Let's break down and solve the given differential equation step by step.

Given:

$\frac{dy}{dx} = y(y - 2)$ where $y_0 > 0$ is the initial condition.

Part (a): Qualitative Analysis (Classifying Equilibrium Solutions)

To classify the equilibrium solutions, we need to set $\frac{dy}{dx} = 0$ and solve for $y$.

$y(y - 2) = 0$

The solutions are:

$y = 0 \quad \text{or} \quad y = 2$

These are the equilibrium solutions. Now, let's classify their stability by analyzing the sign of $\frac{dy}{dx}$ near these points.

1. For $y = 0$:

   • For $y > 0$, we have $y(y - 2) > 0$ (since both $y$ and $y - 2$ are positive).
   • For $y < 0$, we have $y(y - 2) < 0$ (since $y$ is negative and $y - 2$ is negative for $y < 2$).

   Conclusion: For $y = 0$, solutions move away from $y = 0$ on either side. Hence, $y = 0$ is an unstable equilibrium.

2. For $y = 2$:

   • For $y > 2$, we have $y(y - 2) > 0$ (since both $y > 2$ and $y - 2 > 0$).
   • For $y < 2$, we have $y(y - 2) < 0$ (since $y > 0$ and $y - 2 < 0$).

   Conclusion: For $y = 2$, solutions tend to move towards $y = 2$. Hence, $y = 2$ is a stable equilibrium.

Part (b): Solve the Differential Equation

This is a separable differential equation. Let's solve it by separating the variables.

$\frac{1}{y(y - 2)} dy = dx$

Now, we use partial fraction decomposition on the left-hand side:

$\frac{1}{y(y - 2)} = \frac{A}{y} + \frac{B}{y - 2}$

Multiplying through by $y(y - 2)$ to clear the denominator:

$1 = A(y - 2) + By$

Expanding and collecting like terms:

$1 = (A + B)y - 2A$

This gives two equations:

1. $A + B = 0$
2. $-2A = 1$

Solving for $A$ and $B$: $A = -\frac{1}{2}, \quad B = \frac{1}{2}$

Thus, we can rewrite the integral as:

$\frac{1}{y(y - 2)} = -\frac{1}{2y} + \frac{1}{2(y - 2)}$

Now, integrating both sides:

$-\frac{1}{2} \int \frac{1}{y} \, dy + \frac{1}{2} \int \frac{1}{y - 2} \, dy = \int dx$

This yields:

$-\frac{1}{2} \ln|y| + \frac{1}{2} \ln|y - 2| = x + C$

Simplifying the logarithms:

$\frac{1}{2} \ln\left|\frac{y - 2}{y}\right| = x + C$

Multiplying both sides by 2:

$\ln\left|\frac{y - 2}{y}\right| = 2(x + C)$

Exponentiating both sides:

$\left|\frac{y - 2}{y}\right| = e^{2(x + C)} = e^{2x}e^{2C}$

Let $e^{2C} = C_1$, so:

$\left|\frac{y - 2}{y}\right| = C_1 e^{2x}$

Removing the absolute value and solving for $y$:

$\frac{y - 2}{y} = \pm C_1 e^{2x}$

Multiplying through by $y$ and solving for $y$:

$y - 2 = \pm C_1 y e^{2x}$ $y(1 \mp C_1 e^{2x}) = 2$ $y = \frac{2}{1 \mp C_1 e^{2x}}$

Confirming Equilibrium Analysis:

• As $x \to \infty$, if $C_1 > 0$, $y \to 2$, which confirms $y = 2$ is a stable equilibrium.
• If $y(0) = 0$, the solution $y = 0$ is an unstable equilibrium as solutions move away from it.

Follow-Up Questions:

1. How does the behavior change if $y_0$ is negative?
2. What are the possible long-term behaviors for different values of $C_1$?
3. How would a phase portrait of this system look like?
4. Can you provide a numerical example of this solution?
5. How would adding a constant term to the differential equation affect the equilibria?

Tip: Stability can often be inferred by checking the sign of the derivative near equilibrium points.