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Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of \(t .\) $$\begin{array}{ll} x_{1}^{\prime}=x_{1}+x_{2}, & x_{1}(0)=1 \\ x_{2}^{\prime}=x_{1}-x_{2}, & x_{2}(0)=0 \end{array}$$

Short Answer

Expert verified
The general solution is \( x_1(t) = c_1 e^{2t} + c_2 e^{-t}, x_2(t) = c_1 e^{2t} - 2c_2 e^{-t} \). The specific solution satisfying the initial conditions is \( x_1(t) = \frac{2}{3}e^{2t} + \frac{1}{3}e^{-t} \), \( x_2(t) = \frac{2}{3}e^{2t} - \frac{2}{3}e^{-t} \)."

Step by step solution

01

Write the System in Matrix Form

The given differential equations can be expressed as a system using a matrix. We can write this system as: \[\mathbf{x}' = \begin{bmatrix} x_1' \ x_2' \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix}\]or, \[\mathbf{x}' = A \mathbf{x}\], where \( A = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \) and \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \).
02

Find Eigenvalues of Matrix A

To solve the system, we need to find the eigenvalues of matrix \( A \). The eigenvalues \( \lambda \) can be found by solving the characteristic equation:\[\det(A - \lambda I) = 0\]Substituting \( A \) and simplifying, we get:\[\begin{vmatrix} 1-\lambda & 1 \ 1 & -1-\lambda \end{vmatrix} = (1-\lambda)(-1-\lambda) - 1 = \lambda^2 - \lambda - 2 = 0\]Solve this quadratic equation to find the eigenvalues: \( \lambda = 2 \) and \( \lambda = -1 \).
03

Find Eigenvectors for Each Eigenvalue

For \( \lambda = 2 \), solve \((A - 2I)\mathbf{v} = 0\):\[\begin{bmatrix} -1 & 1 \ 1 & -3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]This system simplifies to \( v_1 = v_2 \), so a valid eigenvector is \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \).For \( \lambda = -1 \), solve \((A + I)\mathbf{v} = 0\):\[\begin{bmatrix} 2 & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]This system simplifies to \( 2v_1 + v_2 = 0 \), yielding the eigenvector \( \begin{bmatrix} 1 \ -2 \end{bmatrix} \).
04

Construct General Solution

Using the eigenvalues and eigenvectors, the general solution of the system is of the form:\[\mathbf{x}(t) = c_1 e^{2t} \begin{bmatrix} 1 \ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \ -2 \end{bmatrix}\]Thus:\[x_1(t) = c_1 e^{2t} + c_2 e^{-t}\]\[x_2(t) = c_1 e^{2t} - 2c_2 e^{-t}\]
05

Apply Initial Conditions to Find Specific Solution

Substitute the initial conditions \( x_1(0) = 1 \) and \( x_2(0) = 0 \) into the general solution:- For \( x_1(0) = 1 \): \[ 1 = c_1 + c_2 \]- For \( x_2(0) = 0 \): \[ 0 = c_1 - 2c_2 \]Solving this system of equations, we find:\[c_1 = \frac{2}{3}, \quad c_2 = \frac{1}{3}\]
06

Write Specific Solution

Substitute \( c_1 \) and \( c_2 \) back into the general solution to find the specific solution:\[x_1(t) = \frac{2}{3}e^{2t} + \frac{1}{3}e^{-t}\]\[x_2(t) = \frac{2}{3}e^{2t} - \frac{2}{3}e^{-t}\]

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Key Concepts

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

Eigenvalues
When facing a system of differential equations, one of the first steps to solving it involves finding the eigenvalues of the associated matrix. Eigenvalues are special scalars in a mathematical equation, and they reveal important properties about a system. In the context of differential equations, eigenvalues help determine the behavior and stability of each solution component.
For a given matrix, the eigenvalues are found by solving the characteristic equation, which is derived from the matrix minus a scaled identity matrix: \[ \det(A - \lambda I) = 0 \]
In our original problem, the matrix \( A = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \) leads us to solve the equation \( \lambda^2 - \lambda - 2 = 0 \), resulting in eigenvalues \( \lambda = 2 \) and \( \lambda = -1 \).
This means our system has two distinct eigenvalues, which play a critical role in constructing the general solution.
Eigenvectors
Eigenvectors are vectors that, when multiplied by the matrix, do not change direction, only their magnitude, which is scaled by the eigenvalues. Each eigenvalue has a corresponding eigenvector, and these eigenvectors form the basis of the solution space for the system of equations.
For eigenvalue \( \lambda = 2 \) in our problem, the associated eigenvector is found by solving: \((A - 2I)\mathbf{v} = 0\). This results in \( \begin{bmatrix} -1 & 1 \ 1 & -3 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), simplifying to \( v_1 = v_2 \). Thus, a valid eigenvector is \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
Similarly, for \( \lambda = -1 \), the eigenvector \( \begin{bmatrix} 1 \ -2 \end{bmatrix} \) is found through \((A + I)\mathbf{v} = 0\), leading to the system \( 2v_1 + v_2 = 0 \).
These eigenvectors allow us to construct the general solution of the differential equation.
Initial Conditions
Initial conditions are specific values given for the solution at a particular time, usually \( t = 0 \). They help determine the exact trajectory a solution will take out of the infinite solutions in a differential system.
In our problem, initial conditions are \( x_1(0) = 1 \) and \( x_2(0) = 0 \).
To incorporate these into the general solution, we substitute these values to solve for the constants, \( c_1 \) and \( c_2 \), in the general equation: \[ x_1(t) = c_1 e^{2t} + c_2 e^{-t} \] \[ x_2(t) = c_1 e^{2t} - 2c_2 e^{-t} \]
By applying the initial conditions, we set up equations: \[ 1 = c_1 + c_2 \] \[ 0 = c_1 - 2c_2 \] leading us to the specific values \( c_1 = \frac{2}{3} \) and \( c_2 = \frac{1}{3} \). These constants precisely define the specific trajectory of the solution.
Matrix Form
Converting a system of differential equations into matrix form is a crucial step that allows us to utilize linear algebra techniques to find a solution. By expressing the system in terms of matrices, we can work with eigenvalues and eigenvectors to determine the behavior of the system over time.
The original system of equations: \( x_1' = x_1 + x_2 \) and \( x_2' = x_1 - x_2 \) can be written in matrix form as: \[ \mathbf{x}' = A\mathbf{x} \] where \( A = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \) and \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \).
This transformed representation simplifies the system into a more manipulable form, allowing us to systematically approach finding the general and specific solutions. Using matrices also clarifies the impact of each component on the system dynamics, enhancing our understanding of how solutions evolve.

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Most popular questions from this chapter

Draw the Gerschgorin disks for the given matrix. $$\left[\begin{array}{rrr} 2 & -i & 0 \\ 1 & 2 i & 1+i \\ 0 & 1 & -2 i \end{array}\right]$$

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Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of \(t .\) $$\begin{array}{ll} y_{1}^{\prime}=y_{1}-y_{2}, & y_{1}(0)=1 \\ y_{2}^{\prime}=y_{1}+y_{2}, & y_{2}(0)=1 \end{array}$$

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It can be shown that a nonnegative \(n \times n\) matrix is irreducible if and only if \((I+A)^{n-1}>0 .\) Use this criterion to determine whether the matrix \(A\) is irreducible. If \(A\) is reducible, find a permutation of its rows and columns that puts \(A\) into the block form \\[ \left[\begin{array}{ll} B & C \\ O & D \end{array}\right] \\] $$A=\left[\begin{array}{llll} 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}\right]$$

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