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Solve the given initial-value problem. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 2 & 6 \\ 1 & 3 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=-3 \text { and } x_{2}(0)=1 \text { . } $$

Short Answer

Expert verified
The particular solution is \( x_1(t) = -\frac{3}{5}e^{6t} - \frac{12}{5}e^{-t} \) and \( x_2(t) = -\frac{1}{5}e^{6t} + \frac{6}{5}e^{-t} \).

Step by step solution

01

Write the system in matrix form

The system of differential equations can be written in matrix form as: \[ \frac{d\mathbf{x}}{dt} = A\mathbf{x} \]where \( \mathbf{x} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) and \( A = \begin{bmatrix} 2 & 6 \ 1 & 3 \end{bmatrix} \).
02

Determine the eigenvalues of matrix A

To find the eigenvalues of matrix \( A \), solve for \( \lambda \) in the equation: \[ \text{det}(A - \lambda I) = 0 \]where \( I \) is the identity matrix. Calculate:\[ A - \lambda I = \begin{bmatrix} 2-\lambda & 6 \ 1 & 3-\lambda \end{bmatrix} \]Find the determinant:\[ (2-\lambda)(3-\lambda) - 6(1) = \lambda^2 - 5\lambda - 6 = 0 \].Solve for \( \lambda \): The roots of \( \lambda^2 - 5\lambda - 6 = 0 \) are \( \lambda = 6 \) and \( \lambda = -1 \).
03

Find eigenvectors for each eigenvalue

For \( \lambda = 6 \): solve \((A - 6I)\mathbf{v} = 0\). Calculate:\[ A - 6I = \begin{bmatrix} -4 & 6 \ 1 & -3 \end{bmatrix} \]Find a non-zero vector \( \mathbf{v_1} \): The eigenvector can be chosen as \( \mathbf{v_1} = \begin{bmatrix} 3 \ 1 \end{bmatrix} \).For \( \lambda = -1 \): solve \((A + I)\mathbf{v} = 0\). Calculate:\[ A + I = \begin{bmatrix} 3 & 6 \ 1 & 4 \end{bmatrix} \]Find a non-zero vector \( \mathbf{v_2} \): The eigenvector can be chosen as \( \mathbf{v_2} = \begin{bmatrix} -2 \ 1 \end{bmatrix} \).
04

Write the general solution

The general solution of the system is:\[ \mathbf{x}(t) = c_1 e^{6t} \begin{bmatrix} 3 \ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2 \ 1 \end{bmatrix} \] where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.
05

Apply initial conditions

Substitute \( t = 0 \) into the general solution and use the initial conditions \( x_1(0) = -3 \) and \( x_2(0) = 1 \):\[ \begin{bmatrix} -3 \ 1 \end{bmatrix} = c_1 \begin{bmatrix} 3 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} -2 \ 1 \end{bmatrix} \]. This gives the system of equations:1. \( 3c_1 - 2c_2 = -3 \)2. \( c_1 + c_2 = 1 \).Solve these equations for \( c_1 \) and \( c_2 \).
06

Solve system of equations for c1 and c2

From \( c_1 + c_2 = 1 \), we have \( c_2 = 1 - c_1 \). Substitute \( c_2 = 1 - c_1 \) into the first equation:\[ 3c_1 - 2(1 - c_1) = -3 \]which simplifies to:\[ 3c_1 - 2 + 2c_1 = -3 \] \[ 5c_1 = -1 \].Therefore, \( c_1 = -\frac{1}{5} \).Substitute back to find \( c_2 \):\[ c_2 = 1 + \frac{1}{5} = \frac{6}{5} \].
07

Write the particular solution

Substitute \( c_1 = -\frac{1}{5} \) and \( c_2 = \frac{6}{5} \) into the general solution to get:\[ \mathbf{x}(t) = -\frac{1}{5} e^{6t} \begin{bmatrix} 3 \ 1 \end{bmatrix} + \frac{6}{5} e^{-t} \begin{bmatrix} -2 \ 1 \end{bmatrix} \].This can also be expressed as:\[ \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} -\frac{3}{5}e^{6t} - \frac{12}{5}e^{-t} \ -\frac{1}{5}e^{6t} + \frac{6}{5}e^{-t} \end{bmatrix} \].

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Key Concepts

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

Eigenvalues and Eigenvectors
In the context of linear algebra, eigenvalues and eigenvectors are fundamental concepts that can help solve system of differential equations. When dealing with a matrix, eigenvalues are special scalars, while eigenvectors are non-zero vectors associated with these scalars. If you think of a matrix as a function that transforms vectors, then finding an eigenvector means identifying the vector that doesn't change direction under this transformation, only its magnitude, which is scaled by its eigenvalue.

Mathematically, this relationship is expressed as \( A\mathbf{v} = \lambda \mathbf{v} \), where \( A \) is a matrix, \( \mathbf{v} \) is an eigenvector, and \( \lambda \) is its corresponding eigenvalue. To find the eigenvalues of a matrix, one typically solves the characteristic equation \( \det(A - \lambda I) = 0 \). This involves subtracting \( \lambda \) along the diagonal of the matrix and equating its determinant to zero.

In our exercise, we calculated the eigenvalues \( \lambda = 6 \) and \( \lambda = -1 \) for the matrix \( A = \begin{bmatrix} 2 & 6 \ 1 & 3 \end{bmatrix} \). For each eigenvalue, we find a corresponding eigenvector by solving \( (A - \lambda I)\mathbf{v} = 0 \). This gives us the direction in which the system evolves. These steps help us form the general solution to our system of differential equations.
System of Differential Equations
A system of differential equations is a set of equations that involve derivatives of multiple interdependent variables. These systems often model real-world phenomena like population dynamics, electrical circuits, or mechanical systems, where multiple variables influence each other over time. To solve such systems mathematically, we often represent them in matrix form, which succinctly describes the relationships between variables.

Consider the problem at hand, modeled as \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), where \( \mathbf{x} \) is a vector of functions. Here, \( A \) is the matrix that determines how each function interacts with others in the system. Such a representation allows us to apply techniques from linear algebra, like finding eigenvalues and eigenvectors, thus simplifying the task of solving the system.

The matrix approach consolidates the differential equations into one concise equation. It paves the way for utilizing the powerful tools of linear algebra, allowing us to explore solutions that describe how the system evolves over time. The general solution comprises combinations of exponentials with eigenvalues and eigenvectors, representing the system's behavior under various initial conditions.
Initial Value Problems
Initial value problems (IVPs) are a class of differential equations where the solution is conditioned on specific initial values for the variables. These problems are common in applications where you are interested in how a system evolves from a particular starting state, for example, how a population grows from an initial size or how a capacitor charges from an initial voltage level.

The solution process begins by solving the general differential equation, often resulting in a solution with constants. The given initial conditions then provide specific values for these constants. In our exercise, we have the initial conditions \( x_1(0) = -3 \) and \( x_2(0) = 1 \). These initial conditions guide us in calculating the constants in the solution's expression.

By substituting these values into the general solution, we solve for the specific constants \( c_1 \) and \( c_2 \). This transforms the general solution into a particular solution, which fulfills the initial conditions and precisely describes the system's state over time. Solving IVPs gives a concrete outcome to abstract systems of equations, revealing the distinct path a system follows from its initial state.

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Most popular questions from this chapter

Use the mass action law to translate each chemical reaction into a system of differential equations. \(\mathrm{A}+\mathrm{B} \stackrel{k}{\longrightarrow} \mathrm{C}\)

Transform the second-order differential equation $$ \frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}=x $$ into a system of first-order differential equations.

A very simple two-compartment model for gap dynamics in a forest assumes that gaps are created by disturbances (wind, fire, etc.) and that gaps revert to forest as trees grow in the gaps. We denote by \(x_{1}(t)\) the area occupied by gaps and by \(x_{2}(t)\) the area occupied by adult trees. We assume that the dynamics are given by $$ \begin{array}{l} \frac{d x_{1}}{d t}=-0.2 x_{1}+0.1 x_{2} \\ \frac{d x_{2}}{d t}=0.2 x_{1}-0.1 x_{2} \end{array} $$ (a) Find the corresponding compartment diagram. (b) Show that \(x_{1}(t)+x_{2}(t)\) is a constant. Denote the constant by \(A\) and give its meaning. [Hint: Show that \(\frac{d}{d t}\left(x_{1}+x_{2}\right)=0 .\) ] (c) Let \(x_{1}(0)+x_{2}(0)=20\). Use your answer in (b) to explain why this equation implies that \(x_{1}(t)+x_{2}(t)=20\) for all \(t>0\). (d) Use your result in (c) to replace \(x_{2}\) in (11.44) by \(20-x_{1}\), and show that doing so reduces the system \((11.44)\) and \((11.45)\) to $$ \frac{d x_{1}}{d t}=2-0.3 x_{1} $$ with \(x_{1}(t)+x_{2}(t)=20\) for all \(t \geq 0\). (e) Solve the system (11.44) and (11.45), and determine what fraction of the forest is occupied by adult trees at time \(t\) when \(x_{1}(0)=2\) and \(x_{2}(0)=18\). What happens as \(t \rightarrow \infty\) ?

\- Iranstorm the second-order dutferentral equation $$ \frac{d^{2} x}{d t^{2}}=3 x $$ into a system of first-order differential equations.

Find the general solution of each given system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line the direction in which the solution would move if it starts on that line. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} -2 & 0 \\ -3 & 1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

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