Problem 13 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). $$\left[\begin{array}{rrr} 2 & -2 & 1 \\ 1 & -4 & 1 \\ 2 & 2 & -3 \end{array}\right]$$ [Hint: The eigenvalues of \(A \text { are } \lambda=2,-2,-5 .]\)

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x'}=A\mathbf{x}\) with the given matrix \(A\) is: \[ \mathbf{x}(t) = C_1 e^{2t} \begin{bmatrix}1 \\ 1 \\ 4\end{bmatrix} + C_2 e^{-2t} \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix} + C_3 e^{-5t} \begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix} \] where \(C_1\), \(C_2\), and \(C_3\) are constants determined by initial conditions.

Step 1: Eigenvalues and eigenvectors

We are given the eigenvalues as \(\lambda_1=2, \lambda_2=-2,\) and \(\lambda_3=-5\). Now let's find the eigenvectors corresponding to each eigenvalue.

Step 1.1: Find the eigenvector for \(\lambda_1\)

For eigenvalue \(\lambda_1 = 2\), we need to solve the equation \((A - 2I)\mathbf{v}_1 = 0\): \[ \left[\begin{array}{rrr} 0 & -2 & 1 \\ 1 & -6 & 1 \\ 2 & 2 & -5 \end{array}\right] \begin{bmatrix} v_{11} \\ v_{12} \\ v_{13} \end{bmatrix} = 0 \] One possible eigenvector is \(\mathbf{v}_1 = \begin{bmatrix}1 \\ 1 \\ 4\end{bmatrix}\).

Step 1.2: Find the eigenvector for \(\lambda_2\)

For eigenvalue \(\lambda_2 = -2\), we need to solve the equation \((A + 2I)\mathbf{v}_2 = 0\): \[ \left[\begin{array}{rrr} 4 & -2 & 1 \\ 1 & -2 & 1 \\ 2 & 2 & -1 \end{array}\right] \begin{bmatrix} v_{21} \\ v_{22} \\ v_{23} \end{bmatrix} = 0 \] One possible eigenvector is \(\mathbf{v}_2 = \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}\).

Step 1.3: Find the eigenvector for \(\lambda_3\)

For eigenvalue \(\lambda_3 = -5\), we need to solve the equation \((A + 5I)\mathbf{v}_3 = 0\): \[ \left[\begin{array}{rrr} 7 & -2 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right] \begin{bmatrix} v_{31} \\ v_{32} \\ v_{33} \end{bmatrix} = 0 \] One possible eigenvector is \(\mathbf{v}_3 = \begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix}\).

Step 2: Solutions for each eigenvalue

Now, we build a solution for each eigenvalue: \(\mathbf{x}_1(t) = \mathbf{v}_1 e^{\lambda_1 t} = e^{2t} \begin{bmatrix}1 \\ 1 \\ 4\end{bmatrix}\) \(\mathbf{x}_2(t) = \mathbf{v}_2 e^{\lambda_2 t} = e^{-2t} \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}\) \(\mathbf{x}_3(t) = \mathbf{v}_3 e^{\lambda_3 t} = e^{-5t} \begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix}\)

Step 3: General solution for the linear system

Finally, we can combine the three solutions to create the general solution for the linear system: \[ \mathbf{x}(t) = C_1\mathbf{x}_1(t) + C_2\mathbf{x}_2(t) + C_3\mathbf{x}_3(t) \\ = C_1 e^{2t} \begin{bmatrix}1 \\ 1 \\ 4\end{bmatrix} + C_2 e^{-2t} \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix} + C_3 e^{-5t} \begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix} \] Here, \(C_1\), \(C_2\), and \(C_3\) are constants determined by initial conditions.

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial when dealing with systems of linear differential equations. These concepts are the foundation for solving such systems, especially when they can be represented in matrix form.

An eigenvalue, denoted as \(\lambda\), is a scalar that satisfies the characteristic equation of a matrix \(A\), given by \(\det(A - \lambda I) = 0\), where \(I\) represents the identity matrix of the same size as \(A\). The corresponding eigenvector, \(\mathbf{v}\), is a non-zero vector that undergoes a scale transformation by the matrix. In other words, when the matrix \(A\) acts on \(\mathbf{v}\), the vector is scaled by \(\lambda\), formally expressed as \(A\mathbf{v} = \lambda\mathbf{v}\).

In the given exercise, we are provided with the eigenvalues of the matrix \(A\) and the task is to find the eigenvectors for each eigenvalue. These eigenvectors will later form the fundamental solutions to the system of differential equations, allowing us to express the general solution as a combination of these fundamental solutions.

Linear Differential Equations

Linear differential equations describe the relationship between a function and its derivatives. When these equations are assembled into a system, they can often be expressed in the form \(\mathbf{x}' = A\mathbf{x}\), where \(A\) is a matrix and \(\mathbf{x}\) is a vector of functions. Solving these equations usually involves finding functions that satisfy both the individual differential equations and the system as a whole.

The general solution of a linear differential equation system is a combination of solutions that correspond to the eigenvalues and eigenvectors of matrix \(A\). For each eigenvalue, we find a particular solution that involves the corresponding eigenvector multiplied by an exponential function with the eigenvalue as the exponent. In the context of the provided exercise, each distinct eigenvalue \(\lambda_i\) leads to an exponential solution \(\mathbf{x}_i(t)\) associated with its eigenvalue. The overall solution is then a linear combination of these particular solutions with arbitrary constants.

Matrix Operations

Matrix operations, such as addition, multiplication, and scaling, are essential tools in linear algebra, particularly when solving systems of equations. Key operations used in the context of eigenvalues and eigenvectors include subtracting a scalar multiple of an identity matrix from a matrix and solving homogeneous systems.

To find an eigenvector corresponding to a given eigenvalue, we perform the operation \(A - \lambda I\), where \(A\) is our original matrix and \(\lambda\) is the eigenvalue of interest. This process forms a new matrix whose columns are linearly dependent, resulting in a homogeneous system with non-trivial solutions—these solutions are the eigenvectors. As shown in the solution steps for the exercise, we resolve this system for each eigenvalue to find the associated eigenvectors.

Homogeneous Systems

A homogeneous system of linear equations is one in which all the constant terms are zero, typically written as \(A\mathbf{x} = 0\). The trivial solution to this system is always \(\mathbf{x} = 0\), but we are often interested in finding non-trivial solutions that require the matrix \(A\) to have dependent columns, implying that \(\det(A) = 0\).

In the context of eigenvalues, finding non-trivial solutions to the homogeneous system is equivalent to finding eigenvectors. These are vectors that will not be altered besides being scaled when multiplied by the matrix \(A\). During the steps to solve the exercise, we looked for non-zero vectors that satisfy the equation \( (A - \lambda I)\mathbf{v} = 0 \) for each eigenvalue \(\lambda\). Through this process, we obtained the eigenvectors that contribute to the formation of the general solution for the system of linear differential equations.