Question

Plot trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -4 & -3 \\ 3 & 2 \end{array}\right] \mathbf{y} $$

Short answer

Question: Determine the general solution for the following system of ordinary differential equations (ODEs) using the given matrix and plot several trajectories for different initial conditions: $$ A = \left[\begin{array}{rr} -4 & -3 \\\ 3 & 2 \end{array}\right] $$ Answer: The general solution for the given system of ODEs is: $$ \mathbf{y}(t) = c_1 e^{-t}\left[\begin{array}{c}1\\-1\end{array}\right] + c_2 e^{t}\left[\begin{array}{c}1\\1\end{array}\right] $$ To plot trajectories, select several initial conditions, find the corresponding values of \(c_1\) and \(c_2\), and plot the resulting solutions \(\mathbf{y}(t)\) in the \((y_1, y_2)\) plane. Alternatively, use software or an online tool like Desmos or Wolfram Alpha to plot the trajectories by inputting the general solution.

Step-by-step solution

Identify the system of ODEs

First, let's identify the system of ODEs by using the given matrix: $$ \begin{cases} y_1' = -4y_1 - 3y_2 \\ y_2' = 3y_1 + 2y_2 \end{cases} $$

Find the eigenvalues and eigenvectors of the matrix

Now, let's find the eigenvalues and eigenvectors of the matrix. The matrix is given by: $$ A = \left[\begin{array}{rr} -4 & -3 \\\ 3 & 2 \end{array}\right] $$ To find the eigenvalues, we solve the characteristic equation: $$ \text{det}(A-\lambda I) = \left|\begin{array}{cc} -4-\lambda & -3 \\ 3 & 2-\lambda \end{array}\right| = \lambda^2+2\lambda-1 = 0 $$ The eigenvalues are: $$ \lambda_1=-1, \quad \lambda_2=1 $$ Next, we find the eigenvectors corresponding to the eigenvalues. For \(\lambda_1=-1\), we solve \((A+\mathbf{I})\mathbf{v}_1 = \mathbf{0}\): $$ \left[\begin{array}{rr} -3 & -3 \\ 3 & 3 \end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right] $$ One eigenvector is \(\mathbf{v}_1 = \left[\begin{array}{c}1\\-1\end{array}\right]\) For \(\lambda_2=1\), we solve \((A-\mathbf{I})\mathbf{v}_2 = \mathbf{0}\): $$ \left[\begin{array}{rr} -5 & -3 \\ 3 & 1 \end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right] $$ One eigenvector is \(\mathbf{v}_2 = \left[\begin{array}{c}1\\1\end{array}\right]\)

Express the general solution in terms of the eigenvalues and eigenvectors

The general solution for this system can be written as: $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 = c_1 e^{-t}\left[\begin{array}{c}1\\-1\end{array}\right] + c_2 e^{t}\left[\begin{array}{c}1\\1\end{array}\right] $$ where \(c_1\) and \(c_2\) are constants.

Plot several trajectories using the general solution

To plot trajectories, select several initial conditions and find the corresponding values of \(c_1\) and \(c_2\). Place the initial point \((y_1(0),y_2(0))\) in the general solution and solve for the constants \(c_1\) and \(c_2\). Then, plot the resulting solutions \(\mathbf{y}(t)\) in the \((y_1, y_2)\) plane. Alternatively, you can use software or an online tool like Desmos or Wolfram Alpha to plot the trajectories of the system of ODEs by inputting the general solution derived in Step 3. This step will give you a visual representation of the behavior of the system for different initial conditions.

Key concepts

Eigenvalues and Eigenvectors

The concepts of eigenvalues and eigenvectors are central to understanding systems of differential equations. An eigenvalue is a special scalar that, when multiplied by an eigenvector, changes only the scale of the eigenvector and not its direction. To find them for a given square matrix, we solve the characteristic equation, which is created by subtracting a scalar multiple of the identity matrix from our original matrix and setting the determinant to zero.

Eigenvectors provide us with a powerful tool: they allow us to uncouple a system of differential equations. In the case of our exercise, the eigenvalues found were \( \lambda_1=-1 \) and \( \lambda_2=1 \). Each eigenvalue has a corresponding eigenvector, which represents the direction in which a transformation acts. Specifically, when we have the eigenvalues, we can then solve for the eigenvectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \), which tell us in which direction the system expands or contracts.

Characteristic Equation

The characteristic equation of a square matrix is a polynomial equation derived from the determinant of an altered version of the matrix in question. This alteration involves subtracting a scalar \( \lambda \) times the identity matrix \(I\) from the matrix. The characteristic equation for our matrix \(A\) was \( \lambda^2 + 2\lambda - 1 = 0 \).

By solving this quadratic equation, we get the system's eigenvalues. The characteristic equation can reveal a lot about the system's behavior without requiring us to compute the actual eigenvectors. For example, if all eigenvalues are negative, the system will generally converge to a stable state over time. This ties into the phase plane analysis as the eigenvalues help determine the nature of the trajectories on the phase plane.

Phase Plane Trajectories

Phase plane trajectories are a graphical representation of the paths that solutions to a system of differential equations take over time. They are plotted in a plane where each axis represents one of the variables in the system, allowing us to visualize the system's behavior. Depending on the eigenvalues and their associated eigenvectors, these trajectories can look very different: spiral inwards, spiral outwards, or follow straight lines, among other patterns.

In our provided exercise, the phase plane would consist of points \( (y_1, y_2) \) representing the state variables. By computing the general solution, we can plot these trajectories to see how the system evolves from different starting points. The patterns we see are directly linked to the nature of the eigenvalues; real and distinct eigenvalues suggest trajectories that move away or towards equilibrium points, describing the long-term behavior of the system.

General Solution of ODEs

Determining the general solution of a system of ordinary differential equations (ODEs) typically involves finding an expression that encompasses all possible solutions. The general solution reflects how the system behaves for any initial condition, usually by including arbitrary constants that are determined by these conditions.

For the system given in the exercise, the general solution was expressed as a linear combination of the eigenvectors scaled by the exponentiated eigenvalues. The constants \(c_1\) and \(c_2\) are tailored based on the initial conditions of the problem to provide the specific solution that matches them. The formula \( \mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \) encapsulates every possible solution trajectory for the system, demonstrating the power of leveraging eigenvalues and eigenvectors to solve systems of ODEs.