Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 3 & 4 \\ -1 & 7 \end{array}\right] \mathbf{y} $$

Short answer

Question: Find the general solution of the given first-order linear homogeneous system of ordinary differential equations with the matrix A: $$ A=\left[\begin{array}{rr} 3 & 4 \\\ -1 & 7 \end{array}\right] $$ Answer: The general solution of the given system is: $$ \mathbf{y}(t)=e^{5t}\left(c_1+\frac{t c_2}{2}\right)\left[\begin{array}{r} 2 \\\ 1 \end{array}\right] $$ where \(c_1\) and \(c_2\) are arbitrary constants.

Step-by-step solution

Calculate the eigenvalues and eigenvectors of the given matrix

First, we need to find the eigenvalues of the given matrix. We calculate the eigenvalues by finding the roots of the characteristic equation, which is given by: $$ \text{det}(A-\lambda I)=(3-\lambda)(7-\lambda)-(-1\cdot 4)=\lambda^2-10\lambda+25 $$ Now, we find the roots (eigenvalues) of the characteristic equation: $$ \lambda^2-10\lambda+25=(\lambda-5)^2\Rightarrow \lambda_1=\lambda_2=5 $$ Since we have two repeated eigenvalues, we need to find the eigenvectors corresponding to the eigenvalue 5. We get the eigenvector by solving the following equation: $$ (A-\lambda I)\mathbf{v}=0\Rightarrow\left[\begin{array}{rr} -2 & 4 \\\ -1 & 2 \end{array}\right] \mathbf{v}=0 $$ We solve the linear system and obtain one linearly independent eigenvector corresponding to \(\lambda=5\): $$ \mathbf{v}_1=\left[\begin{array}{r} 2 \\\ 1 \end{array}\right] $$

Construct the general solution

Since we have repeated eigenvalues, we need to find the generalized eigenvector corresponding to \(\lambda=5\). To find the generalized eigenvector, we solve the following equation: $$ (A-\lambda I)^2\mathbf{v}=0\Rightarrow\left[\begin{array}{rr} 0 & 0 \\\ 0 & 0 \end{array}\right] \mathbf{v}=0 $$ We can see that the solution space of this equation is the same as that of \((A-\lambda I)\mathbf{v}=0\) since it is the zero matrix. Therefore, we can now construct the general solution of the given system using the eigenvalue, eigenvector, and generalized eigenvector: $$ \mathbf{y}(t)=c_1 e^{5t} \mathbf{v}_1 + c_2 t e^{5t} \mathbf{v}_1 = e^{5t}\left(c_1+\frac{t c_2}{2}\right)\mathbf{v}_1 $$ Where \(c_1\) and \(c_2\) are arbitrary constants. So, the general solution of the system is: $$ \mathbf{y}(t)=e^{5t}\left(c_1+\frac{t c_2}{2}\right)\left[\begin{array}{r} 2 \\\ 1 \end{array}\right] $$

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is fundamental to solving systems of differential equations. In linear algebra, an eigenvalue is a scalar that reflects how a linear transformation scales a vector, which is known as the eigenvector. A vector is an eigenvector of a matrix if it satisfies the equation \(A\mathbf{v} = \lambda\mathbf{v}\), where \(A\) is a matrix, \(\mathbf{v}\) is the eigenvector, and \(\lambda\) is the eigenvalue. These concepts become particularly valuable when we're dealing with systems where we can transform a complex problem into a simpler one by diagonalizing the matrix.

Eigenvectors point in a direction that is unaffected by the transformation (other than by a scalar factor, the eigenvalue), and eigenvalues indicate the factor by which the eigenvectors are scaled. In differential equations, these concepts enable us to find solutions that grow or decay at rates indicated by the eigenvalues, which greatly simplifies the process of finding a general solution to the system.

Characteristic Equation

The characteristic equation plays a pivotal role in finding the eigenvalues of a matrix. It is derived from the matrix by subtracting \(\lambda I\) from the matrix, where \(I\) is the identity matrix and \(\lambda\) is a scalar. Setting the determinant of \(A - \lambda I\) to zero gives us the characteristic equation. Solving this polynomial equation for \(\lambda\) yields the eigenvalues of the matrix.

The characteristic polynomial is a fundamental bridge connecting linear algebra to differential equations as it encapsulates the behavior of a linear system. When you come across a system of differential equations, the characteristic polynomial can reveal the nature of its solutions - whether they oscillate, decay, or grow exponentially over time - based on the eigenvalues you derive from it.

Linear Algebra

Linear algebra is the branch of mathematics that deals with vectors, vector spaces, linear transformations, and systems of linear equations. In the context of differential equations, it provides a set of tools for analyzing and solving systems where the variables change continuously. Matrices and determinants, which are part of linear algebra, are used to compactly represent and work with linear systems.

For example, the matrix \( \left[\begin{array}{ll}1 & 2 \ 2 & 1\end{array}\right] \) represents a linear transformation in a two-dimensional space. The study of such matrices and solving the systems they represent are essential in applications ranging from engineering to economics. Understanding the principles of linear algebra is crucial for anyone looking to explore any field that involves quantitative analysis.

System of Differential Equations

A system of differential equations consists of multiple equations that define relationships involving rates of change. When dealing with a linear system, as we have in the exercise, the solutions often involve exponential functions of time. These systems can describe a wide variety of phenomena in physics, biology, economics, and engineering.

The general solution to such systems typically involves two parts: the homogeneous solution and the particular solution. In the given exercise, the focus is on the homogeneous solution, which is associated with the matrix's eigenvalues and eigenvectors. The complex behaviors of dynamic systems, such as oscillations and exponential growth or decay, can be understood through these solutions. Hence, mastering systems of differential equations is not only about solving equations but also about interpreting their solutions to understand the dynamics they represent.