2. Consider the following differential equation: \( \quad y=0, y=2, y=4 \) \[ \frac{d y}{d t}=y(y-2)(y-4) \] Which of the following is true about a solution \( y(t) \) which satisfies the initial condition \( y(0)=4 \) ? (a) \( 24 \) for all \( t \) (d) \( y(t)<2 \) for all \( t \)

To analyze the differential equation: \[ \frac{d y}{d t} = y(y - 2)(y - 4) \] with the initial condition \( y(0) = 4 \), let's identify the equilibrium solutions and determine their stability. ### Equilibrium Points: The equilibrium points are found by setting \( \frac{d y}{d t} = 0 \): \[ y(y - 2)(y - 4) = 0 \quad \Rightarrow \quad y = 0, \, y = 2, \, y = 4 \] ### Stability of the Equilibrium at \( y = 4 \): To determine the stability, we examine the derivative of the right-hand side function with respect to \( y \): \[ f(y) = y(y - 2)(y - 4) \] \[ f'(y) = \frac{d}{dy}[y(y - 2)(y - 4)] = 3y^2 - 12y + 8 \] Evaluating at \( y = 4 \): \[ f'(4) = 3(4)^2 - 12(4) + 8 = 48 - 48 + 8 = 8 > 0 \] Since \( f'(4) > 0 \), the equilibrium at \( y = 4 \) is **unstable**. This means that any slight perturbation from \( y = 4 \) would cause the solution to move away from 4. ### Solution with \( y(0) = 4 \): However, because the initial condition is exactly at the equilibrium point (\( y(0) = 4 \)), the solution remains at this point for all time. There is no perturbation to push it away. ### Conclusion: The solution \( y(t) \) that satisfies \( y(0) = 4 \) remains constant at \( y = 4 \) for all \( t \). **Answer:** **(b) \( y(t)=4 \) for all \( t \)**