Problem 4 - Differential Equations and Linear Algebra

Question

Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\) $$\left[\begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the system $\mathbf{x}^{\prime}=A\mathbf{x}$ for the given matrix \(A=\left[\begin{array}{cc} 1 & -2 \\\ 5 & -5 \end{array}\right]\) is: \[ \mathbf{x}(t) = c_1 e^{t}\begin{pmatrix} 2\\ 1 \end{pmatrix} + c_2 e^{-5t}\begin{pmatrix} 1\\ 5 \end{pmatrix}\] where c₁ and c₂ are arbitrary constants.

Step 1: Find the eigenvalues of matrix A

First, we need to find the eigenvalues of the matrix A. To do this, we have to solve the following characteristic equation: \[ |A-\lambda I|=0\] where A is the given matrix and λ is the eigenvalue. Let's substitute the matrix A in this equation and find its determinant: \[ |\left[\begin{array}{cc} 1-\lambda & -2 \\\ 5 & -5-\lambda \end{array}\right]|=0\] \[ (1-\lambda)(-5-\lambda) - (-2)\cdot5 =0 \Longrightarrow \lambda^2+4\lambda-5=0\] Now, we need to find the roots of this quadratic equation, which will be our eigenvalues.

Step 2: Find the eigenvectors corresponding to each eigenvalue

To find the eigenvectors corresponding to the eigenvalues we have found, we need to solve the following equation for each eigenvalue: \[ (A-\lambda I) \cdot \vec{v} = \vec{0} \] Let's find the roots of the characteristic equation from step 1: \[ \lambda^2+4\lambda-5=0 \] This equation factors to: \[ (\lambda - 1)(\lambda + 5)=0\] So our eigenvalues are λ₁ = 1 and λ₂ = -5. Now, we need to find the eigenvectors for these eigenvalues. For λ₁ = 1: \[(A-\lambda_1 I) \cdot \vec{v} = \left[\begin{array}{cc} 0 & -2 \\\ 5 & -6 \end{array}\right] \cdot \vec{v} = \vec{0}\] The eigenvector for λ₁ = 1 is found to be \(v_1=\begin{pmatrix} 2\\ 1 \end{pmatrix}\). For λ₂ = -5: \[(A-\lambda_2 I) \cdot \vec{v} = \left[\begin{array}{cc} 6 & -2 \\\ 5 & 0 \end{array}\right] \cdot \vec{v} = \vec{0}\] The eigenvector for λ₂ = -5 is found to be \(v_2=\begin{pmatrix} 1\\ 5 \end{pmatrix}\).

Step 3: Determine the general solution

Now we have the eigenvalues and their corresponding eigenvectors. We can write the general solution for the given linear system of differential equations as: \[ \mathbf{x}(t) = c_1 e^{\lambda_1 t}\vec{v}_1 + c_2 e^{\lambda_2 t}\vec{v}_2\] Substitute the eigenvalues and eigenvectors we found to get: \[ \mathbf{x}(t) = c_1 e^{(1)t}\begin{pmatrix} 2\\ 1 \end{pmatrix} + c_2 e^{(-5)t}\begin{pmatrix} 1\\ 5 \end{pmatrix}\] where c₁ and c₂ are arbitrary constants.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in the study of linear algebra and are essential for solving systems of differential equations. They help simplify complex systems by transforming them into a form that is easier to analyze.

To find the eigenvalues of a matrix, you need to solve its characteristic equation. This equation is formed by subtracting \(\lambda I\) (where \(I\) is the identity matrix) from the matrix \(A\), and setting the determinant of the resulting matrix to zero. The roots of this equation are the eigenvalues.

• Eigenvalues (\

Linear Systems

Linear systems are collections of linear equations that can be used to mathematically model real-world phenomena. These systems can be written in matrix form, making it easier to manipulate and solve them using algebraic rules. Understanding linear systems and their solutions is key to many areas of science and engineering.

A linear system can be expressed in the form \(A\vec{x} = \vec{b}\), where \(A\) is a coefficient matrix, \(\vec{x}\) is the vector of unknowns, and \(\vec{b}\) is the vector of constants. In the context of differential equations, as in our original problem, the system is given as \(\vec{x}^{\prime} = A\vec{x}\). Here, the goal is to find a vector function \(\vec{x}(t)\) that satisfies this relationship.

The general solution to a linear system of differential equations often involves using the eigenvalues and eigenvectors of the coefficient matrix to express the solution in terms of exponential functions. This transformation can convert the problem into one where each part is easier to solve and analyze.

Matrix Solutions

Matrix solutions offer a structured and organized method for solving linear systems through the use of matrices. By employing matrices, complex systems of equations can be simplified and solved systematically using algebraic operations.

In our original exercise, the solution involved calculating the eigenvalues and eigenvectors of matrix \(A\). With these components at hand, the general solution to the system is expressed as a sum of terms, each involving an exponential and the corresponding eigenvector.

• This makes use of the property that matrices can represent linear transformations, simplifying operations and relationships between different vectors.
• Expressing equations in matrix form allows for identifying invariant directions (given by eigenvectors) in the system.'

In many applications, this structured solution approach helps provide insight into how systems behave over time and under different conditions, making matrices a powerful tool in both theoretical and applied problems.