Sure! Let’s break down the problem step by step and provide explanations for each. --- ### Step 1: **Find the x-nullclines** **Given system:** \[ \frac{dx}{dt} = xy - 4x + x^3 \] \[ \frac{dy}{dt} = -xy - y \] **x-nullclines** are where \(\frac{dx}{dt} = 0\). Set the first equation to zero: \[ xy - 4x + x^3 = 0 \] Factor \(x\) out: \[ x(y - 4 + x^2) = 0 \] So the x-nullclines are: 1. \(x = 0\) 2. \(y - 4 + x^2 = 0 \implies y = 4 - x^2\) **Explanation:** x-nullclines are the locations in the phase plane where the rate of change of \(x\) is zero (i.e., the direction field for \(x\) is horizontal). --- ### Step 2: **Find the y-nullclines** **y-nullclines** are where \(\frac{dy}{dt} = 0\). Set the second equation to zero: \[ -xy - y = 0 \] \[ -y(x + 1) = 0 \] So the y-nullclines are: 1. \(y = 0\) 2. \(x + 1 = 0 \implies x = -1\) **Explanation:** y-nullclines are the locations in the phase plane where the rate of change of \(y\) is zero (i.e., the direction field for \(y\) is vertical). --- ### **Final Answers:** - **x-nullclines:** \(x = 0\) and \(y = 4 - x^2\) - **y-nullclines:** \(y = 0\) and \(x = -1\) Let me know if you'd like to proceed with the next parts or a phase plane sketch!

**Step 1: Find the x-nullclines** *Explanation:* x-nullclines are where \(\frac{dx}{dt} = 0\). The given differential equation for \(x\) is: \[ \frac{dx}{dt} = xy - 4x + x^3 \] Factoring out \(x\): \[ x(y - 4 + x^2) = 0 \] which implies: \[ x = 0 \quad \text{or} \quad y - 4 + x^2 = 0 \] or equivalently: \[ \boxed{ x = 0 \quad \text{and} \quad y = 4 - x^2 } \] This gives the x-nullclines as the vertical line \(x=0\) and the parabola \(y=4 - x^2\). --- **Step 2: Find the y-nullclines** *Explanation:* y-nullclines are where \(\frac{dy}{dt} = 0\). The given equation for \(y\) is: \[ \frac{dy}{dt} = -xy - y \] Factor out \(y\): \[ -y(x + 1) = 0 \] which implies: \[ \boxed{ y=0 \quad \text{or} \quad x = -1 } \] These are the y-nullclines: the horizontal line \(y=0\) and the vertical line \(x=-1\). --- ### **Final answer:** - **x-nullclines:** \(x=0\) and \(y=4 - x^2\) - **y-nullclines:** \(y=0\) and \(x=-1\) This completes the nullcline analysis, which is essential for understanding equilibrium points and phase plane behavior.

**Step 1: Find the equilibrium points (nullclines intersection points).** *Explanation:* Equilibria occur where both \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\) simultaneously. From previous steps, the nullclines are: - \(x=0\) or \(y=4 - x^2\) (x-nullclines) - \(y=0\) or \(x=-1\) (y-nullclines) To find equilibria, we check intersections of these nullclines: 1. **At \(x=0\):** - \(y=0\) (from y-nullcline): intersection point \((0,0)\) - \(y=4 - 0^2 = 4\) (from x-nullcline): point \((0,4)\) 2. **At \(x=-1\):** - \(y=0\) (from y-nullcline): point \((-1,0)\) - \(y=4 - (-1)^2 = 4 - 1=3\) (from x-nullcline): point \((-1,3)\) **Final equilibrium points are:** \[ \boxed{ (0, 0), \quad (0, 4), \quad (-1, 0), \quad (-1, 3) } \] --- **Step 2: Summary of equilibrium points** *Explanation:* These points are where the populations do not change, i.e., the system is at rest. They will serve as critical points in the phase plane analysis, indicating potential stable or unstable equilibria depending on the Jacobian analysis (not asked here). **Final answer:** The equilibria are at \(\boxed{(0,0),\, (0,4),\, (-1,0),\, (-1,3)}\). --- **Summary:** - Nullclines: - \(x=0\), \(y=4 - x^2\) - \(y=0\), \(x=-1\) - Equilibria: \(\boxed{(0,0),\, (0,4),\, (-1,0),\, (-1,3)}\) Let me know if you'd like assistance with the phase plane analysis!

Let's proceed step-by-step with each part of the problem: --- ### **Part (a): Why does it make sense that \(x > 0\) and \(y > 0\) in this system?** **Explanation:** In the context of population models, variables \(x\) and \(y\) typically represent species populations, which are inherently non-negative quantities. Negative populations are biologically meaningless, and the model is often valid only when populations are positive. Therefore, assuming \(x > 0\) and \(y > 0\) makes sense because: - They represent actual populations or densities. - The biological interpretation requires non-negative values. - The dynamics near zero (extinction or invasion thresholds) are often of interest, but the model generally assumes positivity for meaningful analysis. **Final Answer:** Since \(x\) and \(y\) represent species populations, it makes biological sense to consider \(x > 0\) and \(y > 0\) because populations cannot be negative, ensuring the model remains meaningful and consistent with real-world scenarios. --- ### **Part (b): Find the nullclines** **Given system:** \[ \frac{dx}{dt} = xy - 4x + x^3 \] \[ \frac{dy}{dt} = -xy - y \] **Step 1: Find the \(x\)-nullclines (\(\frac{dx}{dt} = 0\))** Set: \[ xy - 4x + x^3 = 0 \] Factor out \(x\): \[ x(y - 4 + x^2) = 0 \] which gives: \[ x = 0 \quad \text{or} \quad y - 4 + x^2 = 0 \implies y = 4 - x^2 \] **Step 2: Find the \(y\)-nullclines (\(\frac{dy}{dt} = 0\))** Set: \[ -xy - y = 0 \] Factor out \(y\): \[ -y(x + 1) = 0 \] which gives: \[ y = 0 \quad \text{or} \quad x = -1 \] **Final Nullclines:** - **x-nullclines:** \(x=0\) and \(y=4 - x^2\) - **y-nullclines:** \(y=0\) and \(x=-1\) --- ### **Part (c): Find the equilibrium points** **Equilibria occur at intersections of nullclines:** - From \(x=0\) nullcline: - Intersect with \(y=0\): \(\boxed{(0,0)}\) - Intersect with \(y=4 - x^2\): At \(x=0\): \[ y = 4 - 0^2 = 4 \] So: \(\boxed{(0,4)}\) - From \(x=-1\): - Intersect with \(y=0\): \(\boxed{(-1,0)}\) - Intersect with \(y=4 - x^2\): At \(x=-1\): \[ y=4 - (-1)^2= 4 - 1=3 \] So: \(\boxed{(-1,3)}\) **Final equilibrium points:** \[ \boxed{ (0,0), \quad (0,4), \quad (-1,0), \quad (-1,3) } \] --- ### **Summary:** - **(a)** It makes sense that \(x > 0\) and \(y > 0\) because they represent populations, which are non-negative in real biological systems. - **(b)** Nullclines: - \(x\)-nullclines: \(x=0\) and \(y=4 - x^2\) - \(y\)-nullclines: \(y=0\) and \(x=-1\) - **(c)** Equilibria: \[ (0,0), \quad (0,4), \quad (-1,0), \quad (-1,3) \] Let me know if you'd like to analyze stability or proceed with the phase plane!

Let's proceed with parts (d) and (e): --- ### **Part (d): Find the equilibria using any method** **From previous parts:** - Equilibrium points are the intersections of nullclines: - \(x=0\) & \(y=0\) → \((0,0)\) - \(x=0\) & \(y=4 - x^2\) → \((0,4)\) - \(x=-1\) & \(y=0\) → \((-1,0)\) - \(x=-1\) & \(y=4 - x^2\): at \(x=-1\): \[ y=4 - (-1)^2= 4 - 1=3 \] → \((-1,3)\) **Therefore, the equilibrium points are:** \[ \boxed{ (0,0), \quad (0,4), \quad (-1,0), \quad (-1,3) } \] **Method used:** - Find nullclines: where derivatives are zero. - Find intersections of nullclines. - Equilibrium points occur at these intersections. --- ### **Part (e): Complete a phase plane analysis** **Approach:** - **Plot nullclines:** - \(x=0\) (vertical line), - \(y=4 - x^2\) (parabola opening downward), - \(y=0\) (horizontal line), - \(x=-1\) (vertical line). - **Mark equilibria:** points found above. - **Determine flow directions in each region:** 1. **Region 1:** \(x>0, y>4 - x^2\) (above the parabola, right of \(x=0\)) - \(\frac{dx}{dt} = xy - 4x + x^3\): - Since \(x>0\), \(x^3>0\). - \(y>4 - x^2\), so \(xy - 4x + x^3\) is likely positive → \(x\) increasing. - \(\frac{dy}{dt} = -xy - y\): - \(x>0\), \(y>0\), so \(\frac{dy}{dt}<0\) → \(y\) decreasing. 2. **Region 2:** \(x>0, y<4 - x^2\) - \(x>0\), \(y<4 - x^2\), so \(xy - 4x + x^3\) may be negative → \(x\) decreasing. - \(\frac{dy}{dt}<0\), so \(y\) decreasing. 3. **Region 3:** \(x<-1, y>0\) - \(x<0\), \(y>0\): - \(\frac{dy}{dt} = -xy - y\): - \(x<0\), \(y>0\), so \(-xy\) is positive → \(\frac{dy}{dt}>0\); \(y\) increasing. - \(\frac{dx}{dt} = xy - 4x + x^3\): - \(x<0\), \(y>0\), \(xy<0\). - \(-4x > 0\) (since \(x<0\)), and \(x^3<0\). - Overall, sign depends, but typically the flow indicates \(x\) increasing or decreasing depending on the region. 4. **Region 4:** \(x>-1, y>0\) - Similar analysis, flow directions depend on the signs. - **Arrows:** Draw arrows indicating flow directions based on signs of derivatives in each region. - **Stability of equilibria:** - At each equilibrium, compute the Jacobian matrix and analyze eigenvalues or use qualitative reasoning to determine stability. --- ### **Summary of phase plane analysis:** - Mark equilibrium points: \((0,0), (0,4), (-1,0), (-1,3)\). - Draw nullclines as solid lines. - Use arrows to indicate the direction of flow in each region based on derivative signs. - Identify stable/unstable points by further analysis (eigenvalues, not requested here). --- **Final note:** Due to the complexity, a detailed sketch involves plotting the nullclines and arrows. The flow will generally move toward stable equilibria or away from unstable ones, depending on the Jacobian eigenvalues. --- **In conclusion:** - The phase plane includes four equilibria. - Nullclines divide the plane into regions with predictable flow directions. - The system's behavior near each equilibrium can be inferred qualitatively. Let me know if you'd like me to perform the Jacobian analysis or prepare a detailed phase portrait!