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Chapter 6: Problem 13

Rewrite the given scalar differential equation as a first order system, and find all equilibrium points of the resulting system. $$ y^{\prime \prime}+\frac{2 y^{\prime}}{1+y^{4}}+y^{2}=1 $$

Short Answer

Expert verified
Answer: The equilibrium points are (1, 0) and (-1, 0).

Step by step solution

01

Introduce new variables

Let's introduce the following variables: - \(y_1(t) = y(t)\) - \(y_2(t) = y_1'(t) = y'(t)\) We will rewrite the given second-order differential equation in terms of these new variables.
02

Rewrite the given equation in terms of new variables

Express the given equation in terms of \(y_1(t)\) and \(y_2(t)\): $$ y_2'(t) + \frac{2y_2(t)}{1+y_1(t)^4} + y_1(t)^2 = 1 $$ Now we can write down a system of two first-order differential equations: 1. \(y_1'(t) = y_2(t)\) 2. \(y_2'(t) = 1 - y_1(t)^2 - \frac{2y_2(t)}{1+y_1(t)^4}\) So, the first-order system is: $$ \begin{cases} y_1'(t) = y_2(t), \\ y_2'(t) = 1 - y_1(t)^2 - \frac{2y_2(t)}{1+y_1(t)^4}. \end{cases} $$
03

Find the equilibrium points

Equilibrium points occur when both \(y_1'(t)\) and \(y_2'(t)\) are equal to zero. We can write this as: $$ \begin{cases} 0 = y_2(t), \\ 0 = 1 - y_1(t)^2 - \frac{2y_2(t)}{1+y_1(t)^4}. \end{cases} $$ Solving these equations will give: \begin{cases} y_2(t)=0, \\ 1-y_1(t)^2 = 0. \end{cases} Solve the second equation for \(y_1(t)\): $$ y_1(t)^2 = 1 \Rightarrow y_1(t) = \pm 1 $$ Thus, we have the following equilibrium points: 1. \((y_1(t),y_2(t)) = (1, 0)\) 2. \((y_1(t),y_2(t)) = (-1, 0)\) These are the equilibrium points for the first-order system obtained from the given scalar differential equation.

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Key Concepts

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

Equilibrium Points
Understanding equilibrium points is fundamental when analyzing differential systems, as these are the points where the system does not change, representing steady states or long-term behavior. In the context of a first-order system, equilibrium points are located where the derivative(s) of the system are zero.

To find equilibrium points for our first-order system converted from the second-order differential equation, we set the derivative functions equal to zero. This process helps us determine the system's behavior and stability at different points. In the given exercise, equilibrium points were found by setting both derivatives to zero, leading to the solutions \((y_1(t), y_2(t)) = (1, 0)\) and \((y_1(t), y_2(t)) = (-1, 0)\). These solutions indicate that the dynamic system will come to rest at these points, and any small movement away from them will determine the system's stability in their vicinity.
Second-Order Differential Equation
A second-order differential equation involves the second derivative of a function and is crucial in fields such as physics and engineering, as it models systems with acceleration, such as oscillations. To analyze these equations, one often converts them into a system of first-order differential equations, as first-order systems are generally easier to solve and interpret dynamically.

The initial equation from the exercise, \(y^{\backprime \backprime} + \frac{2 y^{\backprime}}{1+y^{4}} + y^{2} = 1 \), is a second-order differential equation. The conversion into a first-order system requires introducing new variables, effectively breaking the second-order equation into two first-order equations that capture both the position and velocity/derivative of the system, allowing us to use phase plane analysis methods to study system behavior.
System of Differential Equations
A system of differential equations consists of multiple equations that involve derivatives of several functions. Systems can describe more complex scenarios, such as predator-prey interactions in biology or multiple interacting components in mechanical systems. When faced with a second-order differential equation, converting it into a system allows us to use powerful tools and methods for analysis, such as matrix exponential or numerical solvers.

The restructured first-order differential system from our exercise is essential for investigating the system's dynamics. The system composed of equations \(y_1'(t) = y_2(t)\) and \(y_2'(t) = 1 - y_1(t)^2 - \frac{2y_2(t)}{1+y_1(t)^4}\) provides insights into how one state variable changes with respect to another. This offers a pathway to visualizing the system's trajectory in a phase plane and evaluating the local behavior near equilibrium points.

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Most popular questions from this chapter

A linear system is given in each exercise. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at \(\mathbf{y}=\mathbf{0}\). (c) The given linear system is a Hamiltonian system. Derive the conservation law for this system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 2 & 1 \\ 0 & -2 \end{array}\right] \mathbf{y} $$

Assume the given autonomous system models the population dynamics of two species, \(x\) and \(y\), within a colony. (a) For each of the two species, answer the following questions. (i) In the absence of the other species, does the remaining population continuously grow, decline toward extinction, or approach a nonzero equilibrium value as time evolves? (ii) Is the presence of the other species in the colony beneficial, harmful, or a matter of indifference? (b) Determine all equilibrium points lying in the first quadrant of the phase plane (including any lying on the coordinate axes). (c) The given system is an almost linear system at the equilibrium point \((x, y)=(0,0)\). Determine the stability properties of the system at \((0,0)\). $$ \begin{aligned} &x^{\prime}=x-x^{2}+x y \\ &y^{\prime}=y-y^{2}+x y \end{aligned} $$

In each exercise, the eigenpairs of a \((2 \times 2)\) matrix \(A\) are given where both eigenvalues are real. Consider the phase-plane solution trajectories of the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\), where $$ \mathbf{y}(t)=\left[\begin{array}{l} x(t) \\ y(t) \end{array}\right] $$ (a) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at \(\mathbf{y}=\mathbf{0}\). (b) Sketch the two phase-plane lines defined by the eigenvectors. If an eigenvector is \(\left[\begin{array}{l}u_{1} \\ u_{2}\end{array}\right]\), the line of interest is \(u_{2} x-u_{1} y=0\). Solution trajectories originating on such a line stay on the line; they move toward the origin as time increases if the corresponding eigenvalue is negative or away from the origin if the eigenvalue is positive. (c) Sketch appropriate direction field arrows on both lines. Use this information to sketch a representative trajectory in each of the four phase- plane regions having these lines as boundaries. Indicate the direction of motion of the solution point on each trajectory. $$ \lambda_{1}=2, \quad \mathbf{x}_{1}=\left[\begin{array}{l} 2 \\ 0 \end{array}\right] ; \quad \lambda_{2}=1, \quad \mathbf{x}_{2}=\left[\begin{array}{l} 0 \\ 2 \end{array}\right] $$

In each exercise, the eigenpairs of a \((2 \times 2)\) matrix \(A\) are given where both eigenvalues are real. Consider the phase-plane solution trajectories of the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\), where $$ \mathbf{y}(t)=\left[\begin{array}{l} x(t) \\ y(t) \end{array}\right] $$ (a) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at \(\mathbf{y}=\mathbf{0}\). (b) Sketch the two phase-plane lines defined by the eigenvectors. If an eigenvector is \(\left[\begin{array}{l}u_{1} \\ u_{2}\end{array}\right]\), the line of interest is \(u_{2} x-u_{1} y=0\). Solution trajectories originating on such a line stay on the line; they move toward the origin as time increases if the corresponding eigenvalue is negative or away from the origin if the eigenvalue is positive. (c) Sketch appropriate direction field arrows on both lines. Use this information to sketch a representative trajectory in each of the four phase- plane regions having these lines as boundaries. Indicate the direction of motion of the solution point on each trajectory. $$ \lambda_{1}=-2, \quad \mathbf{x}_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] ; \quad \lambda_{2}=-1, \quad \mathbf{x}_{2}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] $$

Locate the equilibrium point of the given nonhomogeneous linear system \(\mathbf{y}^{\prime}=A \mathbf{y}+\mathbf{g}_{0}\). [Hint: Introduce the change of dependent variable \(\mathbf{z}(t)=\mathbf{y}(t)-\mathbf{y}_{0}\), where \(\mathbf{y}_{0}\) is chosen so that the equation can be rewritten as \(\mathbf{z}^{\prime}=A \mathbf{z}\).] Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point. $$ \begin{aligned} &x^{\prime}=-x+2 \\ &y^{\prime}=2 y-4 \end{aligned} $$
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