/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Solve the given initial-value pr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the given initial-value problem. \(\left[\begin{array}{l}\frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t}\end{array}\right]=\left[\begin{array}{ll}-3 & 4 \\ -1 & 2\end{array}\right]\left[\begin{array}{l}x_{1}(t) \\\ x_{2}(t)\end{array}\right]\) with \(x_{1}(0)=1\) and \(x_{2}(0)=2\).

Short Answer

Expert verified
Solve the matrix system, find eigenvalues and eigenvectors, then apply initial conditions to find constants for the particular solution.

Step by step solution

01

Write the system as a matrix equation

The system of differential equations can be expressed in matrix form as \( \frac{d}{dt} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} -3 & 4 \ -1 & 2 \end{bmatrix} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \). Let \( X(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) and \( A = \begin{bmatrix} -3 & 4 \ -1 & 2 \end{bmatrix} \), so \( \frac{dX}{dt} = AX \).
02

Find the eigenvalues of the matrix A

Solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \) where \( I \) is the identity matrix. Substitute \( A = \begin{bmatrix} -3 & 4 \ -1 & 2 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). The determinant is computed as follows: \[ \begin{vmatrix} -3 - \lambda & 4 \ -1 & 2 - \lambda \end{vmatrix} = (-3 - \lambda)(2 - \lambda) - (4)(-1) \]. This simplifies to \( \lambda^2 + \lambda + 2 = 0 \).
03

Solve the eigenvalue equation

The characteristic polynomial \( \lambda^2 + \lambda + 2 = 0 \) does not have real roots. Use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = 2 \). Calculate \( \lambda_1, \lambda_2 = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2} = \frac{-1 \pm i\sqrt{7}}{2} \). The eigenvalues are complex conjugates.
04

Find the eigenvectors

For \( \lambda_1 = \frac{-1 + i\sqrt{7}}{2} \), solve \( (A - \lambda_1 I)\mathbf{v} = 0 \). Form the matrix \( A - \lambda_1 I \) = \( \begin{bmatrix} \frac{-5 - i\sqrt{7}}{2} & 4 \ -1 & \frac{5 - i\sqrt{7}}{2} \end{bmatrix} \) and solve for the vector \( \mathbf{v} = \begin{bmatrix} v_1 \ v_2 \end{bmatrix} \). Choose \( v_1 = 1 \) and solve for \( v_2 \) to get an eigenvector.
05

Write the general solution

The general solution is formed using the eigenvectors and eigenvalues: \( X(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \). Using \( \lambda_1 = \frac{-1 + i\sqrt{7}}{2} \) and associated eigenvector \( \mathbf{v}_1 \), along with its conjugate, express \( x_1(t) \) and \( x_2(t) \) in terms of real sinusoidal functions using Euler's formula.
06

Apply initial conditions

Use the initial conditions \( x_1(0) = 1 \) and \( x_2(0) = 2 \) to solve for the constants \( c_1 \) and \( c_2 \). Substitute \( t=0 \) into the general solution and solve the resulting system of linear equations for \( c_1 \) and \( c_2 \). This provides the particular solution satisfying the initial condition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Eigenvalues and Eigenvectors
In systems of differential equations, eigenvalues and eigenvectors play a crucial role in determining the behavior of solutions. These concepts originate from linear algebra and are essential for simplifying complex systems.

To find the eigenvalues of a matrix, you solve the characteristic equation, which is derived from the determinant of the matrix minus an identity matrix multiplied by a scalar \( \lambda \). For instance, given a matrix \( A \), the characteristic equation is \( \text{det}(A - \lambda I) = 0 \). Solving this gives the eigenvalues \( \lambda \).

Once the eigenvalues are determined, the corresponding eigenvectors can be found. Each eigenvalue has at least one associated eigenvector, and these are found by solving \( (A - \lambda I)\mathbf{v} = 0 \) for the vector \( \mathbf{v} \). These eigenvectors form the basis for expressing the solution of the differential equation. They define the direction of stretching or compressing in the system as it evolves over time.
Complex Numbers in Differential Equations
Complex numbers come into play in differential equations when the roots of the characteristic polynomial are not real. This situation often leads to solutions involving sine and cosine functions.

In our example, the characteristic polynomial \( \lambda^2 + \lambda + 2 = 0 \) yields complex roots in the form \( \lambda = \frac{-1 \pm i\sqrt{7}}{2} \). These complex eigenvalues indicate that the system has oscillatory behavior. The presence of \( i \) (the imaginary unit) suggests that the solutions will involve periodic functions.

Using Euler's formula, \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), we can express the complex exponential solutions in terms of sine and cosine. This conversion is crucial for interpreting the behavior of the system and transforming the solution into a real function since real-world systems cannot manifest complex results directly.
Initial Value Problems
An initial value problem in differential equations provides a specific starting point for the solution. The uniqueness of the solution is guaranteed by specifying the initial conditions at a particular point.

For instance, given the initial conditions \( x_1(0) = 1 \) and \( x_2(0) = 2 \), the system of differential equations adopts a unique solution that passes through this point. These conditions allow us to solve for the constants in the general solution, tuning the equation to fit the particulars of the situation described.

To refine our solution to a specific initial value problem, we substitute these initial conditions into the general solution. This helps us solve for any constants mixed into the solution functions. By doing this, we translate a general behavior into something practically useful by adapting it to known system starting values.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Forest Succession Forest succession can be modeled by a three-compartment model. We assume that gaps in a forest are created by disturbances just as in Problem 21 . These gaps are initially filled by fast-growing, early colonizing plants, which are then replaced by slower growing species, a process known as succession. We denote by \(x_{1}(t)\) the total area occupied by gaps at time \(t\), by \(x_{2}(t)\) the total area occupied by fast growing species at time \(t\), and by \(x_{3}(t)\) the total area occupied by slow growing species at time \(t\). The dynamics are given by $$ \begin{array}{l} \frac{d x_{1}}{d t}=0.2 x_{2}+x_{3}-2 x_{1} \\ \frac{d x_{2}}{d t}=2 x_{1}-0.7 x_{2} \\ \frac{d x_{3}}{d t}=0.5 x_{2}-x_{3} \end{array} $$ (a) Draw the corresponding compartment diagram. (b) Show that $$ x_{1}(t)+x_{2}(t)+x_{3}(t)=A $$ where \(A\) is a constant, and give the meaning of \(A\).

Assume that \(a>0\). Find all point equilibria of $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{2}\left(x_{1}-a\right) \\ \frac{d x_{2}}{d t}=x_{2}^{2}-x_{1} \end{array} $$ and characterize their stability.

We consider differential equations of the form \(\frac{d \mathbf{x}}{d t}=A \mathbf{x}(t)\) where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilib\mathrm{\\{} r i u m ~ a c c o r d i n g ~ t o ~ w h e t h e r ~ i t ~ i s ~ a ~ s i n k , ~ a ~ s o u r c e , ~ o r ~ a ~ s a d d l e ~ point. \(A=\left[\begin{array}{rr}-2 & -3 \\ 1 & 3\end{array}\right]\)

Find all equilibria of each system of differential equations and determine the stability of each equilibrium. \(\frac{d x_{1}}{d t}=4 x_{1}\left(1-x_{1}\right)-2 x_{1} x_{2}\) \(\frac{d x_{2}}{d t}=x_{2}\left(2-x_{2}\right)-x_{2}\)

Find all equilibria of each system of differential equations and determine the stability of each equilibrium. \(\frac{d x_{1}}{d t}=2 x_{1}-x_{1}^{2}-2 x_{2} x_{1}\) \(\frac{d x_{2}}{d t}=x_{2}-2 x_{2}^{2}-x_{1} x_{2}\)

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.