Solving an autonomous system finding critical points

[Rubik]

Homework Statement

Find all critical points and derive the linearised system about each critical point

\dot{y} = [3y₁ + y₁y₂; y₁ + y₂ - y₂²]

Homework Equations

The Attempt at a Solution

Setting

0 = 3y₁ + y₁y₂ so y₂ = -3
0 = y₁ + y₂ - y₂² so y₁ = 12

= [0, 1; 12, -3]
so p = -3 (<0)
q = -12 (<0)
\Delta = 57 (>0)

Is this right that my only critical point is (12,-3) and is that how I determine the nature of the critical point and whether my critical point is stable or unstable with my values for p, q and \Delta?

 

[fzero]

You have quadratic equations, so there are more solutions, and so more critical points.

The nature of the critical points can be determined by studying the linearized systems asked for in the problem. Around a critical point (y_1^{(c)},y_2^{(c)}), expand

y_1 = y_1^{(c)} + \epsilon_1,~~y_2=y_2^{(c)}+\epsilon_2,

and keep the terms linear in \epsilon_{1,2} (and derivatives). Solving the linear system tells you how the complete solution must behave in the neighborhood of the critical points.
Perhaps your p,q,\Delta are somehow relevant to the linear system but you haven't explained how they're defined.

 

[HallsofIvy]

Homework Statement

Find all critical points and derive the linearised system about each critical point

\dot{y} = [3y₁ + y₁y₂; y₁ + y₂ - y₂²]

Homework Equations

The Attempt at a Solution

Setting

0 = 3y₁ + y₁y₂ so y₂ = -3

Either y₂= -3 or y₁= 0.

0 = y₁ + y₂ - y₂² so y₁ = 12

If y₂= -3, then y₁= 12, but if y₁= 0, then y₂- y₂²= 0.

= [0, 1; 12, -3]
so p = -3 (<0)
q = -12 (<0)
\Delta = 57 (>0)

Is this right that my only critical point is (12,-3) and is that how I determine the nature of the critical point and whether my critical point is stable or unstable with my values for p, q and \Delta?