we can set both equations equal to zero and solve for x and y:
- x^2 - 3xy + 2xy = 0
- x + y - 1 = 0
Simplifying the first equation, we get:
x^2 - xy = 0
Factoring out x, we have:
x(x - y) = 0
So, either x = 0 or x - y = 0.
For x = 0, substitute this value into the second equation to solve for y:
0 + y - 1 = 0
y = 1.
Therefore, one critical point is (0, 1).
For x - y = 0, substitute x = y into the second equation:
y + y - 1 = 0
2y - 1 = 0
y = 1/2.
Therefore, the other critical point is (1/2, 1/2).
Now let's determine the type and stability of each critical point. To do this, we need to find the eigenvalues of the Jacobian matrix at each critical point. The Jacobian matrix is given by:
Jacobian = [[2x - y, -3x + 2y], [1, 1]].
At the critical point (0, 1), the Jacobian matrix evaluated is:
Jacobian = [[-1, 2], [1, 1]].
The eigenvalues of this matrix are -1 and 1, which indicates a saddle point.
At the critical point (1/2, 1/2), the Jacobian matrix evaluated is:
Jacobian = [[0, -3/2], [1, 1]].
The eigenvalues of this matrix are approximately -0.37 and 1.87, which also indicate a saddle point.
Therefore, both critical points are saddle points.