## Step 1: Analyze the Differential Equation Given: \[ \frac{dy}{dx} = y - y^3 \] **Find the equilibrium points:** Set \(\frac{dy}{dx} = \): \[ y - y^3 = \implies y(1 - y^2) = \implies y = , \pm1 \] - These are the equilibrium solutions. **Stability Analysis:** - For \(y > 1\): \(y - y^3 < \) (since \(y^3 > y\)), so \(y\) decreases. - For \( < y < 1\): \(y - y^3 > \), so \(y\) increases. - For \(-1 < y < \): \(y - y^3 < \), so \(y\) decreases. - For \(y < -1\): \(y - y^3 > \), so \(y\) increases. **Explanation:** We determined the equilibrium points \(y = -1, , 1\), and the direction of the solution's movement depending on the value of \(y_\). --- ## Step 2: Sketch and Describe Typical Solution Behaviors ### (a) \(y_ > 1\) - Solution **decreases toward** \(y=1\) but never crosses it. ### (b) \( < y_ < 1\) - Solution **increases toward** \(y=1\). ### (c) \(-1 < y_ < \) - Solution **decreases toward** \(y=-1\). ### (d) \(y_ < -1\) - Solution **increases toward** \(y=-1\). **Explanation:** - \(y=1\) and \(y=-1\) are stable equilibrium points (attractors). - \(y=\) is an unstable equilibrium (repellor). --- ## Final Answer - **For \(y_ > 1\):** Solution decreases, approaching \(y=1\) from above. - **For \( < y_ < 1\):** Solution increases, approaching \(y=1\) from below. - **For \(-1 < y_ < \):** Solution decreases, approaching \(y=-1\) from above. - **For \(y_ < -1\):** Solution increases, approaching \(y=-1\) from below. **Graphically:** All solutions tend toward \(y=1\) (if they start above ) or toward \(y=-1\) (if they start below ). The line \(y=\) is unstable.

# Analysis of Autonomous First-Order Differential Equation The given differential equation is: \[ \frac{dy}{dx} = y - y^3 \] ## Step 1: Find Equilibrium Points and Stability ### Equilibrium Points Set \(\frac{dy}{dx} = 0\): \[ y - y^3 = 0 \implies y(1 - y^2) = 0 \] The solutions are: - \(y = 0\) - \(y = 1\) - \(y = -1\) ### Stability Analysis - **For \(y > 1\)**: \(y - y^3 < 0\) (decreases). - **For \(0 < y < 1\)**: \(y - y^3 > 0\) (increases). - **For \(-1 < y < 0\)**: \(y - y^3 < 0\) (decreases). - **For \(y < -1\)**: \(y - y^3 > 0\) (increases). ### Explanation - The equilibrium points are classified as: - **Stable:** \(y = 1\) and \(y = -1\) (solutions approach these points). - **Unstable:** \(y = 0\) (solutions move away from this point). --- ## Step 2: Sketch Typical Solution Behaviors Based on Initial Conditions ### (a) \(y_0 > 1\) - The solution **decreases** toward \(y = 1\). ### (b) \(0 < y_0 < 1\) - The solution **increases** toward \(y = 1\). ### (c) \(-1 < y_0 < 0\) - The solution **decreases** toward \(y = -1\). ### (d) \(y_0 < -1\) - The solution **increases** toward \(y = -1\). ### Explanation - Solutions starting above \(y = 1\) will decrease, while those starting between \(0\) and \(1\) will increase toward \(y = 1\). Solutions starting between \(-1\) and \(0\) will decrease toward \(y = -1\), and those starting below \(-1\) will increase toward \(y = -1\). --- ## Final Answer Summary - **For \(y_0 > 1\)**: Decreases toward \(y = 1\). - **For \(0 < y_0 < 1\)**: Increases toward \(y = 1\). - **For \(-1 < y_0 < 0\)**: Decreases toward \(y = -1\). - **For \(y_0 < -1\)**: Increases toward \(y = -1\). This behavior highlights the stability of the equilibrium points \(y = 1\) and \(y = -1\) and the instability at \(y = 0\).