Question

In Exercises \(1-16\) find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 3 & 2 \\ -5 & 1 \end{array}\right] \mathbf{y} $$

Short answer

Question: Find the general solution for the first order linear homogeneous system of differential equations: \(\mathbf{y'} = A\mathbf{y}\), where \(A = \begin{bmatrix} 3 & 2 \\ -5 & 1 \end{bmatrix}\). Answer: The general solution is given by \(\mathbf{y}(t) = c_1e^{(2 + 3i)t} \left[\begin{array}{c} 1 \\ \frac{1}{2}(3i-1) \end{array}\right] + c_2e^{(2 - 3i)t} \left[\begin{array}{c} 1 \\ \frac{1}{2}(-1 - 3i) \end{array}\right]\), where \(c_1\) and \(c_2\) are arbitrary constants.

Step-by-step solution

Find the eigenvalues of the coefficient matrix A

To find the eigenvalues, we need to solve the following characteristic equation for the matrix \(A\): $$ \text{det}(A - \lambda I) = 0, $$ where \(\lambda\) denotes the eigenvalues and \(I\) is the identity matrix. Let's evaluate the determinant for the matrix \(A - \lambda I\): $$ \text{det}(A - \lambda I) = \left | \begin{array}{cc} 3 - \lambda & 2 \\ -5 & 1 - \lambda \end{array} \right | = (3 - \lambda)(1 - \lambda) - (-5) \cdot 2 = \lambda^2 - 4\lambda + 11. $$ Now, we need to find the roots of the quadratic equation \(\lambda^2 - 4\lambda + 11 = 0\). We will use the quadratic formula: $$ \lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4\cdot 11}}{2} = 2 \pm 3i. $$ So, the eigenvalues are \(\lambda_1 = 2 + 3i\) and \(\lambda_2 = 2 - 3i\).

Determine the eigenvectors for each eigenvalue

For each eigenvalue, we will need to solve the following linear system of equations: $$ (A - \lambda_i I)\mathbf{v}_i = \mathbf{0}. $$ For \(\lambda_1 = 2 + 3i\), we get: $$ \left[\begin{array}{cc} 1 - 3i & 2 \\ -5 & -1 - 3i \end{array}\right] \left[\begin{array}{c} x\\ y\end{array}\right] = \left[\begin{array}{c} 0\\ 0\end{array}\right]. $$ Taking the first equation: \((1 - 3i)x + 2y = 0\). Let's solve for \(y\): $$ y = \frac{1}{2}(3i-1)x. $$ Taking \(x = 1\), we get the eigenvector \(\mathbf{v}_1 = \left[\begin{array}{c} 1 \\ \frac{1}{2}(3i-1) \end{array}\right]\). For \(\lambda_2 = 2 - 3i\), we have: $$ \left[\begin{array}{cc} 1 + 3i & 2 \\ -5 & -1 + 3i \end{array}\right] \left[\begin{array}{c} x\\ y\end{array}\right] = \left[\begin{array}{c} 0\\ 0\end{array}\right]. $$ Taking the first equation: \((1 + 3i)x + 2y = 0\). Solving for \(y\): $$ y = \frac{1}{2}(-1 - 3i)x. $$ Taking \(x = 1\), we get the eigenvector \(\mathbf{v}_2 = \left[\begin{array}{c} 1 \\ \frac{1}{2}(-1 - 3i) \end{array}\right]\).

Write the general solution

The general solution for a first order linear homogeneous system of differential equations is given by: $$ \mathbf{y}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2, $$ where \(c_1\) and \(c_2\) are arbitrary constants. Plugging in our computed eigenvalues and eigenvectors, we get: $$ \mathbf{y}(t) = c_1e^{(2 + 3i)t} \left[\begin{array}{c} 1 \\ \frac{1}{2}(3i-1) \end{array}\right] + c_2e^{(2 - 3i)t} \left[\begin{array}{c} 1 \\ \frac{1}{2}(-1 - 3i) \end{array}\right]. $$ This is the general solution for the given system of differential equations.

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in the study of linear algebra and differential equations. They come into play when looking to simplify complex system behaviors, particularly when dealing with linear transformations represented by matrices.

In the context of differential equations, especially those representing systems of linear equations like the one provided in the exercise, eigenvalues help us understand the nature of the system's growth or decay. They tell us if the system diverges, converges, or oscillates over time.

An eigenvector is a non-zero vector that only gets scaled when a linear transformation is applied to it. The amount by which it's scaled is known as the eigenvalue. Mathematically, if we have a square matrix \textbf{A}, an eigenvector \textbf{v}, and an eigenvalue \textlambda, they satisfy the following condition:
\[\begin{equation}\textbf{Av} = \textlambda \textbf{v}\end{equation}\]
The process to find these values and vectors is crucial to solving systems as it reduces a potentially complex system into a more manageable form.

First Order Linear Homogeneous System

A first order linear homogeneous system refers to a set of linear differential equations where each equation involves a first derivative, and no equation contains terms without variables—also known as a nonhomogeneous term. In matrix form, these systems can be written as
\[\begin{equation}\textbf{y}' = \textbf{Ay}\end{equation}\]
where \(\textbf{y}\) is a vector of unknown functions, \(\textbf{A}\) is a matrix of constants, and \(\textbf{y}'\) is the derivative of \(\textbf{y}\).

In our exercise, the goal is to find a general solution to such a system. This entails finding functions that satisfy all equations in the system simultaneously. Eigenvalues and eigenvectors are utilized to construct this solution because they essentially 'decouple' the system into separate equations which are easier to solve.

Characteristic Equation

The characteristic equation is a pivotal step in finding the eigenvalues of a matrix, which subsequently lead to the solution of a system of differential equations. It's derived from the matrix equation involving the identity matrix \textbf{I} and eigenvalue \(\lambda\):
\[\begin{equation}\textbf{det(A} - \lambda\textbf{I)} = 0\end{equation}\]
This equation originates from the principle that for an eigenvector \(\textbf{v}\) to exist, the matrix \(\textbf{A} - \lambda\textbf{I}\) must be singular, meaning it has no inverse, which is true if and only if its determinant is zero. Solving this equation yields the matrix's eigenvalues, which tell us about the long-term behavior of the system associated with the matrix.

Quadratic Formula

When solving for the eigenvalues from the characteristic equation, we often encounter quadratic equations. Fortunately, the quadratic formula provides a straightforward method for finding the roots of any quadratic equation of the form
\[\begin{equation}ax^2 + bx + c = 0\end{equation}\]
where \(a\),\(b\), and \(c\) are constants, and \(a eq 0\). The roots or solutions for \(x\) can be found using:
\[\begin{equation}\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\end{equation}\]
In our case, the characteristic equation resolves into a quadratic equation used to find the eigenvalues of the matrix. By applying the quadratic formula, we are able to determine the complex eigenvalues which are crucial for forming the general solution to our differential equation system.