Let's solve the problem in **two steps** with explanations at each step, as you requested. --- ### **Step 1: Find the x-nullclines and y-nullclines** #### **(b) Find the x-nullclines** **x-nullclines** are where \(\frac{dx}{dt} = 0\): \[ \frac{dx}{dt} = xy - 4x + x^3 = 0 \] Factor out \(x\): \[ x(y - 4 + x^2) = 0 \] So, the **x-nullclines** are: \[ x = 0 \quad \text{or} \quad y - 4 + x^2 = 0 \] \[ x = 0 \quad \text{or} \quad y = 4 - x^2 \] **Explanation:** The x-nullclines are the set of points in the phase plane where the rate of change of \(x\) is zero. That happens either when \(x = 0\) or when \(y = 4 - x^2\). --- #### **(c) Find the y-nullclines** **y-nullclines** are where \(\frac{dy}{dt} = 0\): \[ \frac{dy}{dt} = x - y = 0 \] \[ x = y \] So, the **y-nullcline** is: \[ y = x \] **Explanation:** The y-nullcline is the set of points where the rate of change of \(y\) is zero, which is the line \(y = x\) in the phase plane. --- ### **Step 2: Find the equilibria** #### **(d) Find the equilibria using any method** The equilibria are where **both** derivatives are zero: \[ \frac{dx}{dt} = 0 \quad \text{and} \quad \frac{dy}{dt} = 0 \] From above: - \(x = 0\) or \(y = 4 - x^2\) - \(y = x\) Substitute \(y = x\) into the x-nullcline: \[ x = 0 \implies (y = x \implies y = 0) \] So, one equilibrium: \((0, 0)\) Now try \(y = 4 - x^2\) and \(y = x\): \[ x = 4 - x^2 \implies x^2 + x - 4 = 0 \implies x = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} \] So, two more equilibria: \[ \left(\frac{-1 + \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2}\right) \quad \text{and} \quad \left(\frac{-1 - \sqrt{17}}{2}, \frac{-1 - \sqrt{17}}{2}\right) \] **Explanation:** Equilibria are where both \(dx/dt\) and \(dy/dt\) are zero. By solving the nullcline equations together, we get three equilibrium points. --- ### **Final Answer (Summary):** 1. **x-nullclines**: \( x = 0 \) and \( y = 4 - x^2 \) 2. **y-nullcline**: \( y = x \) 3. **Equilibria**: \( (0,0) \), \( \left(\frac{-1 + \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2}\right) \), \( \left(\frac{-1 - \sqrt{17}}{2}, \frac{-1 - \sqrt{17}}{2}\right) \) Let me know if you want a sketch of the nullclines and equilibria or if you need help with part (e)!

Let's proceed with **Part (e):** *Complete a phase plane analysis including the equilibrium points, nullclines, and direction arrows.* --- ### **Step 3: Phase Plane Analysis** #### **a) Plot nullclines and equilibrium points** - **x-nullclines:** - \(x=0\) (vertical line) - \(y=4 - x^2\) (a downward-opening parabola) - **y-nullcline:** - \(y=x\) (diagonal line through the origin) - **Equilibria:** - \(E_1 = (0,0)\) - \(E_2 = \left(\frac{-1 + \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2}\right)\) - \(E_3 = \left(\frac{-1 - \sqrt{17}}{2}, \frac{-1 - \sqrt{17}}{2}\right)\) --- #### **b) Determine the stability and direction of flow** - **At \(E_1 = (0,0)\):** - **Check the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):** Near \((0,0)\), \[ \frac{dx}{dt} = xy - 4x + x^3 \] For small \(x,y > 0\), \(xy - 4x \approx -4x\) (since \(xy\) is small). \(\frac{dy}{dt} = x - y \approx 0\) when near the origin. - **Flow:** - To the right of \(x=0\), \(\frac{dx}{dt} \approx -4x < 0\), so flow leftward. - Above \(y = x\), \(\frac{dy}{dt} = x - y < 0\), flow downward. - Below \(y=x\), \(\frac{dy}{dt} > 0\), flow upward. - **Conclusion:** The origin behaves as a saddle or unstable node depending on detailed eigenvalue analysis (not shown here), but generally, flow moves away from the origin along some directions. - **At \(E_2\) and \(E_3\):** - These are symmetric with respect to the origin due to the quadratic form. - Their stability can be assessed via linearization (Jacobian), but generally, depending on eigenvalues, these can be stable or saddle points. --- #### **c) Direction arrows in each region** - **For example:** - **Region near \(x > 0, y > 4 - x^2\):** - \(\frac{dx}{dt}\) positive or negative depending on \(xy - 4x + x^3\). - \(\frac{dy}{dt}\) depends on \(x - y\). - **Within the parabola \(y = 4 - x^2\):** \(\frac{dx}{dt} = 0\), nullcline. - **Along \(y = x\):** \(\frac{dy}{dt} = 0\), nullcline. --- ### **Final Summary:** - **Nullclines:** - \(x=0\) and \(y=4 - x^2\) (x-nullclines) - \(y=x\) (y-nullcline) - **Equilibrium points:** - \((0,0)\) (origin) - \(\left(\frac{-1 + \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2}\right)\) - \(\left(\frac{-1 - \sqrt{17}}{2}, \frac{-1 - \sqrt{17}}{2}\right)\) - **Flow directions:** - From the nullclines and sign analysis, arrows should indicate movement away from unstable points and towards stable points, with flow crossing nullclines accordingly. --- **Note:** To fully visualize, sketch the nullclines, mark the equilibrium points, and draw arrows showing the flow direction based on the sign analysis above. --- Would you like me to generate a detailed sketch or further analyze the stability of each equilibrium?