Problem 3 - Differential Equations and Linear Algebra

Question

Solve the given system of differential equations. $$x_{1}^{\prime}=4 x_{1}+2 x_{2}, \quad x_{2}^{\prime}=-x_{1}+x_{2}$$

Step-by-Step Solution

Answer

The general solution for the given system of differential equations is: \(\textbf{x}(t) = C_1e^{3t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} + C_2e^{2t}\begin{pmatrix} 1 \\ -1\end{pmatrix}\) Where \(C_1\) and \(C_2\) are arbitrary constants.

Step 1: Write the system in matrix form

Let's begin by writing the system of differential equations in matrix form: \(\begin{pmatrix} x_{1}' \\ x_{2}' \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}\)

Step 2: Find the eigenvectors and eigenvalues of the coefficient matrix

Now, we need to find the eigenvalues and eigenvectors of the coefficient matrix: \(\begin{pmatrix} 4 & 2 \\ -1 & 1 \end{pmatrix}\) To find the eigenvalues, we need to compute the determinant of the matrix minus the eigenvalue times the identity matrix and set it equal to zero: \(|A - \lambda I| = \begin{vmatrix} 4-\lambda & 2 \\ -1 & 1-\lambda \end{vmatrix} = (4 - \lambda)(1 - \lambda) - (-1)(2) = 0\) Expanding and solving the characteristic equation gives us: \(\lambda^2 - 5\lambda + 6 = (\lambda - 3)(\lambda - 2) = 0\) Therefore, we have two eigenvalues: \( \lambda_1 = 3\) and \( \lambda_2 = 2\). Next, we need to find the eigenvectors associated with each eigenvalue. Let's start with \( \lambda_1 = 3 \): \((A - 3I)\textbf{x} = \begin{pmatrix} 1 & 2 \\ -1 & -2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\) From the above equation, we find that the eigenvector for \( \lambda_1 = 3\) is: \(\textbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\) Now, let's find the eigenvector for \( \lambda_2 = 2\): \((A - 2I)\textbf{x} = \begin{pmatrix} 2 & 2 \\ -1 & -1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\) From the above equation, we find that the eigenvector for \( \lambda_2 = 2\) is: \(\textbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\)

Step 3: Write the general solution using eigenvectors and eigenvalues

Finally, we can write the general solution using our eigenvectors and eigenvalues: \(\textbf{x}(t) = C_1e^{3t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} + C_2e^{2t}\begin{pmatrix} 1 \\ -1\end{pmatrix}\) Where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Eigenvectors

In solving a system of differential equations, one crucial aspect is finding the eigenvectors of the matrix associated with the system. Eigenvectors are vectors that, when multiplied by a matrix, transform only by a scalar multiplier, known as an eigenvalue.
Finding the eigenvectors involves solving the equation \((A - \lambda I)\textbf{x} = \textbf{0}\), where \(A\) is the coefficient matrix, \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix.
The solution to this equation gives us a set of vectors associated with each eigenvalue. For example, once we find that the eigenvalues of a matrix are \(3\) and \(2\), we solve for the vectors related to these values. These vectors provide directions in which the system of equations remains consistent.

• Eigenvectors help in expressing the system's solutions in a simplified form.
• They provide a basis that can simplify complex differential systems.

By determining the eigenvectors, we can proceed to represent the general solution of the system, which is easier to analyze and interpret in practical applications.

Eigenvalues

Finding the eigenvalues of a matrix is a significant step when solving systems of differential equations. Eigenvalues are numbers that give us important information about how a system behaves. They can be found by solving the characteristic equation \(|A - \lambda I| = 0\), where \(A\) is the coefficient matrix, \(\lambda\) is the eigenvalue, and \(I\) is the identity matrix.
In our example, the characteristic equation was derived by calculating the determinant \(\begin{vmatrix} 4-\lambda & 2 \ -1 & 1-\lambda \end{vmatrix}\), leading to the quadratic equation \(\lambda^2 - 5\lambda + 6 = 0\).
Solving this, we found two eigenvalues: \(\lambda_1 = 3\) and \(\lambda_2 = 2\). These values tell us about the system's stability and long-term behavior.

• Eigenvalues determine whether solutions grow, decay, or remain constant over time.
• They are essential in identifying the nature of fixed points in the system.

Once known, eigenvalues are employed along with eigenvectors to construct the system's general solution, producing an efficient and comprehensible description of the differential equations in question.

Matrix Form

In the context of solving differential equations, converting the system to matrix form is an invaluable strategy. Matrix representation simplifies the process of handling multiple equations at once.
Given the equations \(x_1' = 4x_1 + 2x_2\) and \(x_2' = -x_1 + x_2\), we arrange these into matrix form as: \[\begin{pmatrix} x_{1}' \ x_{2}' \end{pmatrix} = \begin{pmatrix} 4 & 2 \ -1 & 1 \end{pmatrix}\begin{pmatrix} x_{1} \ x_{2} \end{pmatrix}\]
This approach uses matrices to encapsulate complex interactions between variables, allowing us to apply linear algebra techniques like finding eigenvalues and eigenvectors.

• Matrices help in applying structured mathematical procedures systematically.
• They enhance the clarity and manageability of multivariable systems.

By integrating the differential equations into matrix form, we can proceed to solve using matrix algebra. This is particularly beneficial in not only simplifying calculations but also providing a clear framework for dynamic analysis of solutions in engineering and physics applications.