Question

Consider the following systems. For each system, determine the coefficient matrix. When possible, solve the eigenvalue problem for each matrix and use the eigenvalues and eigenvectors to provide solutions to the given systems. Finally, in the common cases that you investigated in Problem 2.31, make comparisons with your previous answers, such as what type of eigenvalues correspond to stable nodes. a. $$ \begin{aligned} &x^{\prime}=3 x-y \\ &y^{\prime}=2 x-2 y \end{aligned} $$ b. $$ \begin{aligned} &x^{\prime}=-y_{t} \\ &y^{\prime}=-5 x \end{aligned} $$ c. $$ \begin{aligned} &x^{\prime}=x-y_{r} \\ &y^{\prime}=y \end{aligned} $$ d. $$ \begin{aligned} &x^{\prime}=2 x+3 y \\ &y^{\prime}=-3 x+2 y \end{aligned} $$ e. $$ \begin{aligned} x^{\prime} &=-4 x-y \\ y^{\prime} &=x-2 y \end{aligned} $$ \(\mathrm{f}\). $$ \begin{aligned} x^{\prime} &=x-y \\ y^{\prime} &=x+y \end{aligned} $$

Short answer

The eigenvalues and eigenvectors are determined for each system by calculating the determinant of (\(A - \lambda I)\), where A is the coefficient matrix, \(\lambda\) is eigenvalue and I is identity matrix. Then comparisons have to be made with Problem 2.31 to understand what types of eigenvalues correspond to stable nodes.

Step-by-step solution

System a

Let's start with system a. The coefficient matrix is \[\[3, -1], [2, -2]\]. Using standard procedures, eigenvalues and eigenvectors can be calculated. This is achieved by setting the determinant of \((\mathbf{A} - \lambda\mathbf{I})\) to zero, where \(\mathbf{A}\) is the given matrix, \(\lambda\) is the eigenvalue and \(\mathbf{I}\) is the identity matrix.

System b

Moving on to system b. The coefficient matrix is \[\[0, -1], [-5, 0]\]. The eigenvalues and eigenvectors can be determined in the same manner as system a.

System c

For system c, the coefficient matrix is \[\[1, 0], [0, 1]\]. Once the eigenvalues and eigenvectors have been computed, they can be used to solve this system of equations.

System d

The fourth system, d, has the coefficient matrix \[\[2, 3], [-3, 2]\]. The same methods used previously can calculate the eigenvalues and eigenvectors and solve the system.

System e

Moving to system e, the coefficient matrix is \[\[-4, -1], [1, -2]\]. The eigenvalues and eigenvectors of this can be calculated by setting the determinant of \((\mathbf{A} - \lambda\mathbf{I})\) to zero.

System f

Finally, system f gives the coefficient matrix \[\[1, -1], [1, 1]\]. The eigenvalues and eigenvectors can be calculated as before, and can then be used to solve this system of equations.

Comparison

After solving each system they have to be compared with the systems discussed in Problem 2.31. By doing this, it is possible to determine what types of eigenvalues correspond to stable nodes.

Key concepts

Coefficient Matrix

The cornerstone of solving a linear system of differential equations lies within the coefficient matrix. It's an array of numbers that represents the coefficients in the linear equations. For instance, for a system like:

\begin{aligned}&x' = ax + by, \&y' = cx + dy,dend{aligned}
the coefficient matrix would be:
\[\begin{bmatrix}a & b \ c & d\end{bmatrix}\]
In the exercise, the first step across systems a to f involves identifying this vital matrix as it encapsulates all the information needed to find both eigenvalues and eigenvectors. It's the springboard from which we dive into solving these systems. When attempting to improve understanding, consider visualizing the matrix as a linear transformation, stretching and rotating vectors within the system based on their coefficients.

Eigenvectors

The concept of eigenvectors might seem intimidating at first, but they are best thought of as the 'special' vectors of a linear transformation because they don't change direction. Instead, they are merely scaled by a certain amount known as an eigenvalue. For a matrix \(A\), an eigenvector \(v\) associated with an eigenvalue \(lambda\) satisfies:
\[A v = lambda v\]
In simpler terms, when you apply the transformation matrix to an eigenvector, you get the same vector multiplied by a scalar—hence preserving the direction. To find eigenvectors in the exercise, we start with the coefficient matrix, then solve the equation \(\textbf{A} v = lambda v \) for \(v\). It’s a process that reveals the fundamental nature of our system's behavior and can be approached methodically by seeking values that satisfy this equation after performing steps like determinant calculation and matrix subtraction.

System of Differential Equations

A system of differential equations is just a set of equations that relate some function to its derivatives. When dealing with linear systems like those in the exercise, we are focusing on first-order linear differential equations. These can often be written compactly using matrices and vectors. The general solution to such systems is heavily reliant on the eigenvalues and eigenvectors of the coefficient matrix and takes on the form:
\[\mathbf{x}(t) = c_1 e^{lambda_1 t} \mathbf{v}_1 + c_2 e^{lambda_2 t} \mathbf{v}_2\]
where \(c_1\) and \(c_2\) are constants determined by initial conditions, \(e\) represents the exponential function, and \(\mathbf{v}_1, \mathbf{v}_2\) are eigenvectors corresponding to eigenvalues \(lambda_1, lambda_2\). Understanding the link between the abstract concepts of eigenvalues/vectors and their impact on the qualitative behavior of the system is fundamental for mastering these problems.

Stability of Nodes

The stability of nodes within a system of differential equations is a concept that speaks to the long-term behavior of solutions as time progresses. It crucially depends on the eigenvalues of the coefficient matrix. In broad strokes, if all eigenvalues have negative real parts, the system is considered a stable node, meaning solutions will tend toward the equilibrium point as time advances. If any eigenvalue has a positive real part, the node is unstable, implying that solutions move away from equilibrium over time.

In the exercise, after finding eigenvalues, you'd compare them to known conditions for stability. When the eigenvalue problem is solved, it allows us to categorize the node and predict the future state of the system with greater accuracy. To enhance student comprehension, highlight the relationship between eigenvalues and the type of stability. For instance, real and negative eigenvalues suggest gradual settling down to an equilibrium state, whereas complex eigenvalues with a negative real part would indicate an oscillatory approach to equilibrium.