Let's tackle the problem step by step: --- ### **Step 1: Find the steady state solutions** The ODE is: \[ \frac{dw}{dt} = (w-3)^2 - 1 \] **Steady states** occur when \(\frac{dw}{dt} = 0\): \[ (w-3)^2 - 1 = 0 \] \[ (w-3)^2 = 1 \] \[ w-3 = \pm 1 \] \[ w = 4 \quad \text{or} \quad w = 2 \] **Explanation:** The steady state solutions are found by setting the derivative to zero and solving for \(w\). The solutions are \(w = 2\) and \(w = 4\). --- ### **Step 2: Stability of steady states and long-term behavior** To determine stability, we analyze the sign of \(\frac{dw}{dt}\) around \(w = 2\) and \(w = 4\). - For \(w < 2\): \((w-3)^2 - 1 > 0\) (since the squared term is always positive and increases as we move away from 3) - For \(2 < w < 4\): \((w-3)^2 - 1 < 0\) - For \(w > 4\): \((w-3)^2 - 1 > 0\) Let's check the direction of the flow: - **At \(w = 2\):** - For \(w\) just less than 2, \(\frac{dw}{dt} > 0\) (increasing). - For \(w\) just greater than 2, \(\frac{dw}{dt} < 0\) (decreasing). - This means \(w = 2\) is **stable**. - **At \(w = 4\):** - For \(w\) just less than 4, \(\frac{dw}{dt} < 0\) (decreasing). - For \(w\) just greater than 4, \(\frac{dw}{dt} > 0\) (increasing). - This means \(w = 4\) is **unstable**. Now, **for the initial condition \(w(0) = 3\)**: \[ \frac{dw}{dt}\Big|_{w=3} = (3-3)^2 - 1 = -1 < 0 \] So, \(w\) will decrease and move towards the stable fixed point at \(w = 2\). **Explanation:** - \(w = 2\) is a **stable** steady state; \(w = 4\) is **unstable**. - Starting at \(w = 3\), the solution will approach \(w = 2\) as \(t \to \infty\). --- ### **Final Answers** - **Steady state solutions:** \(w = 2\) (stable), \(w = 4\) (unstable) - **Limiting value as \(t \to \infty\) for \(w(0) = 3\):** \(\boxed{2}\) --- #### **(Optional: Phase Diagram Sketch Explanation)** - Draw a number line with points at \(w = 2\) and \(w = 4\). - Arrows point **towards** \(w = 2\) (stable) and **away from** \(w = 4\) (unstable). - For \(w < 2\) or \(w > 4\), arrows point **away** from steady states.