Question

Solve the given system of equations by the method of Problem 19 of Section \(7.5 .\) Assume that \(t>0 .\) $$ t \mathbf{x}^{\prime}=\left(\begin{array}{cc}{3} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x} $$

Short answer

Answer: The general solution to the given system of linear differential equations is $\mathbf{x}(t) = c_1 t e^t \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, where $c_1$ is an arbitrary constant.

Step-by-step solution

Find the eigenvalues and eigenvectors of the matrix

To find the eigenvalues, we first need to compute the characteristic equation, which is given by the determinant of \((A-\lambda I)\), where \(A\) is the matrix, \(\lambda\) is the eigenvalue, and \(I\) is the identity matrix. $$\mathrm{det}(A-\lambda I) = \left|\begin{array}{cc}{3 - \lambda} & {-4} \\ {1} & {-1 - \lambda} \end{array}\right| = (3 - \lambda)(-1 - \lambda) + 4 = \lambda^2 - 2\lambda +1$$ Next, solve the characteristic equation for \(\lambda\): $$\lambda^2 - 2\lambda + 1 = 0$$ This factors to \((\lambda - 1)^2 = 0\), which has a single eigenvalue \(\lambda = 1\). Now, to find the eigenvectors, we need to solve \((A - \lambda I) v = 0\): $$\left(\begin{array}{cc}{2} & {-4} \\ {1} & {-2} \end{array}\right) \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ Notice that the second equation in the linear system is redundant, since it is a multiplication of the first one. We can solve the system and find the eigenvector \(v = \begin{pmatrix} 2 \\ 1\end{pmatrix}\).

Apply the eigenvector method

Following Problem 19 of Section 7.5, we can write the general solution to the system of differential equations as: $$\mathbf{x} = c_1 t e^{(\lambda t)} \mathbf{v}$$ where \(c_1\) is a constant, \(\lambda\) is the eigenvalue, and \(\mathbf{v}\) is the eigenvector. Substitute the eigenvalue and eigenvector we found into the formula: $$\mathbf{x} = c_1 t e^t \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$ Thus, the general solution to the system is given by $$\mathbf{x}(t) = c_1 t e^t \begin{pmatrix} 2 \\ 1 \end{pmatrix},$$ where \(c_1\) is an arbitrary constant.

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra and are essential for solving many types of differential equations. They provide insight into the properties of matrices and systems associated with them.

- **Eigenvalue (\(\lambda\)):** It is a scalar value that indicates how a corresponding eigenvector is scaled during a linear transformation represented by a matrix.- **Eigenvector:** It is a non-zero vector that only changes by the scalar factor (\(\lambda\)) during the associated linear transformation.
To determine these, we begin by solving the characteristic equation derived from the matrix. In our provided example, the matrix \(A\) was:\[A = \begin{pmatrix} 3 & -4 \ 1 & -1 \end{pmatrix}\]The next step involved calculating \(\lambda\) (eigenvalue), by setting up and solving the determinant \(\text{det}(A-\lambda I) = 0\). This yielded an eigenvalue of \(\lambda = 1\). Solving \((A - \lambda I)v = 0\) then gave us the associated eigenvector \(v = \begin{pmatrix} 2 \ 1 \end{pmatrix}\), which describes the direction of the transformation.

Characteristic Equation

The characteristic equation is formed as part of finding the eigenvalues of a matrix. It starts with \(A - \lambda I\), where \(\lambda\) is a scalar, and \(I\) is the identity matrix.
The equation involves determining where the determinant of this new matrix equals zero. Mathematically, it's stated as:\[\text{det}(A - \lambda I) = 0\]The solution to this equation gives us the eigenvalues of the matrix.
In the example solution provided, we had \(A = \begin{pmatrix} 3 & -4 \ 1 & -1 \end{pmatrix}\). The characteristic equation \(\lambda^2 - 2\lambda + 1 = 0\) was derived and factored to yield \((\lambda - 1)^2 = 0\), providing us the single eigenvalue \(\lambda = 1\). By solving the characteristic equation, we discover key properties about matrix \(A\). This critical step leads us toward calculating the eigenvectors needed to resolve the system.

System of Differential Equations

A system of differential equations involves multiple equations that must be solved simultaneously. These systems can describe a variety of dynamic phenomena, such as motion, growth, or changes in state across time.
In simpler terms, it's about finding unknown functions that satisfy all equations within the system. In this case, we look at the system described by \[t \mathbf{x}^{\prime}= \begin{pmatrix} 3 & -4 \ 1 & -1 \end{pmatrix} \mathbf{x}\]where \(\mathbf{x}\) is the vector function we are trying to solve for.Usually, the system may not be directly integrable, so we leverage eigenvalues and eigenvectors to understand transformation behaviors of matrix \(A\) and solve the system.
Such systems often appear in scientific and engineering contexts, modeling real-world interactions, and behaviors over time, like heat conduction, population dynamics, or economic systems. Breaking them down into smaller, more manageable parts using linear algebra techniques enables us to study these complex systems.

General Solution

The general solution of a system of differential equations consolidates all potential solutions into a singular expression. It considers both the homogeneous and particular solutions of the system.
In the context of our provided solution, we followed a standard approach leveraging eigenvectors:1. **Calculate the eigenvalue and eigenvector:** Which involved identifying \(\lambda = 1\) and \(\mathbf{v} = \begin{pmatrix} 2 \ 1 \end{pmatrix}\).2. **Formulate the general solution:** Using the eigenvector method to align the solution with the system's structure. The formula applied:\[\mathbf{x} = c_1 t e^{(\lambda t)} \mathbf{v}\] This results in a solution of \[\mathbf{x}(t) = c_1 t e^t \begin{pmatrix} 2 \ 1 \end{pmatrix}\], where \(c_1\) is a constant adjusted based on initial conditions.By crafting a general solution, we provide a complete and adaptable representation, capable of adjusting to specific external conditions. This helps in determining particular solutions when more information (like initial conditions) is added.