/*! 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 19 Solve the initial value problem ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the initial value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(A=\left[ \begin{array}{rr}{1} & {0} \\ {4} & {-1}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{l}{3} \\ {2}\end{array}\right]\)

Short Answer

Expert verified
\( \mathbf{x}(t) = \begin{pmatrix} 3 e^{t} \\ 2 e^{-t} \end{pmatrix} \) is the solution.

Step by step solution

01

Introduction to the Problem

We are given a differential equation \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) with an initial condition \( \mathbf{x}(0) = \mathbf{x}_0 \). The matrix \( A \) and vector \( \mathbf{x}_0 \) are specified, and we aim to find \( \mathbf{x}(t) \).
02

Find the Eigenvalues of A

The matrix \( A \) is \( \left[ \begin{array}{rr}{1} & {0} \ {4} & {-1}\end{array}\right] \). To find the eigenvalues, solve \( \text{det}(A - \lambda I) = 0 \). Here, the calculation gives:\[\text{det}\left( \begin{array}{cc} 1-\lambda & 0 \ 4 & -1-\lambda \end{array}\right) = (1-\lambda)(-1-\lambda) - 0 = \lambda^2 - 1\]Setting this equal to zero, we solve \( \lambda^2 - 1 = 0 \), which results in the eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \).
03

Find Eigenvectors for Each Eigenvalue

For \( \lambda_1 = 1 \), solve \( (A - I) \mathbf{v}_1 = 0 \). Setting up the matrix equation:\[\left( \begin{array}{cc} 0 & 0 \ 4 & -2 \end{array}\right) \begin{pmatrix} x_1 \ y_1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\]This results in dependencies giving \( y_1 = 2 \) with \( x_1 \) free, so \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 0 \end{pmatrix} \).For \( \lambda_2 = -1 \), solve \( (A + I) \mathbf{v}_2 = 0 \). The matrix setup is:\[\left( \begin{array}{cc} 2 & 0 \ 4 & 0 \end{array}\right) \begin{pmatrix} x_2 \ y_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\]Which gives \( y_2 = 0 \) and \( x_2 \) free, so \( \mathbf{v}_2 = \begin{pmatrix} 0 \ 1 \end{pmatrix} \).
04

General Solution of the Differential Equation

The general solution for \( \mathbf{x}(t) \) is a linear combination of the eigenvectors based on their eigenvalues:\[\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 = c_1 e^{t} \begin{pmatrix} 1 \ 0 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 0 \ 1 \end{pmatrix}\]Thus, \( \mathbf{x}(t) = \begin{pmatrix} c_1 e^{t} \ c_2 e^{-t} \end{pmatrix} \).
05

Apply Initial Conditions

Use the initial condition \( \mathbf{x}(0) = \mathbf{x}_0 = \begin{pmatrix} 3 \ 2 \end{pmatrix} \). At \( t = 0 \):\[\mathbf{x}(0) = \begin{pmatrix} c_1 e^0 \ c_2 e^0 \end{pmatrix} = \begin{pmatrix} c_1 \ c_2 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \end{pmatrix}\]Hence, \( c_1 = 3 \) and \( c_2 = 2 \).
06

Final Solution

Substitute \( c_1 \) and \( c_2 \) into the general solution:\[\mathbf{x}(t) = \begin{pmatrix} 3 e^{t} \ 2 e^{-t} \end{pmatrix}\]This function solves the initial value problem given.

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.

Initial Value Problem
An Initial Value Problem (IVP) is a type of differential equation paired with specified conditions at the starting point, typically time zero. These problems are foundational in understanding how systems evolve over time.
To solve an IVP, you often start with a differential equation like \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) where \( A \) is a matrix, and \( \mathbf{x}(0) = \mathbf{x}_0 \) is the initial condition. The goal is to find the function \( \mathbf{x}(t) \) that satisfies both the equation and the initial condition.
To solve IVPs:
  • Identify the type of differential equation and its order.
  • Apply relevant techniques, such as finding eigenvalues and eigenvectors in case of systems of equations.
  • Use initial conditions to determine constants in the general solution.
This kind of problem is very common in physics and engineering, where it is essential to forecast the future state of systems based on their current conditions.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are key concepts in linear algebra, especially when dealing with matrix equations in differential equations. They provide insights into the behavior of systems described by these equations.
For a given square matrix \( A \), an eigenvalue \( \lambda \) and a non-zero vector \( \mathbf{v} \) satisfy the equation \( A \mathbf{v} = \lambda \mathbf{v} \). This means multiplying \( \mathbf{v} \) by the matrix \( A \) results in a vector that is a scaled version of \( \mathbf{v} \), with \( \lambda \) as the scaling factor.
To find eigenvalues:
  • Set up the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix.
  • Solve this polynomial equation for \( \lambda \).
Once you have the eigenvalues, find the corresponding eigenvectors by solving \( (A - \lambda I) \mathbf{v} = 0 \).
Eigenvalues can indicate stability in a system, with positive eigenvalues suggesting growth, and negative eigenvalues indicating decay. These concepts are crucial in understanding the dynamics of differential equations.
General Solution of Differential Equations
The general solution of a differential equation represents the family of all possible solutions of the equation. When dealing with linear systems like \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \), the solution is expressed as a combination of exponential functions derived from the eigenvalues and eigenvectors of the matrix \( A \).
The general solution is often written as:\[\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 + \ldots\]where \( c_1, c_2, \ldots \) are constants determined by initial conditions, \( \lambda_1, \lambda_2, \ldots \) are eigenvalues, and \( \mathbf{v}_1, \mathbf{v}_2, \ldots \) are their corresponding eigenvectors.
Finding the general solution involves:
  • Solving the characteristic equation for eigenvalues.
  • Finding the eigenvectors for each eigenvalue.
  • Combining these components with arbitrary constants to represent all solutions.
Initially, the solution is general, meaning it includes all potential cases that fit the differential equation. Applying any specific conditions narrows down the solution to a particular case. This method is widely used in solving systems of linear differential equations, especially in modeling dynamic systems in science and engineering.

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

8\. Metapopulations Consider a simple metapopulation in which subpopulation \(\mathrm{A}\) grows at a per capita rate of \(r_{\mathrm{A}}=1\) and subpopulation \(\mathrm{B}\) declines at a per capita rate of \(r_{\mathrm{B}}=-1 .\) Suppose the per capita rate of movement between subpopulation patches is \(m\) in both directions. This gives $$\begin{aligned} \frac{d x_{\mathrm{A}}}{d t} &=(1-m) x_{\mathrm{A}}+m x_{\mathrm{B}} \\ \frac{d x_{\mathrm{B}}}{d t} &=-(1+m) x_{\mathrm{B}}+m x_{\mathrm{A}} \end{aligned}$$ where \(x_{\text { A }}\) and \(x_{\mathrm{B}}\) are the numbers of individuals in patches \(\mathrm{A}\)and \(\mathrm{B},\) respectively. $$\begin{array}{l}{\text { (a) Classify the equilibrium at the origin. }} \\\ {\text { (b) Find the general solution. }} \\ {\text { (c) What is the solution to the initial-value problem if }} \\ {x_{\mathrm{A}}(0)=1 \text { and } x_{\mathrm{B}}(0)=0 ?}\end{array}$$

9\. Suppose a glass of cold water is sitting in a warm room and you place a coin at room temperature \(R\) into the glass. The coin gradually cools down while, at the same time, the glass of water warms up. Newton's law of cooling suggests the following system of differential equations to describe the process $$\frac{d w}{d t}=-k_{m}(w-R) \quad \frac{d p}{d t}=-k_{p}(p-w)$$ where \(w\) and \(p\) are the temperatures of the water and coil (in "C), respectively, and the \(k\) 's are positive constants. $$\begin{array}{l}{\text { (a) Explain the form of the system of differential equations }} \\ {\text { and the assumptions that underlie them. }} \\\ {\text { (b) Use a change of variables to obtain a homogeneous }} \\ {\text { system. }} \\ {\text { (c) What is the general solution to the system you found in }} \\ {\text { part (b)? }} \\ {\text { (d) What is the solution to the original initial-value problem }} \\ {\text { if } w(0)=w_{0} \text { and } p(0)=p_{0} ?}\end{array}$$

2\. Hemodialysis is a process by which a machine is used to filter urea and other waste products from a patient's blood if the kidneys fail. The amount of urea within a patient during dialysis is sometimes modeled by supposing there are two compartments within the patient: the blood, which is directly filtered by the dialysis machine, and another com- partment that cannot be directly filtered but that is con- nected to the blood. A system of two differential equations describing this is $$\frac{d c}{d t}=-\frac{K}{V} c+a p-b c \quad \frac{d p}{d t}=-a p+b c$$ where \(c\) and \(p\) are the urea concentrations in the blood and the inaccessible pool (in \(\mathrm{mg} / \mathrm{mL} )\) and all constants are positive (see also Exercise 14 in the Review Section of this chapter). Suppose that \(K / V=1, a=b=\frac{1}{2},\) and the initial urea concentration is \(c(0)=c_{0}\) and \(p(0)=c_{0} \mathrm{mg} / \mathrm{mL}\) $$\begin{array}{l}{\text { (a) Classify the equilibrium of this system. }} \\\ {\text { (b) Solve this initial-value problem. }}\end{array}$$

A Jacobian matrix and two equlibria are given. Determine if each is locally stable, unstable, or if the analysis is inconclusive. $$\begin{array}{l}{J=\left[ \begin{array}{cc}{1-\cos x_{2}} & {\left(x_{1}-1\right) \sin x_{2}} \\ {\cos x_{1}} & {-\sin 1}\end{array}\right]} \\ {\text { (i) } \hat{x}_{1}=0, \hat{x}_{2}=0} \\\ {\text { (ii) } \hat{x}_{1}=1, \hat{x}_{2}=1}\end{array}$$

Specify whether each system is autonomous or nonautonomous, and whether it is linear or nonlinear. If it is linear, specify whether it is homogeneous or nonhomogeneous. \(d y / d t=3 y z-2 z, \quad d z / d t=2 z+5 y\)

See all solutions

Recommended explanations on Math 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.