Let's break down and solve the question step by step: --- ### (a) **Find the fixed points.** Given: \[ \frac{dx}{dt} = (2 - x)(y - x) \] \[ \frac{dy}{dt} = y[2 - x - x^2] \] A fixed point occurs when both derivatives are zero: \[ (2 - x)(y - x) = 0 \] \[ y(2 - x - x^2) = 0 \] Let's solve these equations: #### From the first equation: - \(2 - x = 0 \implies x = 2\) - \(y - x = 0 \implies y = x\) #### From the second equation: - \(y = 0\) - \(2 - x - x^2 = 0 \implies x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0 \implies x = -2, 1\) Combine these conditions: ##### **Case 1:** \(x = 2\) - Plug into the second equation: \(y(2 - 2 - 4) = y(-4) = 0 \implies y = 0\) - So, \(\boxed{(2, 0)}\) is a fixed point. ##### **Case 2:** \(y = x\) - Plug into the second equation: \(x[2 - x - x^2] = 0\), so either \(x = 0\) or \(2 - x - x^2 = 0\). - If \(x = 0\), then \(y = 0\). So \(\boxed{(0, 0)}\) is a fixed point. - If \(2 - x - x^2 = 0\), then \(x = 1\) or \(x = -2\), so \((1, 1)\) and \((-2, -2)\) are fixed points. ##### **Case 3:** \(y = 0\) - Plug into the first equation: \((2 - x)(0 - x) = -x(2 - x) = 0\) - \(x = 0\) or \(x = 2\), both already found. **So the fixed points are:** \[ \boxed{(0,0),\, (2,0),\, (1,1),\, (-2,-2)} \] --- ### (b) **Jacobian, Eigenvalues, and Stability** The Jacobian matrix is: \[ J = \begin{bmatrix} \frac{\partial}{\partial x}\left[(2-x)(y-x)\right] & \frac{\partial}{\partial y}\left[(2-x)(y-x)\right] \\ \frac{\partial}{\partial x}\left[y(2-x-x^2)\right] & \frac{\partial}{\partial y}\left[y(2-x-x^2)\right] \end{bmatrix} \] Calculate the partial derivatives: #### For \(dx/dt = (2-x)(y-x)\): - \(\frac{\partial}{\partial x} = -(y-x) + (2-x)(-1) = -(y-x) - (2-x) = x - y - 2 + x = 2x - y - 2\) - \(\frac{\partial}{\partial y} = (2-x)\) #### For \(dy/dt = y(2-x-x^2)\): - \(\frac{\partial}{\partial x} = y(-1-2x)\) - \(\frac{\partial}{\partial y} = 2-x-x^2\) So, \[ J = \begin{bmatrix} 2x - y - 2 & 2-x \\ y(-1-2x) & 2-x-x^2 \end{bmatrix} \] #### Evaluate at each fixed point: --- #### **1. At (0, 0):** \[ J = \begin{bmatrix} 2(0) - 0 - 2 & 2-0 \\ 0(-1-2\cdot0) & 2-0-0 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 0 & 2 \end{bmatrix} \] **Eigenvalues:** \[ \det\begin{bmatrix} -2-\lambda & 2 \\ 0 & 2-\lambda \end{bmatrix} = (-2-\lambda)(2-\lambda) \] So, \(\lambda_1 = -2\), \(\lambda_2 = 2\). - **Saddle point** (unstable). --- #### **2. At (2, 0):** \[ J = \begin{bmatrix} 2(2) - 0 - 2 & 2-2 \\ 0(-1-2\cdot2) & 2-2-4 \end{bmatrix} = \begin{bmatrix} 4 - 2 & 0 \\ 0 & -4 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & -4 \end{bmatrix} \] **Eigenvalues:** \(2, -4\). - **Saddle point** (unstable). --- #### **3. At (1, 1):** \[ J = \begin{bmatrix} 2(1) - 1 - 2 & 2-1 \\ 1(-1-2\cdot1) & 2-1-1 \end{bmatrix} = \begin{bmatrix} 2 - 1 - 2 & 1 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -3 & 0 \end{bmatrix} \] **Eigenvalues:** Solve \(\lambda^2 + \lambda + 3 = 0\), \[ \lambda = \frac{-1 \pm i\sqrt{11}}{2} \] - **Spiral/Focus (unstable)** (since real part is negative, it's a **spiral sink**). --- #### **4. At (-2, -2):** \[ J = \begin{bmatrix} 2(-2) - (-2) - 2 & 2 - (-2) \\ -2(-1-2\cdot(-2)) & 2 - (-2) - (-2)^2 \end{bmatrix} = \begin{bmatrix} -4 + 2 - 2 & 4 \\ -2(-1+4) & 2 + 2 - 4 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2(3) & 0 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -6 & 0 \end{bmatrix} \] **Eigenvalues:** Solve \(\lambda^2 + 4\lambda + 24 = 0\), \[ \lambda = \frac{-4 \pm i\sqrt{80}}{2} = -2 \pm i\sqrt{20} = -2 \pm i 4.472 \] - **Spiral sink** (since real part is negative). --- ### (c) **Compare with direction field, locations and separatrices** - **Fixed points:** Mark (0,0), (2,0), (1,1), (-2,-2) on the plot. - **Saddles:** (0,0) and (2,0). Show separatrices (stable and unstable manifolds) through these points. - **Spiral sinks:** (1,1) and (-2,-2). Show trajectories spiraling into these points. **Sketch on the direction field:** - At (0,0) and (2,0), draw saddle structure (one stable, one unstable direction). - At (1,1) and (-2,-2), draw spiral arrows into the fixed points. - Draw the stable and unstable manifolds (separatrices) for the saddle fixed points. --- **If you want me to mark these points and sketch the separatrices on the direction field, let me know!**

Let's analyze the nonlinear system step by step, compare the results, and interpret the direction field. --- ### **Given System:** \[ \frac{dx}{dt} = (2 - x)(y - x) \] \[ \frac{dy}{dt} = y(2 - x - x^2) \] --- ### **Part (a): Find the fixed points** **Step 1:** Set derivatives to zero: \[ (2 - x)(y - x) = 0 \] \[ y(2 - x - x^2) = 0 \] **Step 2:** Solve for fixed points: - From the first equation: - \(2 - x = 0 \implies x = 2\) - \(y - x = 0 \implies y = x\) - From the second equation: - \(y = 0\) - \(2 - x - x^2 = 0 \implies x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0 \implies x = -2, 1\) **Step 3:** Find corresponding \(y\): - When \(2 - x = 0 \implies x=2\), substitute into second equations: - \(y(2 - 2 - 4) = y(-4) = 0 \implies y=0\) - Fixed point: \(\boxed{(2, 0)}\) - When \(y = x\): - For \(x=0\), \(y=0\), fixed point \(\boxed{(0,0)}\) - For \(x=1\), \(y=1\), fixed point \(\boxed{(1,1)}\) - For \(x=-2\), \(y=-2\), fixed point \(\boxed{(-2,-2)}\) **Final Fixed Points:** \[ \boxed{ (0, 0), \quad (1, 1), \quad (2, 0), \quad (-2, -2) } \] --- ### **Part (b): Jacobian, Eigenvalues, and Stability** **Step 1:** Compute the Jacobian matrix: \[ J = \begin{bmatrix} \frac{\partial}{\partial x} [(2 - x)(y - x)] & \frac{\partial}{\partial y} [(2 - x)(y - x)] \\ \frac{\partial}{\partial x} [ y (2 - x - x^2) ] & \frac{\partial}{\partial y} [ y (2 - x - x^2) ] \end{bmatrix} \] Calculate each component: - \(dx/dt = (2 - x)(y - x)\): - \(\partial_x = -(y - x) + (2 - x)(-1) = - (y - x) - (2 - x) = - y + x - 2 + x = 2x - y - 2\) - \(\partial_y = 2 - x\) - \(dy/dt = y(2 - x - x^2)\): - \(\partial_x = y(-1 - 2x)\) - \(\partial_y = 2 - x - x^2\) **Step 2:** Evaluate at each fixed point: --- #### **1. At (0, 0):** \[ J = \begin{bmatrix} 2(0) - 0 - 2 & 2 - 0 \\ 0(-1 - 0) & 2 - 0 - 0 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 0 & 2 \end{bmatrix} \] Eigenvalues: \[ \det(J - \lambda I) = (-2 - \lambda)(2 - \lambda) = 0 \] \[ \Rightarrow \lambda_1 = -2, \quad \lambda_2 = 2 \] **Conclusion:** Saddle point (unstable). --- #### **2. At (1, 1):** \[ J = \begin{bmatrix} 2(1) - 1 - 2 & 2 - 1 \\ 1(-1 - 2\cdot1) & 2 - 1 - 1 \end{bmatrix} = \begin{bmatrix} 2 - 1 - 2 & 1 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -3 & 0 \end{bmatrix} \] Eigenvalues from characteristic polynomial: \[ \lambda^2 + \lambda + 3 = 0 \] \[ \lambda = \frac{-1 \pm i \sqrt{11}}{2} \] **Conclusion:** Spiral (focus), with eigenvalues having negative real part → **stable spiral (sink)**. --- #### **3. At (2, 0):** \[ J = \begin{bmatrix} 2(2) - 0 - 2 & 2 - 2 \\ 0(-1 - 2\cdot 2) & 2 - 2 - 4 \end{bmatrix} = \begin{bmatrix} 4 - 2 & 0 \\ 0 & -4 \end{bmatrix} \] Eigenvalues: \[ \lambda = 2, -4 \] **Conclusion:** Saddle point. --- #### **4. At (-2, -2):** \[ J = \begin{bmatrix} 2(-2) - (-2) - 2 & 2 - (-2) \\ -2(-1 - 2(-2)) & 2 - (-2) - (-2)^2 \end{bmatrix} \] Compute step-by-step: - \(2(-2) = -4\) - \(-4 + 2 - 2 = -4\) - \(2 - (-2) = 4\) - For \(\partial_x\): \[ -2(-1 - 2(-2)) = -2(-1 + 4) = -2(3) = -6 \] - For \(\partial_y\): \[ 2 - (-2) - 4 = 2 + 2 - 4 = 0 \] So, \[ J = \begin{bmatrix} -4 & 4 \\ -6 & 0 \end{bmatrix} \] Eigenvalues: \[ \lambda^2 + 4 \lambda + 24 = 0 \] \[ \lambda = -2 \pm i \sqrt{20} \] **Conclusion:** Spiral sink (stable focus). --- ### **Part (c): Interpretation with the direction field** - **Fixed points are located at:** - \(\boxed{(0,0)}\) and \(\boxed{(2,0)}\): saddle points. - \(\boxed{(1,1)}\) and \(\boxed{(-2,-2)}\): spiral sinks (attractors). - **Separable trajectories:** - Near saddle points: trajectories approach along stable manifolds and depart along unstable manifolds, forming separatrices. - Near spiral sinks: trajectories spiral into these points, indicating local stability. - **Sketching all separatrices:** - For saddle points: draw lines representing stable and unstable manifolds passing through these points. - For the spiral sinks: draw trajectories spiraling inward. - **Comparison with the direction field:** - Confirm that fixed points on the graph match the computed locations. - The saddle points are on the axes or lines where trajectories diverge or converge. - The spiral points are centers of attraction with trajectories spiraling inward. --- ### **Summary:** | Fixed Point | Stability | Eigenvalues | Behavior | |--------------|--------------|--------------|----------| | (0, 0) | Saddle (unstable) | \(-2, 2\) | Saddle with diverging trajectories along one eigenvector and converging along another | | (1, 1) | Stable spiral | \(-0.5 \pm i \sqrt{11}/2\) | Spiral sink, trajectories spiral into (1, 1) | | (2, 0) | Saddle | \(2, -4\) | Saddle, trajectories diverge along one direction and converge along another | | (-2, -2) | Stable spiral | \(-2 \pm i \sqrt{20}\) | Spiral sink, trajectories spiral into (-2, -2) | --- ### **Final note:** - The direction field should visually confirm these fixed points and their stability types. - Drawing the separatrices (for saddles) and spiraling trajectories (for sinks) will complete the qualitative analysis. --- **Let me know if you'd like a detailed sketch or further clarification!**