Question

Plot trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\ -1 & 2 \end{array}\right] \mathbf{y} $$

Short answer

Question: Plot the trajectories (solutions) of the given linear system of differential equations: $$ \begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} $$ Answer: The general solution of the given system of differential equations is: $$ \mathbf{y}(t) = \begin{bmatrix} c_1 \\ c_2 e^{2t} \end{bmatrix} $$ Plot the representative trajectories of solutions using graphing software and varying the values of c₁ and c₂. Some trajectories will be horizontal lines, while others will be exponential curves with a horizontal asymptote.

Step-by-step solution

Find the eigenvalues of the matrix.

To find the eigenvalues of the matrix, we solve the following equation: $$ \operatorname{det}\left(\begin{array}{rr} 0-\lambda & 1 \\ -1 & 2-\lambda \end{array}\right) = 0 $$ Solve for λ: $$ (-\lambda)(2-\lambda) - (-1)(1) = \lambda^2 - 2\lambda -1 + 1= \lambda^2 - 2\lambda = \lambda(\lambda - 2) = 0 $$ So, the eigenvalues are: $$ \lambda_1 = 0, \lambda_2 = 2 $$

Find the eigenvectors for each eigenvalue.

For each eigenvalue, we solve the following vector equation to find their associated eigenvectors: $$ A \mathbf{v} = \lambda \mathbf{v} $$ For λ₁ = 0: $$ \begin{bmatrix} 0 & 1\\ -1 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= 0 \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$ We get the following system of equations: $$ \begin{cases} x_2 = 0 \\ -x_1 + 2x_2 = 0 \end{cases} $$ We can choose x₁ = 1, and we'll have x₂ = 0. Thus, the eigenvector associated with λ₁ = 0 is: $$ \mathbf{v}_{1} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$ For λ₂ = 2: $$ \begin{bmatrix} 0 & 1\\ -1 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= 2 \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$ We get the following system of equations: $$ \begin{cases} -2x_1 + x_2 = 0 \\ -x_1 = 0 \end{cases} $$ We can choose x₁ = 0, and we'll have x₂ = 0. Thus, the eigenvector associated with λ₂ = 2 is: $$ \mathbf{v}_{2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$

Construct the general solution of the system.

Now that we have the eigenvalues and eigenvectors, we can construct the general solution of the system, which is given by: $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 = c_1 e^{0t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{2t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$ Which simplifies to: $$ \mathbf{y}(t) = \begin{bmatrix} c_1 \\ c_2 e^{2t} \end{bmatrix} $$

Plot representative trajectories of solutions.

With the general solution, we can plot representative trajectories for different values of c₁ and c₂. However, since this is a text-based format, we cannot directly draw the trajectories. To plot the trajectories, you can use graphing software like Desmos, GeoGebra, or MATLAB/Octave. Input the general solution components as parametric functions and vary the parameters c₁ and c₂ to see different trajectories. Keep in mind that some trajectories will be horizontal lines (when c₂ = 0) and others will be exponential curves with a horizontal asymptote (when c₁ ≠ 0).

Key concepts

Eigenvalues and Eigenvectors

One of the fundamental concepts in solving linear systems of differential equations is finding eigenvalues and eigenvectors of the system's matrix. The matrix in our exercise is \(\begin{bmatrix} 0 & 1 \ -1 & 2 \end{bmatrix}\). To find the eigenvalues, we solve the characteristic equation, which comes from setting the determinant of \(A - \lambda I\) to zero. This gives us the polynomial equation \(\lambda^2 - 2\lambda = 0\). Solving this equation results in eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 2\).

Each eigenvalue has an associated eigenvector that satisfies the equation \(A\mathbf{v} = \lambda\mathbf{v}\). For \(\lambda_1 = 0\), we solve and find that any vector of the form \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\) is an eigenvector. Similarly, for \(\lambda_2 = 2\), we find \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\) as the eigenvector. Understanding these concepts helps in constructing general solutions to the system of differential equations.

General Solutions

Once we have the eigenvalues and eigenvectors of a matrix corresponding to a linear system, the next step is constructing the general solution. For each eigenvalue and its associated eigenvector, the general solution incorporates terms \(c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\).

In our case, the general solution becomes:

\[ \mathbf{y}(t) = \begin{bmatrix} c_1 \ c_2 e^{2t} \end{bmatrix} \]

This expression summarizes all possible solutions of the system, dependent on constants \(c_1\) and \(c_2\). This formula provides a robust method to understand how various initial conditions will evolve over time.

Trajectory Plotting

Plotting the trajectories of a system of differential equations visually demonstrates the solutions and their behaviors over time. Each trajectory corresponds to different initial conditions determined by varying \(c_1\) and \(c_2\). For our system, plotting the solution \(\mathbf{y}(t) = \begin{bmatrix} c_1 \ c_2 e^{2t} \end{bmatrix}\) gives two distinctive behaviors:

• Horizontal lines when \(c_2 = 0\) since the \(y\)-coordinate remains constant.
• Upward curves when \(c_2 eq 0\) with exponential growth rate in the \(y\)-coordinate.

For more complex visualization, software tools like Desmos or MATLAB can be utilized. These tools help you see the effects of varying constants on the system's behavior.

Matrix Algebra

Matrix algebra forms the backbone for analyzing linear systems of differential equations. Finding eigenvalues and eigenvectors involves operations such as matrix addition, multiplication, and determinant computation. In our task, the matrix \(\begin{bmatrix} 0 & 1 \ -1 & 2 \end{bmatrix}\) functions as our system matrix, and manipulating it through various algebraic steps allows us to gain insights into the system's dynamics.

When solving \(A\mathbf{v} = \lambda\mathbf{v}\), it essentially translates into solving a linear algebraic system for \(\mathbf{v}\) given \(\lambda\). Mastery in matrix algebra techniques not only aids in finding solutions to differential equations but is also applicable across many fields in mathematics and engineering for solving similar linear problems.