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Chapter 9: Problem 26

Find the equilibrium points and assess their stability. \(x^{\prime}(t)=0.8 x(1-x / 24)-0.2 x y\) \(y^{\prime}(t)=-0.6 y+0.1 x y\)

Short Answer

Expert verified
The equilibrium points are (0,0), (24,0), and (6, 6). Stability can be assessed using the Jacobian matrix and eigenvalues at each point.

Step by step solution

01

Set the derivatives to zero

To find the equilibrium points, set the derivatives equal to zero:\(x'(t) = 0\) and \(y'(t) = 0\). This gives the equations:\[0.8 x(1-x / 24) - 0.2 x y = 0\]\[-0.6 y + 0.1 x y = 0\]
02

Solve for y in the second equation

The second equation is \(-0.6 y + 0.1 x y = 0\). Factor out y:\[y(-0.6 + 0.1x) = 0\]This gives two cases: \(y = 0\) or \(-0.6 + 0.1x = 0\)
03

Solve for x in the second case

Solve \(-0.6 + 0.1x = 0\):\[0.1x = 0.6\]\[x = 6\]
04

Substitute x-values into the first equation

Substitute \(y = 0\) and \(x = 6\) into the first equation:For \(y = 0\):\[0.8 x (1 - x / 24) = 0\]\[0.8x - 0.8x^2 / 24 = 0\]\[0.8x(1 - x / 24) = 0\]This gives\(0.8x = 0\) or \(1 - x / 24 = 0\), which equates to two points, \(x = 0\) or \(x = 24\).
05

Determine equilibrium points

The equilibrium points are combinations of (x,y) from solving the equations:\((x, y) = (0,0)\), \((x, y) = (24,0)\), and \((x, y) = (6, 0.6)\)
06

Assess stability

To assess stability, determine the Jacobian matrix at each equilibrium point:\(J = \begin{pmatrix} \frac{d f_1}{d x} & \frac{d f_1}{d y} \ \frac{d f_2}{d x} & \frac{d f_2}{d y} \end{pmatrix}\)Calculate the partial derivatives and the Jacobian matrix for each point.Evaluate the eigenvalues to assess stability (if they have negative real parts, the point is stable).

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Key Concepts

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

stability analysis
Stability analysis is crucial in determining the behavior of equilibrium points in differential equations. Equilibrium points are where the system does not change over time. To check their stability, we often use the Jacobian matrix. This involves checking the eigenvalues of the matrix. If all eigenvalues have negative real parts, the equilibrium is stable. Otherwise, it's unstable. Stability tells us if small perturbations will cause the system to return to equilibrium or diverge away.
Jacobian matrix
The Jacobian matrix is used to study the behavior of a system near its equilibrium points. For a system of differential equations, it's a square matrix of first-order partial derivatives of the system. In this case, the Jacobian matrix is:ewline\(J = \begin{pmatrix} \frac{d f_1}{d x} & \frac{d f_1}{d y} \ \frac{d f_2}{d x} & \frac{d f_2}{d y} \end{pmatrix}\)ewlineBy evaluating the Jacobian matrix at each equilibrium point, we can understand how the system behaves close to those points.
eigenvalues
Eigenvalues are found by solving the characteristic equation of the Jacobian matrix. These values are crucial for determining stability. If all eigenvalues of the Jacobian matrix have negative real parts, the equilibrium point is considered stable. If any eigenvalue has a positive real part, the equilibrium is unstable. For our system, calculating eigenvalues tells us how perturbations will evolve over time.
differential equations
A differential equation relates a function with its derivatives. In our exercise, we have a system of differential equations:ewline\(x'(t) = 0.8 x(1 - x / 24) - 0.2 x y\)ewline\(y'(t) = -0.6 y + 0.1 x y\)ewlineThese equations describe how variables x and y change over time. Finding equilibrium points involves solving these equations when their time derivative is zero.
equilibrium analysis
Equilibrium analysis involves finding points where the system does not change. These points are found by setting the derivatives to zero. For the given system, the equilibrium points were:ewline\((x, y) = (0,0)\), \((x, y) = (24,0)\), and \((x, y) = (6, 0.6)\)ewlineAfter determining these points, the next step is analyzing their stability using the Jacobian matrix and its eigenvalues. This tells us if small perturbations will cause the system to stay near the equilibrium or move away.

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Most popular questions from this chapter

Find the equilibrium points and assess the stability of each. \(x^{\prime}=x-e^{y}, y^{\prime}=2 \ln x+y-6\)

Determine the eigenvalues for the system of differential equations. If the eigenvalues are real and distinct, find the general solution by determining the associated eigenvectors. If the eigenvalues are complex or repeated, solve using the reduction method. \(x^{\prime}=-2 x+4 y, y^{\prime}=-x-7 y\)

The displacement \(x(t)\) of a spring from its rest position after \(t\) seconds follows the differential equation $$ m x^{\prime \prime}+\gamma x^{\prime}+k x=q(t) $$ where \(m\) is the mass of the object attached to the spring, \(q(t)\) is the forcing function, and \(\gamma\) and \(k\) are the stiffness and damping coefficients, respectively. Suppose that the spring starts at rest, so that \(x(0)=0\) and \(x^{\prime}(0)=0 .\) Solve for \(x(t)\) given the following conditions. \(m=1, k=16, \gamma=0, q(t)=1\)

When our vocal folds vibrate under normal speech, their initial movement is governed by a second-order differential equation. Let \(x(t)\) denote the lateral displacement of the lower vocal folds at time \(t\). Then the equation of motion for the lower vocal folds is $$ m x^{\prime \prime}+\frac{1}{c}\left(x-x_{0}\right)=p d $$ In this equation, \(m\) is the mass of the lower vocal folds per unit length, \(c\) is the mechanical compliance per unit length, \(x_{0}\), is the resting position, \(p\) is the subglottal pressure, and \(d\) is the length of the lower vocal folds. a) Find the general solution of this differential equation. b) Assume that the vocal folds are at rest before the outward displacement begins, so that \(x(0)=0\) and \(x^{\prime}(0)=0 .\) Solve this initial-value problem.

Verify by substitution that the given functions solve the system of differential equations. \(\left[\begin{array}{l}x \\\ y\end{array}\right]^{\prime}=\left[\begin{array}{rr}4 & 1 \\ -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]\) \(x=2 e^{3 t}-e^{2 t}, y=-2 e^{3 t}+2 e^{2 t}\)
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