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Chapter 7: Problem 4

Given the differential equation \(x^{\prime}(t)=x^{4}-5 x^{3}-2 x^{2}+24 x+0 .\) List the constant (or equilibrium) solutions to this differential equation in increasing order and indicate whether or not these equations are stable, semi- stable, or unstable. (It helps to sketch the graph. xFunctions will plot functions as well as phase planes.)

Short Answer

Expert verified
The equilibrium solutions are \( x = -2 \) (Unstable), \( x = 0 \) (Stable), \( x = 3 \) (Unstable), and \( x = 4 \) (Stable).

Step by step solution

01

- Identify Equilibrium Solutions

Equilibrium solutions are found by setting the differential equation equal to zero: \[ x^{\'{}}(t) = x^{4} - 5x^{3} - 2x^{2} + 24x = 0 \] Factor the polynomial to find the values of \( x \) where the derivative is zero.
02

- Factor the Polynomial

To solve the equation: \[ x^{4} - 5x^{3} - 2x^{2} + 24x = 0 \] First, factor out \( x \): \[ x(x^{3} - 5x^{2} - 2x + 24) = 0 \] The factors of \( x^{3} - 5x^{2} - 2x + 24 \) can be obtained using trial and error or synthetic division.
03

- Continue Factoring

By synthetic division or trial and error, you find: \[ x^{3} - 5x^{2} - 2x + 24 = (x-4)(x^{2}-x-6) \] Continue factoring: \[ x^{2} - x - 6 = (x-3)(x+2) \] Thus, we have: \[ x(x-4)(x-3)(x+2) = 0 \] The solutions are \( x = 0, 4, 3, -2 \).
04

- Determine Stability

Analyze the stability of each equilibrium solution by considering the first derivative's sign changes across each solution. Sketch the graph of \( x'(t) = x(x-4)(x-3)(x+2) \) to determine the behavior around each equilibrium point.
05

- State Stability of Equilibrium Points

By plotting the graph or examining sign changes: - \( x = -2 \): Unstable - \( x = 0 \): Stable - \( x = 3 \): Unstable - \( x = 4 \): Stable Thus, classify each equilibrium point accordingly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. In this exercise, we have a differential equation of the form \( x^{\prime}(t) = x^4 - 5x^3 - 2x^2 + 24x \).
The prime notation (\(x^{\prime}(t) \)) represents the derivative of \(x(t)\), which indicates the rate of change of \(x\) with respect to time \(t\).
To solve the exercise, we need to find the constant solutions, called equilibrium solutions, where the rate of change is zero. This means we set the differential equation to zero and solve for \(x\).
Stability Analysis
Stability analysis helps us determine if the equilibrium solutions are stable, semi-stable, or unstable.
A solution is considered stable if small perturbations around the equilibrium point lead the system back to the equilibrium.
Conversely, an unstable solution will diverge away from the equilibrium point if perturbed.
Equilibrium points are categorized by examining the sign changes of the first derivative around these points.
  • If the sign of the first derivative changes from positive to negative, the equilibrium point is stable.
  • If it changes from negative to positive, the point is unstable.
  • If the derivative does not change sign across the point, it might be semi-stable.
In our exercise, sketching the graph of \( x^{\prime}(t) = x(x-4)(x-3)(x+2) \) is helpful to visually identify these sign changes.
Polynomial Factorization
Solving the differential equation involves polynomial factorization. Initially, we set the differential equation equal to zero: \[ x^4 - 5x^3 - 2x^2 + 24x = 0 \].
The first step is to factor out common terms. In this case, factor out \(x\): \[ x(x^3 - 5x^2 - 2x + 24) = 0 \]
Next, factorize \( x^3 - 5x^2 - 2x + 24 \) via trial and error or synthetic division to simplify further: \[ x^3 - 5x^2 - 2x + 24 = (x-4)(x^2 - x - 6) \]
Finally, continue factoring \(x^2 - x - 6\), which yields: \[ x(x-4)(x-3)(x+2) = 0 \] The equilibrium solutions are where each factor equals zero, giving us \( x = 0, -2, 3, 4 \).
Behavior Around Equilibrium Points
Understanding the behavior around equilibrium points is crucial to determining stability. For each equilibrium solution:
  • **\( x = -2 \)**: The graph shows the derivative is positive on both sides, indicating instability.
  • **\( x = 0 \)**: The derivative changes from negative to positive, indicating stability.
  • **\( x = 3 \)**: The derivative changes from positive to negative, indicating instability.
  • **\( x = 4 \)**: The derivative changes from negative to positive, indicating stability.
Evaluating the sign changes of the derivatives qualitatively and quantitatively helps classify the stability of each equilibrium point.

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