Problem 7 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). $$\left[\begin{array}{rrr} 3 & 0 & 4 \\ 0 & 2 & 0 \\ -4 & 0 & -5 \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x'}=A\mathbf{x}\) for the given matrix \(A\) is: \[\mathbf{x}(t)=c_1e^{t}\left(\begin{array}{c} 2 \\ 0 \\ -1\end{array}\right)+c_2e^{3t}\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)+c_3e^{4t}\left(\begin{array}{c} 4 \\ 0 \\ 1\end{array}\right)\]

Step 1: Determine the eigenvalues

First, we need to find the determinant of the matrix \(A - \lambda I\), where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix: \[\det\left(\begin{array}{ccc} 3-\lambda & 0 & 4 \\ 0 & 2-\lambda & 0 \\ -4 & 0 & -5-\lambda \end{array}\right)= (3-\lambda)((2-\lambda)(-5-\lambda)) \] Now, solve for \(\lambda\): \[\lambda^3-10\lambda^2+19\lambda-12=(\lambda-1)(\lambda-3)(\lambda-4)\] The eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = 3\), and \(\lambda_3 = 4\).

Step 2: Determine the eigenvectors

For each eigenvalue, we will find the corresponding eigenvectors by plugging \(\lambda\) back into the matrix equation \((A-\lambda I)\mathbf{v}=\mathbf{0}\), where \(\mathbf{v}\) is the eigenvector. For \(\lambda_1=1\): \[\left(\begin{array}{ccc} 2 & 0 & 4 \\ 0 & 1 & 0 \\ -4 & 0 & -6 \end{array}\right)\left(\begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right)=\mathbf{0}\] Solving this system, we get the eigenvector \(\mathbf{v_1} = \left(\begin{array}{c} 2 \\ 0 \\ -1\end{array}\right)\) For \(\lambda_2 = 3\): \[\left(\begin{array}{ccc} 0 & 0 & 4 \\ 0 & -1 & 0 \\ -4 & 0 & -8 \end{array}\right)\left(\begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right)=\mathbf{0}\] Solving this system, we get the eigenvector \(\mathbf{v_2} = \left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)\) For \(\lambda_3 = 4\): \[\left(\begin{array}{ccc} -1 & 0 & 4 \\ 0 & -2 & 0 \\ -4 & 0 & -9 \end{array}\right)\left(\begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right)=\mathbf{0}\] Solving this system, we get the eigenvector \(\mathbf{v_3} = \left(\begin{array}{c} 4 \\ 0 \\ 1\end{array}\right)\)

Step 3: Write the general solution

Now, we can write the general solution to the linear system \(\mathbf{x'}=A\mathbf{x}\): \[\mathbf{x}(t) = c_1e^{\lambda_1t}\mathbf{v_1}+c_2e^{\lambda_2t}\mathbf{v_2}+c_3e^{\lambda_3t}\mathbf{v_3}\] Substitute the eigenvalues and eigenvectors: \[\mathbf{x}(t)=c_1e^{t}\left(\begin{array}{c} 2 \\ 0 \\ -1\end{array}\right)+c_2e^{3t}\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)+c_3e^{4t}\left(\begin{array}{c} 4 \\ 0 \\ 1\end{array}\right)\] This is the general solution to the linear system \(\mathbf{x'}=A\mathbf{x}\).

Key Concepts

Eigenvalues and Eigenvectors

When dealing with linear systems of differential equations, eigenvalues and eigenvectors play a crucial role. Essentially, an eigenvector of a matrix is a non-zero vector that changes only by a scalar factor when that matrix is applied to it.
In simpler terms, if you have a matrix \(A\) and you multiply it by a vector \(\mathbf{v}\), the resulting vector is the same as scaling \(\mathbf{v}\) by a number \(\lambda\) (this number is the eigenvalue). Mathematically, this is expressed as \(A\mathbf{v} = \lambda\mathbf{v}\). In order to find the eigenvalues of a matrix, we solve the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix.
Once eigenvalues are obtained, eigenvectors can be found by solving \((A - \lambda I)\mathbf{v} = \mathbf{0}\), a system of linear equations. It's essential to note that each eigenvalue corresponds to at least one eigenvector, and these vectors are crucial in creating solutions for differential equations.

Matrix Algebra

Matrix algebra is fundamental to understanding linear systems of differential equations. It involves operations such as addition, subtraction, multiplication, and finding inverses. In our context, matrices provide a structured way to handle systems of equations.
Matrices are grids of numbers (or functions) that can represent systems of linear equations. When we analyze linear systems, each system can be expressed as \(\mathbf{x}' = A\mathbf{x}\), where \(\mathbf{x}\) is a vector and \(A\) is a matrix. In this form, solving the system often requires manipulating matrices to find eigenvalues and eigenvectors, which directly lead to the solutions of differential equations.

• Addition and Subtraction: Combine matrices of the same dimension by adding or subtracting corresponding elements.
• Multiplication: Possible only when dimensions align properly, involves taking the dot product of rows and columns.
• Determinants: A scalar value, crucial for finding eigenvalues, calculated using specific rules.

Understanding these operations opens the door to solving complex linear systems efficiently.

General Solution of Differential Equations

The ultimate goal of analyzing linear systems such as \(\mathbf{x}' = A\mathbf{x}\) is to determine the general solution of differential equations. The general solution encapsulates all possible solutions of the system and is crucial for predicting system behavior.
In a system described by a matrix \(A\), the solution generally involves the following steps:

• Identify the eigenvalues and corresponding eigenvectors of \(A\).
• Use these to construct solutions of the form \(e^{\lambda t}\mathbf{v}\), where \(\lambda\) is an eigenvalue and \(\mathbf{v}\) is the corresponding eigenvector.
• Combine these individual solutions into a general solution: \(\mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2 + \cdots\)

Constants \(c_1, c_2, \ldots\) are determined by initial conditions if available. This method reflects how dynamic models evolve over time, making them integral in fields such as physics, engineering, and even economics.