Sure! Let's break down the solution step by step for each part of Questions 2 and 3. --- ## **Question 2** Given the differential equation: \[ \frac{dy}{dt} = y(y-1)(3-y) \] ### **2(a): Identify steady states, plot \(\frac{dy}{dt}\) vs \(y\), and show flow directions** #### **Step 1: Find steady states** Steady states occur when \(\frac{dy}{dt} = 0\): \[ y(y-1)(3-y) = 0 \] So, the steady states are: \[ y = 0, \quad y = 1, \quad y = 3 \] #### **Step 2: Plot \(\frac{dy}{dt}\) vs \(y\) and label steady states** Let's analyze the sign of \(\frac{dy}{dt}\) in each region: - For \(y < 0\): Pick \(y = -1\) \[ (-1)((-1)-1)(3-(-1)) = (-1)(-2)(4) = 8 > 0 \] - For \(0 < y < 1\): Pick \(y = 0.5\) \[ (0.5)(0.5-1)(3-0.5) = (0.5)(-0.5)(2.5) = -0.625 < 0 \] - For \(1 < y < 3\): Pick \(y = 2\) \[ (2)(2-1)(3-2) = (2)(1)(1) = 2 > 0 \] - For \(y > 3\): Pick \(y = 4\) \[ (4)(4-1)(3-4) = (4)(3)(-1) = -12 < 0 \] **Phase line (y-axis):** - At \(y = 0, 1, 3\): Steady states - \(y < 0\): \(\frac{dy}{dt} > 0\) (arrow ↑) - \(0 < y < 1\): \(\frac{dy}{dt} < 0\) (arrow ↓) - \(1 < y < 3\): \(\frac{dy}{dt} > 0\) (arrow ↑) - \(y > 3\): \(\frac{dy}{dt} < 0\) (arrow ↓) #### **Stability:** - \(y = 0\): unstable (arrows away) - \(y = 1\): stable (arrows toward) - \(y = 3\): unstable (arrows away) #### **Plot:** Draw \(\frac{dy}{dt}\) versus \(y\): - Label zeros at \(y=0,1,3\) - Show sign of \(\frac{dy}{dt}\) in each region, and add arrows on the y-axis (phase line) indicating direction of flow. --- ### **2(b): Sketch \(y(t)\) for each steady state for initial conditions** For each steady state \(y = a\) (\(a = 0, 1, 3\)), sketch solutions \(y(t)\) for: - \(y(0) = a - 0.25\) - \(y(0) = a\) - \(y(0) = a + 0.25\) **General behavior:** - Near stable point (\(y=1\)): Solutions approach \(y=1\) as \(t\) increases. - Near unstable points (\(y=0, 3\)): Solutions move away from those points. **Sketch:** - For \(a=1\): All solutions with nearby initial conditions approach \(y=1\) as \(t \to \infty\). - For \(a=0\): If \(y(0) < 0\), solution increases towards 0, but for \(y(0) > 0\), it moves away toward 1. - For \(a=3\): If \(y(0) < 3\), solution increases toward 3 but then diverges toward 1; if \(y(0) > 3\), solution decreases toward 3 but then diverges away. --- ## **Question 3** ### **3(a): Linearize \(f(y)\) near each steady state** Given: \[ f(y) = y(y-1)(3-y) = -y^3 + 4y^2 - 3y \] Linearization at \(y=a\): \[ L(y;a) = f'(a)(y-a) + f(a) \] We already have steady states: \(y = 0, 1, 3\). #### **Find \(f'(y)\):** \[ f'(y) = \frac{d}{dy}\left(-y^3 + 4y^2 - 3y\right) = -3y^2 + 8y - 3 \] #### **At each steady state:** - **At \(y=0\):** - \(f(0) = 0\) - \(f'(0) = -3(0)^2 + 8(0) - 3 = -3\) - Linearized equation: \(\frac{dy}{dt} = -3y\) - **At \(y=1\):** - \(f(1) = 0\) - \(f'(1) = -3(1)^2 + 8(1) - 3 = -3 + 8 - 3 = 2\) - Linearized equation: \(\frac{dy}{dt} = 2(y-1)\) - **At \(y=3\):** - \(f(3) = 0\) - \(f'(3) = -3(9) + 8(3) - 3 = -27 + 24 - 3 = -6\) - Linearized equation: \(\frac{dy}{dt} = -6(y-3)\) #### **Solve the linearized equations for each initial condition:** General solution for \(\frac{dy}{dt} = k(y-a)\): \[ y(t) = a + (y_0 - a) e^{kt} \] For initial conditions \(y(0) = a-0.25, a, a+0.25\): - At \(y=0\): \(k = -3, a=0\) \[ y(t) = 0 + (y_0)e^{-3t} \] - At \(y=1\): \(k=2, a=1\) \[ y(t) = 1 + (y_0-1)e^{2t} \] - At \(y=3\): \(k=-6, a=3\) \[ y(t) = 3 + (y_0-3)e^{-6t} \] Plug in each initial condition for each steady state. --- ### **3(b): Plot and compare** - Plot the linearized solutions and the original nonlinear solutions on the same axes. - For \(y=1\), the nonlinear and linearized solutions will match closely when initial conditions are close to \(a=1\), but differ for larger deviations or as \(t\) becomes large. - For \(y=0\) and \(y=3\), the linearized solutions predict exponential departure from the equilibrium, which matches the nonlinear behavior only close to the equilibrium. **Comment:** - For small initial perturbations, linear and nonlinear solutions match well near stable/unstable points for short times. - As \(t\) increases or for initial conditions far from the equilibrium, the nonlinear solutions may diverge significantly from the linear approximation. --- ## **Final Answers (Summary):** ### 2(a) Steady states: \(y=0, 1, 3\). Plot \(\frac{dy}{dt}\) vs \(y\), mark steady states, indicate flow directions. ### 2(b) Sketch \(y(t)\) for each steady state and initial conditions \(a-0.25, a, a+0.25\). ### 3(a) Linearized equations: - At \(y=0\): \(\frac{dy}{dt} = -3y\), \(y(t) = y_0 e^{-3t}\) - At \(y=1\): \(\frac{dy}{dt} = 2(y-1)\), \(y(t) = 1 + (y_0-1) e^{2t}\) - At \(y=3\): \(\frac{dy}{dt} = -6(y-3)\), \(y(t) = 3 + (y_0-3)e^{-6t}\) ### 3(b) Plot these solutions and compare to nonlinear solutions. Linear and nonlinear solutions agree near steady states for small perturbations and short times, but can diverge for large \(t\) or large perturbations. --- If you need **plots** or **further details/calculations**, let me know!

Certainly! Let's proceed step-by-step, providing detailed calculations and explanations for each part of the question. --- # **Question 2** Given differential equation: \[ \frac{dy}{dt} = y(y - 1)(3 - y) \] --- ## **Part 2(a): Find steady states, plot \(\frac{dy}{dt}\) vs \(y\), and analyze flow** ### **Step 1: Find steady states** Steady states occur when \(\frac{dy}{dt} = 0\): \[ y(y - 1)(3 - y) = 0 \] Set each factor to zero: - \(y = 0\) - \(y - 1 = 0 \Rightarrow y = 1\) - \(3 - y = 0 \Rightarrow y=3\) **Steady states are:** \[ \boxed{ \textbf{Y}_s = 0, \quad 1, \quad 3 } \] --- ### **Step 2: Sign analysis of \(\frac{dy}{dt}\) in intervals** Choose test points in each interval: - \(y < 0\): pick \(y = -1\) - \(0 < y < 1\): pick \(y = 0.5\) - \(1 < y < 3\): pick \(y = 2\) - \(y > 3\): pick \(y=4\) Calculate \(\frac{dy}{dt} = y(y-1)(3 - y)\): 1. **At \(y = -1\):** \[ (-1)((-1)-1)(3 - (-1)) = (-1)(-2)(4) = 8 > 0 \] Flow: **upward** (positive \(\frac{dy}{dt}\)) 2. **At \(y=0.5\):** \[ 0.5(0.5 - 1)(3 - 0.5) = 0.5(-0.5)(2.5) = -0.625 < 0 \] Flow: **downward** 3. **At \(y=2\):** \[ 2(2 - 1)(3 - 2) = 2(1)(1) = 2 > 0 \] Flow: **upward** 4. **At \(y=4\):** \[ 4(4 - 1)(3 - 4) = 4(3)(-1) = -12 < 0 \] Flow: **downward** --- ### **Step 3: Draw phase line with flow directions** - At \(y=0\): **unstable** (arrows pointing away from \(0\)) - At \(y=1\): **stable** (arrows pointing toward \(1\)) - At \(y=3\): **unstable** (arrows pointing away from \(3\)) **Flow directions:** - For \(y < 0\): \(\frac{dy}{dt} > 0\): solutions increase toward 0 - Between 0 and 1: \(\frac{dy}{dt} < 0\): solutions decrease toward 0 - Between 1 and 3: \(\frac{dy}{dt} > 0\): solutions increase toward 3 - For \(y > 3\): \(\frac{dy}{dt} < 0\): solutions decrease toward 3 **Final phase line:** ``` <--0--→1--<--3--> ``` with **stability:** - \(y=0\): unstable - \(y=1\): stable - \(y=3\): unstable --- ## **Part 2(b): Sketch \(y(t)\) for initial conditions near steady states** ### **For \(a=0\):** - Initial \(y(0) = -0.25\): Since \(\frac{dy}{dt} > 0\), solution moves **upward** toward \(0\), approaching \(0\) as \(t \to \infty\). - \(y(0) = 0\): stationary solution, stays at 0. - \(y(0) = 0.25\): \(\frac{dy}{dt} < 0\), solution moves **downward** away from 0, toward negative infinity or possibly toward the lower boundary (but since the phase line indicates solutions nearby tend toward 0 from below, in typical models solutions tend to stay near or move away depending on the stability). **Note:** Actually, because \(0\) is *unstable*, solutions near \(0\) will tend to move away from it. Specifically, for initial \(y(0) = 0.25\), solution moves downward away from 0 towards negative infinity. ### **For \(a=1\):** - \(y(0) = 0.75\): \(\frac{dy}{dt} > 0\), solution moves **upward** toward 1. - \(y(0) = 1\): equilibrium, stays at 1. - \(y(0) = 1.25\): \(\frac{dy}{dt} > 0\), solution moves **away** from 1 (unstable), toward 3 or beyond. ### **For \(a=3\):** - \(y(0) = 2.75\): \(\frac{dy}{dt} > 0\), solution moves **away** from 3, toward larger \(y\). - \(y(0) = 3\): equilibrium. - \(y(0) = 3.25\): solution moves **downward** away from 3, toward lower \(y\). --- # **Question 3** Given the cubic: \[ f(y) = y(y-1)(3-y) = - y^3 + 4 y^2 - 3 y \] --- ## **Part 3(a): Linear approximation of \(f(y)\) at each steady state** ### **Step 1: Compute \(f'(y)\):** \[ f'(y) = \frac{d}{dy}(- y^3 + 4 y^2 - 3 y) = -3 y^2 + 8 y - 3 \] --- ### **Step 2: Find \(f(a)\) at each steady state** Since the steady states satisfy \(f(a)=0\): - \(a=0\): \(f(0)=0\) - \(a=1\): \(f(1)=0\) - \(a=3\): \(f(3)=0\) ### **Step 3: Find \(f'(a)\) at each steady state** - At \(a=0\): \[ f'(0) = -3(0)^2 + 8(0) - 3 = -3 \] - At \(a=1\): \[ f'(1) = -3(1)^2 + 8(1) - 3 = -3 + 8 - 3 = 2 \] - At \(a=3\): \[ f'(3) = -3(9) + 8(3) - 3 = -27 + 24 - 3 = -6 \] --- ### **Step 4: Write the linearized functions** The linear approximation of \(f(y)\) near \(a\): \[ L(y; a) = f(a) + f'(a)(y - a) \] Since \(f(a)=0\) at steady states: - **At \(a=0\):** \[ L(y; 0) = -3(y - 0) = -3 y \] - **At \(a=1\):** \[ L(y; 1) = 2(y - 1) \] - **At \(a=3\):** \[ L(y; 3) = -6(y - 3) \] --- ## **Part 3(b): Write the linearized differential equations and solutions** The linearized ODE near \(a\): \[ \frac{dy}{dt} = L(y; a) \] **Solutions:** - Near \(a=0\): \[ \frac{dy}{dt} = -3 y \implies y(t) = y_0 e^{-3 t} \] - Near \(a=1\): \[ \frac{dy}{dt} = 2(y - 1) \implies y(t) = 1 + (y_0 - 1) e^{2 t} \] - Near \(a=3\): \[ \frac{dy}{dt} = -6(y - 3) \implies y(t) = 3 + (y_0 - 3) e^{-6 t} \] ### **Initial conditions:** For each steady state \(a\), initial conditions: - \(y(0) = a - 0.25\) - \(y(0) = a\) - \(y(0) = a + 0.25\) **Corresponding solutions:** 1. **At \(a=0\):** - \(y(0) = -0.25\): \[ y(t) = -0.25 e^{-3 t} \] - \(y(0) = 0\): \[ y(t) = 0 \quad \text{(trivial solution, stays at 0)} \] - \(y(0) = 0.25\): \[ y(t) = 1 + (0.25 - 1) e^{2 t} = 1 - 0.75 e^{2 t} \] 2. **At \(a=1\):** - \(y(0) = 0.75\): \[ y(t) = 1 + (0.75 - 1) e^{2 t} = 1 - 0.25 e^{2 t} \] - \(y(0) = 1\): \[ y(t) = 1 + (1 - 1) e^{2 t} = 1 \] - \(y(0) = 1.25\): \[ y(t) = 1 + (1.25 - 1) e^{2 t} = 1 + 0.25 e^{2 t} \] 3. **At \(a=3\):** - \(y(0) = 2.75\): \[ y(t) = 3 + (2.75 - 3) e^{-6 t} = 3 - 0.25 e^{-6 t} \] - \(y(0) = 3\): \[ y(t) = 3 \] - \(y(0) = 3.25\): \[ y(t) = 3 + (0.25) e^{-6 t} \] --- # **Summary of solutions** | Steady state \(a\) | Initial \(y(0)\) | Solution \(y(t)\) | |---------------------|-------------------|------------------| | 0 | \(-0.25\) | \(-0.25 e^{-3 t}\) | | | \(0\) | \(0\) | | | \(0.25\) | \(1 - 0.75 e^{2 t}\) | | 1 | \(0.75\) | \(1 - 0.25 e^{2 t}\) | | | \(1\) | \(1\) | | | \(1.25\) | \(1 + 0.25 e^{2 t}\) | | 3 | \(2.75\) | \(3 - 0.25 e^{-6 t}\) | | | \(3\) | \(3\) | | | \(3.25\) | \(3 + 0.25 e^{-6 t}\) | --- # **Final notes:** - The **linearized solutions** approximate the behavior of the nonlinear system **near the steady states** for small deviations. - **Comparison:** The solutions near stable equilibria tend to approach the equilibrium (exponentials decay), while solutions near unstable equilibria diverge away exponentially. - For large \(t\), the exponential terms tend to zero, confirming that solutions approach the steady state if stable, or diverge if unstable. --- If needed, I can help you sketch these solutions or interpret their behaviors further!