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Magazine Chapter 7: Problem 18
Find the general solution of each of the linear systems in Exercises \(1-22\) $$ \begin{aligned} &\frac{d x}{d t}=6 x-5 y \\ &\frac{d y}{d t}=x+2 y \end{aligned} $$
Short Answer
Expert verified
The general solution of the given linear system is:
$$
\begin{bmatrix}
x(t) \\
y(t)
\end{bmatrix}
= c_1 e^{(4 + i)t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{(4 - i)t} \begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
where \(c_1\) and \(c_2\) are constants determined by the initial conditions.
Step by step solution
01
Convert to matrix form
Rewrite the linear system as a matrix equation:
$$
\begin{bmatrix}
\frac{dx}{dt} \\
\frac{dy}{dt}
\end{bmatrix}
= A
\begin{bmatrix}
x \\
y
\end{bmatrix}
$$
where the matrix A is
$$
A = \begin{bmatrix}
6 & -5 \\
1 & 2
\end{bmatrix}
$$
02
Find the eigenvalues of the matrix A
Compute the characteristic equation of the matrix A:
$$
\text{det}(A - \lambda I) = \text{det}
\begin{bmatrix}
6-\lambda & -5 \\
1 & 2-\lambda
\end{bmatrix}
= (6-\lambda)(2-\lambda) - (-5)(1)
$$
Solve for the eigenvalues:
$$
(6-\lambda)(2-\lambda) - (-5)(1) = \lambda^2 - 8\lambda + 17 = (\lambda - 4)^2 + 1
$$
The eigenvalues are \(\lambda_1 = 4 + i\) and \(\lambda_2 = 4 - i\).
03
Find the eigenvectors for the eigenvalues
First, find the eigenvector corresponding to \(\lambda_1 = 4 + i\):
$$
(A - (4 + i) I) \textbf{v}_1 = 0
$$
with
$$
\begin{bmatrix}
2 - i & -5 \\
1 & -2 - i
\end{bmatrix}
\begin{bmatrix}
v_{11}\\
v_{12}
\end{bmatrix}
= 0
$$
Row reduce the augmented matrix to echelon form:
$$
\left[
\begin{array}{cc|c}
2 - i & -5 & 0 \\
1 & -2 - i & 0 \\
\end{array}
\right]
\sim
\left[
\begin{array}{cc|c}
1 & -2 - i & 0 \\
0 & (3 + 2i) & 0 \\
\end{array}
\right]
$$
From the reduced matrix, we have
$$
(3 + 2i)v_{12} = 0
\implies
v_{12} = 0
$$
Thus, we can choose \(v_{11} = 1\), and the eigenvector is \(\textbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\). Then, find the eigenvector corresponding to \(\lambda_2 = 4 - i\):
$$
(A - (4 - i) I) \textbf{v}_2 = 0
$$
with
$$
\begin{bmatrix}
2 + i & -5 \\
1 & -2 + i
\end{bmatrix}
\begin{bmatrix}
v_{21}\\
v_{22}
\end{bmatrix}
= 0
$$
Row reduce the augmented matrix to echelon form:
$$
\left[
\begin{array}{cc|c}
2 + i & -5 & 0 \\
1 & -2 + i & 0 \\
\end{array}
\right]
\sim
\left[
\begin{array}{cc|c}
1 & -2 + i & 0 \\
0 & (3 - 2i) & 0 \\
\end{array}
\right]
$$
From the reduced matrix, we have
$$
(3 - 2i)v_{22} = 0
\implies
v_{22} = 0
$$
Thus, we can choose \(v_{21} = 1\), and the eigenvector is \(\textbf{v}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\).
04
Construct the general solution
Using the eigenvalues and eigenvectors, construct the general solution as follows:
$$
\begin{bmatrix}
x(t) \\
y(t)
\end{bmatrix}
= c_1 e^{(4 + i)t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{(4 - i)t} \begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
where \(c_1\) and \(c_2\) are constants determined by the initial conditions.
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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 differential equations, eigenvalues and eigenvectors play a crucial role when solving systems of equations. Let's break down these concepts to understand why they are so important.
**Eigenvalues** are numbers associated with a square matrix that determine certain characteristics of systems represented by the matrix, such as stability and oscillatory behavior. In our problem, the matrix is \[A = \begin{bmatrix}6 & -5 \1 & 2\end{bmatrix}\]We determine the eigenvalues of this matrix using the characteristic equation: \[\text{det}(A - \lambda I) = 0\]where \(I\) is the identity matrix. This equation ultimately gives us the eigenvalues \(\lambda_1 = 4 + i\) and \(\lambda_2 = 4 - i\).
**Eigenvectors** are non-zero vectors that only scale (not change direction) when a linear transformation, defined by the matrix, is applied to them. For each eigenvalue, there is one or more corresponding eigenvectors, which are found by solving \[(A - \lambda I)\textbf{v} = 0\]In this example, each eigenvalue has the same eigenvector, \(\textbf{v} = \begin{bmatrix} 1 \ 0 \end{bmatrix}\).
**Eigenvalues** are numbers associated with a square matrix that determine certain characteristics of systems represented by the matrix, such as stability and oscillatory behavior. In our problem, the matrix is \[A = \begin{bmatrix}6 & -5 \1 & 2\end{bmatrix}\]We determine the eigenvalues of this matrix using the characteristic equation: \[\text{det}(A - \lambda I) = 0\]where \(I\) is the identity matrix. This equation ultimately gives us the eigenvalues \(\lambda_1 = 4 + i\) and \(\lambda_2 = 4 - i\).
**Eigenvectors** are non-zero vectors that only scale (not change direction) when a linear transformation, defined by the matrix, is applied to them. For each eigenvalue, there is one or more corresponding eigenvectors, which are found by solving \[(A - \lambda I)\textbf{v} = 0\]In this example, each eigenvalue has the same eigenvector, \(\textbf{v} = \begin{bmatrix} 1 \ 0 \end{bmatrix}\).
- Finding both eigenvalues and eigenvectors helps to map the behavior of linear systems over time.
- Eigenvectors provide directions of vector stability, while eigenvalues describe how vectors scale along these directions.
Matrix Form of a System of Equations
The matrix form of a system of equations is a succinct and organized way to represent and solve linear systems, especially when dealing with differential equations. This form lets us use powerful matrix algebra techniques.
To convert a system of differential equations into matrix form, we represent the system as a vector equation. For example, given the equations:\[\frac{dx}{dt}=6x-5y\\frac{dy}{dt}=x+2y\]These can be rewritten using matrices as:
\[\begin{bmatrix}\frac{dx}{dt} \\frac{dy}{dt}\end{bmatrix}=\begin{bmatrix}6 & -5 \1 & 2\end{bmatrix}\begin{bmatrix}x \y\end{bmatrix}\]
To convert a system of differential equations into matrix form, we represent the system as a vector equation. For example, given the equations:\[\frac{dx}{dt}=6x-5y\\frac{dy}{dt}=x+2y\]These can be rewritten using matrices as:
\[\begin{bmatrix}\frac{dx}{dt} \\frac{dy}{dt}\end{bmatrix}=\begin{bmatrix}6 & -5 \1 & 2\end{bmatrix}\begin{bmatrix}x \y\end{bmatrix}\]
- The matrix \(A\) encapsulates the coefficients of the variables, providing a clear picture of how each variable relies on others.
- This matrix form is beneficial because methods like finding eigenvalues and eigenvectors, which are key to solving the system, become straightforward.
- It also simplifies the process of deriving general solutions to such systems.
General Solution of Differential Equations
The general solution of differential equations gives us a comprehensive formulation of all possible solutions of a system, considering any initial conditions.
After finding the eigenvalues \( 4 + i \) and \( 4 - i \), and the eigenvectors \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \), we construct the solution in the matrix form:\[\begin{bmatrix}x(t) \y(t)\end{bmatrix}= c_1 e^{(4+i)t} \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 e^{(4-i)t} \begin{bmatrix} 1 \ 0 \end{bmatrix}\]
This formula represents the general solution, and \( c_1 \) and \( c_2 \) are constants determined by initial conditions. The inclusion of \( e^{(4+i)t} \) and \( e^{(4-i)t} \) indicates an oscillatory solution due to the complex parts \( i \), and an exponential trend because of the real part 4.
After finding the eigenvalues \( 4 + i \) and \( 4 - i \), and the eigenvectors \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \), we construct the solution in the matrix form:\[\begin{bmatrix}x(t) \y(t)\end{bmatrix}= c_1 e^{(4+i)t} \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 e^{(4-i)t} \begin{bmatrix} 1 \ 0 \end{bmatrix}\]
This formula represents the general solution, and \( c_1 \) and \( c_2 \) are constants determined by initial conditions. The inclusion of \( e^{(4+i)t} \) and \( e^{(4-i)t} \) indicates an oscillatory solution due to the complex parts \( i \), and an exponential trend because of the real part 4.
- The general solution is complete and covers every possible scenario by superimposing the effect of different homogeneous solutions.
- It shows how any point in time the states could be captured depending on initial states.
- Understanding the general solution involves grasping how the matrix and its properties like eigenvalues drive the solutions.
