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Chapter 6: Problem 14

\(x^{\prime}=2 x-y, y^{\prime}=-4 x-2 y\)

Short Answer

Expert verified
The solutions for the system are expressed in terms of eigenvalues and eigenvectors.

Step by step solution

01

- Write the System of Equations as a Matrix

Express the given differential equations as a matrix equation. The system can be written in matrix form as:
02

- Find Eigenvalues

Calculate the determinant of the matrix
03

- Calculate Eigenvalues

Solve the characteristic equation
04

- Find Eigenvectors

For each eigenvalue, find the corresponding eigenvector by solving
05

- Write the General Solution

Combine the eigenvalues and eigenvectors to express the general solution

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Key Concepts

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

Matrix Representation
Systems of differential equations can be complex if written as individual equations, so representing them in matrix form simplifies the process.

To convert the system \( x^{\prime}=2 x-y \, y^{\prime}=-4 x-2 y \) into a matrix form:
  • Write the system as \( \begin{bmatrix} x^{\prime} \ y^{\prime} \end{bmatrix} = \begin{bmatrix} 2 & -1 \ -4 & -2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} \)
  • This compact notation is helpful for analysis and solving.
Eigenvalues
Eigenvalues help us understand the behavior of the system over time. To find them:
  • Form the characteristic equation by subtracting \( \lambda \) times the identity matrix from the coefficient matrix.
  • In our case, the characteristic equation becomes \[ det(\begin{bmatrix} 2 & -1 \ -4 & -2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}) = 0 \] which simplifies to \[ (2 - \lambda)(-2 - \lambda) - (-1)(-4) = 0 \]
  • Solving this equation gives the eigenvalues (\( \lambda \)).
Eigenvectors
After finding eigenvalues, use them to find eigenvectors, which indicate directions associated with each eigenvalue.
  • Substitute each \( \lambda \) back into the matrix equation.
  • Solve \[ (A - \lambda I)v = 0 \] for each \( \lambda \) to get eigenvectors.
  • For example, if one eigenvalue is \( \lambda = -2\), plug it into \(\begin{bmatrix} 2 & -1 \ -4 & -2 \end{bmatrix} - (-2)\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0 \) and solve.
General Solution
The general solution combines all eigenvalues and eigenvectors to describe the system's overall behavior.

Using the formula:
  • General solution is \[ \mathbf{x}(t) = c_1 \mathbf{v_1} e^{\lambda_1 t} + c_2 \mathbf{v_2} e^{\lambda_2 t} \].
  • Here, \( c_1 \) and \( c_2 \) are constants determined by initial conditions, and \( \lambda_1, \lambda_2 \) are eigenvalues with corresponding eigenvectors \( \mathbf{v_1}, \mathbf{v_2} \).
  • For our system, this approach gives insights into how variables evolve over time, predicting future states effectively.

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Most popular questions from this chapter

\(\mathbf{X}^{\prime}=\left(\begin{array}{ll}3 & -1 \\ 4 & -1\end{array}\right) \mathbf{X}+\left(\begin{array}{c}\cos t \\ \sin t\end{array}\right), \mathbf{X}(0)=\left(\begin{array}{l}0 \\\ 0\end{array}\right)\)

\(x^{\prime}=y, y^{\prime}=-13 x-4 y, x(0)=-1, y(0)=1\), \(0 \leq t \leq 10\)

Solve the systems (a) \(\left\\{\begin{array}{l}x^{\prime}=2 x-y \\\ y^{\prime}=-x+3 y\end{array}\right.\); (b) \(\left\\{\begin{array}{l}x^{\prime}=2 x \\ y^{\prime}=3 x+2 y\end{array} ;\right.\) (c) \(\left\\{\begin{array}{l}x^{\prime}=x+4 y \\ y^{\prime}=-2 x-y\end{array}\right.\) subject to \(x(0)=1\) and \(y(0)=1\). In each case, graph the solution parametrically and individually.

\(x^{\prime}=-x+2 y+5, y^{\prime}=2 x-y+4, x(0)=1\), \(y(0)=0, t=1\)

\(\left\\{\begin{array}{l}d x / d t=-5 x+6 y+1 \\ d y / d t=-7 y+t \\ x(0)=1, y(0)=-1\end{array}\right.\)
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