Let’s analyze the given system of differential equations and determine the equilibrium points.

System:

$\dot{x} = x^3 - x - 3y$ $\dot{y} = x^3 - 4y$

Equilibrium points:

Equilibrium points occur when both derivatives $\dot{x}$ and $\dot{y}$ are equal to zero: $x^3 - x - 3y = 0 \quad \text{(1)}$ $x^3 - 4y = 0 \quad \text{(2)}$

Step 1: Solve for $y$ in equation (2)

From equation (2): $y = \frac{x^3}{4}$

Step 2: Substitute $y = \frac{x^3}{4}$ into equation (1)

Substitute into equation (1): $x^3 - x - 3\left(\frac{x^3}{4}\right) = 0$ Simplify: $x^3 - x - \frac{3x^3}{4} = 0$ Combine terms: $\frac{4x^3}{4} - \frac{3x^3}{4} - x = 0$ $\frac{x^3}{4} - x = 0$ Factor out $x$: $x\left(\frac{x^2}{4} - 1\right) = 0$ $x\left(\frac{x^2 - 4}{4}\right) = 0$ $x\left(x^2 - 4\right) = 0$

Step 3: Solve for $x$

$x = 0 \quad \text{or} \quad x^2 - 4 = 0$ $x = 0, \quad x = 2, \quad x = -2$

Step 4: Solve for $y$ corresponding to each $x$

From $y = \frac{x^3}{4}$:

• For $x = 0$: $y = \frac{0^3}{4} = 0$
• For $x = 2$: $y = \frac{2^3}{4} = \frac{8}{4} = 2$
• For $x = -2$: $y = \frac{(-2)^3}{4} = \frac{-8}{4} = -2$

Equilibrium points:

The equilibrium points are: $(0, 0), \quad (2, 2), \quad (-2, -2)$

Final Answer:

The system has three equilibrium points.

Would you like me to provide a detailed explanation of the stability of these equilibrium points?

Related Questions:

1. What is the method to classify equilibrium points (e.g., node, saddle)?
2. How do you linearize a system of differential equations around an equilibrium point?
3. What is the significance of equilibrium points in the phase plane of a system?
4. How do you determine the Jacobian matrix for this system at a given point?
5. Can you find the nullclines of this system and interpret their behavior?

Tip:

When analyzing equilibrium points, always substitute the solutions back into the original equations to verify correctness!