4. (Short Answer) Consider the following differential equation: $$ \frac{d y}{d t}=y(y-4) $$ Which of the following are equilibrium solutions? $$ \times $$ $$ -4-4) \quad y(t)=-4, \quad y(t)=2, \quad y(t)=0, \quad y(t)=4 $$

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To determine the equilibrium solutions for the differential equation:

$$
\frac{dy}{dt} = y(y - 4)
$$

**Equilibrium solutions** occur where $$\frac{dy}{dt} = 0$$. Setting the right-hand side equal to zero:

$$
y(y - 4) = 0
$$

This equation holds true when:

$$
y = 0 \quad \text{or} \quad y = 4
$$

Therefore, the equilibrium solutions are:

$$
y(t) = 0 \quad \text{and} \quad y(t) = 4
$$

**Incorrect Options:**

- $$ y(t) = -4 $$: Substituting $$ y = -4 $$ into the equation gives $$\frac{dy}{dt} = (-4)(-4 - 4) = (-4)(-8) = 32 \neq 0$$, so it is **not** an equilibrium solution.
  
- $$ y(t) = 2 $$: Substituting $$ y = 2 $$ gives $$\frac{dy}{dt} = (2)(2 - 4) = (2)(-2) = -4 \neq 0$$, so it is **not** an equilibrium solution.

**Final Answer:**

Only the constant functions $$ y(t) = 0 $$ and $$ y(t) = 4 $$ are equilibrium solutions.
The equilibrium solutions are $$ y(t) = 0 $$ and $$ y(t) = 4 $$.

(Short Answer) Consider the following differential equation:

Which of the following are equilibrium solutions?

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(Short Answer) Consider the following differential equation: Which of the following are equilibrium solutions?