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This image shows several differential equations on the left, each of the form \(\frac{dy}{dt}\), and eight phase-line diagrams labeled A through H on the right. The task is to match each differential equation with the corresponding phase-line diagram, which visually represents the behavior of the solutions of the differential equations in terms of equilibrium points (where \(\frac{dy}{dt} = 0\)) and stability.

### Let's analyze the equations and diagrams:

1. **Equation 1: \(\frac{dy}{dt} = 4y - y^3\)**  
   This is a cubic equation with three roots: \(y = 0, \pm 2\). To find the phase diagram, we need to analyze the behavior between and around these equilibrium points.

2. **Equation 2: \(\frac{dy}{dt} = y(2 - y)^2\)**  
   The right-hand side has a repeated root at \(y = 2\) and another root at \(y = 0\). The repeated root suggests that at \(y = 2\), the solution will neither grow nor decay rapidly, leading to a specific pattern in the phase line.

3. **Equation 3: \(\frac{dy}{dt} = 2y - y^2\)**  
   This quadratic equation has two roots at \(y = 0\) and \(y = 2\). The phase-line diagram will reflect this behavior with changes in the direction of \(y\) near these points.

4. **Equation 4: \(\frac{dy}{dt} = y^2 |y - 2|\)**  
   This equation has a non-differentiable term due to the absolute value, and it has a root at \(y = 2\). The behavior around this root and the power of \(y^2\) will lead to distinctive behavior on the phase line.

5. **Equation 5: \(\frac{dy}{dt} = y^3 - 4y\)**  
   This is another cubic equation, which has three roots: \(y = 0, \pm 2\). The behavior around these equilibrium points can be analyzed to determine the direction of flow.

6. **Equation 6: \(\frac{dy}{dt} = y(y - 2)\)**  
   This equation has roots at \(y = 0\) and \(y = 2\). It's a simple quadratic equation, and the phase-line diagram will show how the solution changes around these points.

7. **Equation 7: \(\frac{dy}{dt} = y^2 - 3y\)**  
   Factoring gives \(\frac{dy}{dt} = y(y - 3)\), so the roots are at \(y = 0\) and \(y = 3\). The phase line will show changes around these points.

8. **Equation 8: \(\frac{dy}{dt} = 3y - y^2\)**  
   This is a quadratic equation with roots at \(y = 0\) and \(y = 3\). Similar to Equation 7, but scaled differently.

Now, based on the behavior of the equations (such as signs of derivatives between the roots, direction of flow), we can match each equation to the correct phase-line diagram.

### Do you want me to match each equation with its corresponding diagram for you?

Match each differential equation with the corresponding phase-line diagram.

Learn how to match various differential equations with their corresponding phase-line diagrams. This problem involves analyzing equilibrium points and stability, using concepts such as cubic and quadratic equations. Suitable for undergraduate students studying differential equations.

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