Pregunta
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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 \( t \)

Ask by Moran West. in the United States
Jan 25,2025

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The solution \( y(t) \) remains constant at \( y = 4 \) for all \( t \).

Solución

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 \( t \)**

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If we consider the differential equation \( \frac{d y}{d t} = y(y-2)(y-4) \), we can analyze the behavior of the solutions based on the initial condition \( y(0) = 4 \). When \( y = 4 \), the right-hand side of the equation becomes zero, which means that at this point, the solution will not change over time. So, the only option that holds true based on this analysis is: (b) \( y(t) = 4 \) for all \( t \). Now, if we think about some real-world applications, this kind of differential equation could model a population or resource that stabilizes at a certain level when reaching a threshold, say, the population of a species in a controlled environment. When the initial population starts at a stable maximum (like 4), it simply remains that way unless external factors intervene. Furthermore, a common mistake in solving such differential equations is failing to recognize the equilibrium points where \( \frac{dy}{dt} = 0 \). Remember, if your initial condition starts at an equilibrium point (in this case, at 4), the solution remains constant, highlighting the importance of where initial conditions place the solution.

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