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Chapter 10: Problem 46

In Problems 35-46, find the general solution of the given system. $$ X^{\prime}=\left(\begin{array}{rrr} 2 & 4 & 4 \\ -1 & -2 & 0 \\ -1 & 0 & -2 \end{array}\right) X $$

Short Answer

Expert verified
The general solution is a combination of eigenvectors multiplied by exponential factors of eigenvalues: \( X(t) = c_1 \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + c_3 \begin{pmatrix} 0 \\ ? \end{pmatrix} \).

Step by step solution

01

Understand the Problem

We need to find the general solution of the linear system of differential equations described by \( X' = AX \), where \( A \) is a 3x3 matrix given in the problem. The solution generally involves finding the eigenvectors and eigenvalues of matrix \( A \).
02

Find Eigenvalues of Matrix A

Calculate the eigenvalues of the matrix \( A \). First, set up the characteristic equation \( \det(A - \lambda I) = 0 \), where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix of the same order as \( A \). For this problem, solve \( \det\begin{pmatrix}2-\lambda & 4 & 4\-1 & -2-\lambda & 0\-1 & 0 & -2-\lambda\end{pmatrix} = 0 \). Solving this yields the eigenvalues.
03

Calculate the Characteristic Polynomial

Calculate the determinant for \( \det(A - \lambda I) \) to get the characteristic polynomial \( (-\lambda)^3+2\lambda^2=0 \). Simplifying gives the polynomial equation \( \lambda(\lambda^2+2\lambda)=0\).
04

Solve for Eigenvalues

Solve \( \lambda(\lambda^2+2\lambda)=0 \). The solutions are \( \lambda_1 = 0, \lambda_2 = -2, \lambda_3 = 0 \).
05

Find Eigenvectors Corresponding to Eigenvalues

For each eigenvalue, find the corresponding eigenvector by solving the system \( (A - \lambda I)\mathbf{v} = 0 \), where \( \mathbf{v} \) is the eigenvector. Solve for each eigenvalue.
06

Eigenvectors for \( \lambda_1 \) and \( \lambda_3 \) (Zero Eigenvalue)

For \( \lambda_1 = 0 \), solve the system \( A\mathbf{v} = 0 \). The system reduces to \(\begin{pmatrix} 2 & 4 & 4\ -1 & -2 & 0\ -1 & 0 & -2\end{pmatrix} \mathbf{v} = 0\). This typically leads to dependent equations, suggesting a free parameter that allows determination of an eigenvector. Finding a particular solution could give eigenvectors such as \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} \) and another independent one for zero.
07

Eigenvector for \( \lambda_2 = -2 \)

For \( \lambda_2 = -2 \), the system becomes \( (A + 2I)\mathbf{v} = 0 \). The system \(\begin{pmatrix} 4 & 4 & 4\ -1 & 0 & 0\ -1 & 0 & 0\end{pmatrix} \mathbf{v} = 0 \) solve to find an eigenvector such as \( \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} \).
08

Formulate the General Solution

Use the eigenvalues and eigenvectors to form the general solution \( X(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 + c_3 e^{\lambda_3 t} \mathbf{v}_3 \). With the found eigenvalues and eigenvectors, the solution is \( X(t) = c_1 \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} + c_3 \begin{pmatrix} 0 \ ? \end{pmatrix} \) and substitute for any valid \( \mathbf{v}_3 \).
09

Verify the Solution

Check if the obtained solution satisfies the original differential equation by substituting it back into \( X' = AX \). Make sure it holds for arbitrary constants \( c_1, c_2, \) and \( c_3 \).

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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 world of linear differential equations, eigenvalues and eigenvectors play a critical role in finding solutions to systems. These terms might sound complex, but they simplify the process of solving differential equations involving matrices.

**What are Eigenvalues?**
  • An eigenvalue is a special number associated with a matrix that, when multiplied by its eigenvector, stretches or compresses it but doesn't change its direction.
  • To find an eigenvalue, solve the characteristic equation, which is derived from the determinant of the matrix minus a scalar times the identity matrix, expressed mathematically as \( \det(A - \lambda I) = 0 \).
Knowing the eigenvalues helps reveal characteristics of the matrix, such as its stability and the nature of its solutions.

**What about Eigenvectors?**
  • An eigenvector is a non-zero vector that changes only by a scalar factor when the matrix is applied to it.
  • For each eigenvalue, the corresponding eigenvector is found by solving the system \( (A - \lambda I)\mathbf{v} = 0 \).
Eigenvectors provide the direction in which the transformation happens and are vital in constructing solutions to differential equations.
Characteristic Polynomial
To find the eigenvalues of a matrix, one key tool is the characteristic polynomial. This polynomial is derived from the determinant of the matrix equation \( A - \lambda I \), where \( A \) is the original matrix and \( \lambda \) is the eigenvalue
.
**Formulating the Polynomial**
  • The process begins by arranging the equation \( \det(A - \lambda I) = 0 \).
  • The resulting determinant, when expanded, provides a polynomial in terms of \( \lambda \).
  • This polynomial is known as the characteristic polynomial.
For the given matrix in the problem, the derived characteristic polynomial was \[ (-\lambda)^3 + 2\lambda^2 = 0 \].

**Solving the Polynomial**
  • After deriving the polynomial, solve it by factoring or using algebraic methods to find the roots, which are the eigenvalues.
  • For instance, solving \( \lambda(\lambda^2+2\lambda)=0 \) provides the eigenvalues \( \lambda_1 = 0, \lambda_2 = -2, \lambda_3 = 0 \).
The characteristic polynomial not only aids in finding eigenvalues but also gives insight into the structure and properties of the matrix.
General Solution of Systems
Once the eigenvalues and corresponding eigenvectors of a system are determined, constructing the general solution to the differential equation becomes straightforward.

**Using the Eigenvalues and Eigenvectors**
  • The general solution involves combining the eigenvectors, each multiplied by an exponential function of time scaled by the eigenvalue.
  • In mathematical terms, the solution can be expressed as \[ X(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 + c_3 e^{\lambda_3 t} \mathbf{v}_3 \],where \( c_1, c_2, \text{ and } c_3 \) are constants determined by initial conditions.
In the original problem, this results in an expression like \[ X(t) = c_1 \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} + c_3 \begin{pmatrix} 0 \ ? \end{pmatrix} \], with \( \mathbf{v}_3 \) being a valid eigenvector for \( \lambda_3 \).

**Verifying the Solution**
  • To ensure correctness, substitute the general solution back into the original system equation \( X' = AX \).
  • Check if it holds true for any suitable values of the constants, confirming that the derived solution satisfies the system.
This way, the general solution encapsulates the behavior of the system over time, providing invaluable insight into its dynamics.

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Most popular questions from this chapter

Use variation of parameters to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{rr}3 & 2 \\ -2 & -1\end{array}\right) \mathbf{X}+\left(\begin{array}{c}2 e^{-1} \\\ e^{-t}\end{array}\right)\)

Find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=2 x-7 y \\ &\frac{d y}{d t}=5 x+10 y+4 z \\ &\frac{d z}{d t}=5 y+2 z \end{aligned} $$

Use variation of parameters to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{lll}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 3\end{array}\right) \mathbf{X}+\left(\begin{array}{c}e^{t} \\ e^{2 t} \\ t e^{3 t}\end{array}\right)\)

Solve the given initial-value problem. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} \frac{1}{2} & 0 \\ 1 & -\frac{1}{2} \end{array}\right) \mathbf{x}, \quad \mathbf{X}(0)=\left(\begin{array}{l} 3 \\ 5 \end{array}\right) $$

Find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=4 x+5 y \\ &\frac{d y}{d t}=-2 x+6 y \end{aligned} $$
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