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The scalar differential equation \(y^{\prime \prime}-y^{\prime}+2 y^{2}=\alpha\), when rewritten as a first order system, results in a system having an equilibrium point at \((x, y)=(2,0)\). Determine the constant \(\alpha\).

Short Answer

Expert verified
Answer: The constant α is 2.

Step by step solution

01

Rewrite the scalar differential equation as a first-order system

Let \(v = y'\). Then the given second-order equation becomes a first-order system of two equations: \begin{align*} y' &= v \\ v' &= y' - v + 2y^2. \end{align*}
02

Find the equilibrium point

The equilibrium point occurs when both \(y'\) and \(v'\) are equal to 0. Using the first-order system derived in Step 1, we get: \begin{align*} 0 &= v \\ 0 &= y' - v + 2y^2. \end{align*} In this case, the equilibrium point is given as \((x, y) = (2, 0)\). Therefore, \(y = 0\) and \(v = 2\).
03

Substitute the equilibrium point into the first-order system

To find the value of the constant α, we need to substitute the equilibrium point into the first-order system: \begin{align*} 0 &= 2 \\ 0 &= y' - 2 + 2y^2 = (\alpha - 2) + 2(0)^2 = \alpha - 2. \end{align*}
04

Solve for the constant α

Now, we just need to solve the equation for α: \begin{align*} 0 &= \alpha - 2 \\ \alpha &= 2. \end{align*} Therefore, the constant \(\alpha\) is 2.

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Key Concepts

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

Equilibrium Solution
An equilibrium solution in differential equations is a solution where the system's state does not change as time progresses. In mathematical terms, it's a constant solution to a differential equation, where the derivative (or derivatives) equates to zero, indicating that there is no change transpiring over time.

For a first-order system, you typically have a pair of equations. In the exercise, we are told that the equilibrium point is at \( (x, y) = (2, 0) \). To find an equilibrium solution, we look for where the derivatives of all variables are zero. In simpler words, it’s like finding conditions where the system is completely at rest or in a state of balance.

Thinking of it like a still pendulum, regardless of the forces that might act on it, the pendulum remains motionless. Similarly, for the given differential equation, the equilibrium solution corresponds to a state where the variables' rates of change are both zero. Thus, substituting the equilibrium point into the system of equations allows us to find the constant \(\alpha\) that makes this balance possible.
First-order System
A first-order system in the realm of differential equations is characterized by equations that involve only the first derivatives of the function. The given exercise converts a second-order differential equation into a first-order system by introducing a new variable to represent the first derivative of the original function. This is a common technique used to simplify the complexity of higher-order equations.

In our case, we create a variable \(v = y'\), and thereby transform the original second-order equation into a system involving \(y'\) and \(v'\). This compact representation allows us to analyze the original second-order behavior through a system of two first-order equations. By working with first-order systems, we can apply a variety of powerful mathematical tools and techniques that are well-established for analyzing and solving first-order differential equations.

These systems often describe a multitude of real-world phenomena, from chemical reaction rates to population dynamics, and understanding the methodology to convert and solve these systems is a fundamental skill in mathematics and applied sciences.
Second-order Differential Equation
A second-order differential equation is an equation that involves the second derivative of a function. These types of equations are important in physics and engineering as they often model phenomena with acceleration, like the motion of springs or electric circuits. The scalar differential equation provided in the exercise is a second-order equation because it includes \(y''\), the second derivative of \(y\) with respect to \(x\).

To solve such equations or to analyze their behavior, one common approach is to reduce them to first-order systems, which is what we did with the exercise. By defining \(v = y'\) and rewriting the equation in terms of \(v\) and \(y\), we simplify the original equation into a form that's typically easier to handle.

Understanding second-order differential equations is crucial because they can describe more complex dynamics than first-order equations, which generally represent processes involving only rate of change. By mastering these equations, students can tackle a wide range of scientific and engineering problems that model the world around us.

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Most popular questions from this chapter

In each exercise, the given system is an almost linear system at each of its equilibrium points. (a) Find the (real) equilibrium points of the given system. (b) As in Example 2, find the corresponding linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\) at each equilibrium point. (c) What, if anything, can be inferred about the stability properties of the equilibrium point(s) by using Theorem \(6.4\) ? $$ \begin{aligned} &x^{\prime}=1-x^{2} \\ &y^{\prime}=x^{2}+y^{2}-2 \end{aligned} $$

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (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 the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 2 & -3 \\ 3 & 2 \end{array}\right] \mathbf{y} $$

These exercises explore the question "When one of two species in a colony is desirable and the other is undesirable, is it better to use resources to nurture the growth of the desirable species or to harvest the undesirable one?" Let \(x(t)\) and \(y(t)\) represent the populations of two competing species, with \(x(t)\) the desirable species. Assume that if resources are invested in promoting the growth of the desirable species, the population dynamics are given by $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x+\mu x \\ &y^{\prime}=r(1-\alpha y-\beta x) y \end{aligned} $$ If resources are invested in harvesting the undesirable species, the dynamics are $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x \\ &y^{\prime}=r(1-\alpha y-\beta x) y-\mu y \end{aligned} $$ In (10), \(r, \alpha, \beta\), and \(\mu\) are positive constants. For simplicity, we assume the same parameter values for both species. For definiteness, assume that \(\alpha>\beta>0\). Consider system (9), which describes the strategy in which resources are invested into nurturing the desirable species. (a) Determine the four equilibrium points for the system. (b) Show that it is possible, by investing sufficient resources (that is, by making \(\mu\) large enough), to prevent equilibrium coexistence of the two species. (c) Assume that \(\mu\) is large enough to preclude equilibrium coexistence of the two species. Compute the linearized system at each of the three physically relevant equilibrium points. Determine the stability characteristics of the linearized system at each of these equilibrium points. (d) System (9) can be shown to be an almost linear system at each of the equilibrium points. Use this fact and the results of part (c) to infer the stability properties of system (9) at each of the three equilibrium points of interest. (e) Sketch the direction field. Will a sufficiently aggressive nurturing of species \(x\) ultimately drive undesirable species \(y\) to extinction? If so, what is the limiting population of species \(x\) ?

In each exercise, an initial value problem for a first order nonlinear system is given. Rewrite the problem as an equivalent initial value problem for a higher order nonlinear scalar differential equation. $$ \frac{d}{d t}\left[\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \end{array}\right]=\left[\begin{array}{c} y_{2} \\ y_{3} \\ \sqrt{y_{2} y_{3}+t^{2}} \end{array}\right], \quad\left[\begin{array}{l} y_{1}(1) \\ y_{2}(1) \\ y_{3}(1) \end{array}\right]=\left[\begin{array}{l} 1 \\ \frac{1}{2} \\ 3 \end{array}\right] $$

Each exercise lists a nonlinear system \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\), where \(A\) is a constant ( \(2 \times 2\) ) invertible matrix and \(\mathbf{g}(\mathbf{z})\) is a \((2 \times 1)\) vector function. In each of the exercises, \(\mathbf{z}=\mathbf{0}\) is an equilibrium point of the nonlinear system. (a) Identify \(A\) and \(\mathbf{g}(\mathbf{z})\). (b) Calculate \(\|\mathbf{g}(\mathbf{z})\|\). (c) Is \(\lim _{\mid \mathbf{z} \| \rightarrow 0}\|\mathbf{g}(\mathbf{z})\| /\|\mathbf{z}\|=0\) ? Is \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\) an almost linear system at \(\mathbf{z}=\mathbf{0}\) ? (d) If the system is almost linear, use Theorem \(6.4\) to choose one of the three statements: (i) \(\mathbf{z}=\mathbf{0}\) is an asymptotically stable equilibrium point. (ii) \(\mathbf{z}=\mathbf{0}\) is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=9 z_{1}+5 z_{2}+z_{1} z_{2} \\ &z_{2}^{\prime}=-7 z_{1}-3 z_{2}+z_{1}^{2} \end{aligned} $$

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