Question

Find the general solution of each of the linear systems in Exercises 1-26. $$ \begin{aligned} &x^{\prime}=5 x+4 y \\ &y^{\prime}=-x+y \end{aligned} $$

Short answer

The general solution of the given linear system is \(x(t) = -2c_1 e^{3t}\) and \(y(t) = c_1 e^{3t}\), where \(c_1\) is an arbitrary constant.

Step-by-step solution

Write equations in matrix form

To begin with, we will write the given system of differential equations as a matrix equation. \(x^{\prime} = 5x + 4y\) and \(y^{\prime} = -x + y\). Now, let \(\vec{X} = \begin{pmatrix} x \\ y \end{pmatrix}\) & \(\vec{X'} = \begin{pmatrix} x^{\prime} \\ y^{\prime} \end{pmatrix}\) Let's write the corresponding matrix A: \(A = \begin{pmatrix} 5 & 4 \\ -1 & 1 \end{pmatrix}\) The system can now be written as: \(\vec{X'} = A\vec{X}\)

Find Eigenvalues

Now, we find the eigenvalues of matrix \(A\). We do that by finding the determinant of the matrix \(|A-\lambda I|\), where \(\lambda\) is an eigenvalue and I is the Identity matrix. \(|A - \lambda I| = \begin{vmatrix} 5-\lambda & 4 \\ -1 & 1-\lambda \end{vmatrix} = (5-\lambda)(1-\lambda) - 4(-1) = \lambda^2 - 6\lambda + 9\) Now find the roots of \(\lambda^2 - 6\lambda + 9 = 0\). The roots are \(\lambda_1 = 3\) and \(\lambda_2 = 3\) (which means there is only one distinct eigenvalue: a repeated eigenvalue).

Find Eigenvectors

Now we need to find the eigenvectors of the matrix \(A\). We do this by solving the following equation: \((A - \lambda I)\vec{v} = 0\) For eigenvalue \(\lambda_1 = 3\), let's compute \(A - 3I\): \((A - \lambda_1 I) = \begin{pmatrix} 5-3 & 4 \\ -1 & 1-3 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ -1 & -2 \end{pmatrix}\) Now solve \(\begin{pmatrix} 2 & 4 \\ -1 & -2 \end{pmatrix}\vec{v} = 0\) for \(\vec{v}\). We notice that the second row is simply the first row multiplied by a constant, so the system reduces to one equation: \(2v_1 + 4v_2 = 0\). Let \(v_2 = 1\), then \(v_1 = -2\). So, the eigenvector for eigenvalue \(\lambda_1 = 3\) is \(\vec{v_1} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}\).

Find General Solution

Since we only have one eigenvalue, \(\lambda_1 = 3\), and its corresponding eigenvector, \(\vec{v_1} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}\), the general solution is given by: \(\vec{X}(t) = c_1 \cdot \vec{v_1} \cdot e^{\lambda_1 t}\) Substituting the eigenvalue and eigenvector, we have: \(\vec{X}(t) = c_1 \cdot \begin{pmatrix} -2 \\ 1 \end{pmatrix} \cdot e^{3t}\) So the general solution is: \(\vec{X}(t) = \begin{pmatrix} -2 \\ 1 \end{pmatrix}c_1 e^{3t} = \begin{pmatrix} -2c_1 e^{3t} \\ c_1 e^{3t} \end{pmatrix}\) As a final step, split the general solution into their x and y parts: \(x(t) = -2c_1 e^{3t}\) and \(y(t) = c_1 e^{3t}\). And that is the general solution to the given linear system of equations.

Key concepts

Matrix Representation of Linear Systems

Linear systems can be elegantly represented using matrices, which allows for compact and powerful manipulations. The matrix representation is particularly useful for solving systems of linear differential equations.

Consider the system \[x^{\'prime}=5x+4y\] and \[y^{\'prime}=-x+y,\] which can be rewritten in matrix form if we let \[vec{X} = \begin{pmatrix}x \ y \end{pmatrix}\] and \[vec{X^{\'prime}} = \begin{pmatrix} x^{\'prime} \ y^{\'prime}\end{pmatrix}.\] The system now becomes \[vec{X^{\'prime}} = Acdot vec{X},\] where matrix \[A = \begin{pmatrix} 5 & 4 \ -1 & 1 \end{pmatrix}\] is called the coefficient matrix. This compact representation is fundamental for computing solutions through methods that involve eigenvalues, eigenvectors, and exponentiation of matrices.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are critical in solving systems of linear equations, especially when dealing with linear differential equations. An eigenvalue \(\lambda\) of a matrix is a scalar such that there is a non-zero vector \(\vec{v}\), known as an eigenvector, which satisfies the equation \(A\vec{v} = \lambda\vec{v}\).

For our example, we start by finding the determinant of \(A - \lambda I\) to compute the eigenvalues, where \(I\) is the identity matrix. After some algebra, we determine that the eigenvalues for our coefficient matrix \(A\) are \(\lambda_1 = \lambda_2 = 3\).

These eigenvalues suggest that the matrix has a specific linear transformation property and will play a pivotal role in forming the general solution to our system.

Characteristic Polynomial

The characteristic polynomial of a matrix is a specific polynomial whose roots are the eigenvalues of the matrix. It is obtained by taking the determinant of \(A - \lambda I\), where \(I\) denotes the identity matrix and \(\lambda\) represents a scalar eigenvalue.

In the given exercise, the characteristic polynomial is \(\lambda^2 - 6\lambda + 9\). By finding the roots of this polynomial, which in this case are \(\lambda_1 = 3\) and \(\lambda_2 = 3\), we have determined the eigenvalues of matrix \(A\). This step is fundamental in the analysis of the system, allowing us to progress towards finding the eigenvectors and eventually the general solution.

Homogeneous Linear Differential Equations

Homogeneous linear differential equations form a special class where the solution space is spanned by eigenvectors. They are called homogeneous because the system can be expressed as \(\vec{X'} = A\vec{X}\), with no extra constants or functions added—just a matrix multiplication involving the function \(\vec{X}\).

The general solution often takes the form of \(\vec{X}(t) = c_1 \cdot \vec{v}_1 \cdot e^{\lambda_1 t}\), where \(\vec{v}_1\) is an eigenvector associated with the eigenvalue \(\lambda_1\), and \(c_1\) is an arbitrary constant. In the given problem, because we have a repeated eigenvalue, we only have one eigenvector, which simplifies the solution to a single term exponentially scaled by that eigenvalue.