Question

Please draw a phase diagram of the system:

x' = 6x - x^2 - xy

y' = -2y + xy

The critical points are and their type are:

(0,0) unstable saddle

(6,0) unstable saddle

(2,4) stable/asympotically satble spiral

Answer 1

actually it is not possible to draw but i will give some points how to draw this phase curve

The phase portrait shows trajectories either moving away from the critical point to infinite-distant away (when r > 0), or moving directly toward, and converge to the critical point (when r < 0). The trajectories that are the eigenvectors move in straight lines. The rest of the trajectories move, initially when near the critical point, roughly in the same direction as the eigenvector of the eigenvalue with the smaller absolute value. Then, farther away, they would bend toward the direction of the eigenvector of the eigenvalue with the larger absolute value The trajectories either move away from the critical point to infinite-distant away (when r are both positive), or move toward from infinite-distant out and eventually converge to the critical point (when r are both negative). This type of critical point is called a node. It is asymptotically stable if r are both negative, unstable if r are both positive.

2

When r1 and r2 have opposite signs (say r1 > 0 and r2 < 0) In this type of phase portrait, the trajectories given by the eigenvectors of the negative eigenvalue initially start at infinite-distant away, move toward and eventually converge at the critical point

Every other trajectory starts at infinite-distant away, moves toward but never converges to the critical point, before changing direction and moves back to infinite-distant away. All the while it would roughly follow the 2 sets of eigenvectors. This type of critical point is called a saddle point. It is always unstable.