Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} 0 & 2 & -2 \\ -1 & 5 & -3 \\ 1 & 1 & 1 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The general solution for the system of first-order linear differential equations is given by \(\mathbf{y}(t) = c_1e^{-t}\begin{bmatrix} 6 \\ 3 \\ 5 \end{bmatrix} + c_2e^{2t}\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} + c_3e^{4t}\begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix}\), where \(c_1, c_2,\) and \(c_3\) are constants determined by the initial conditions.

Step-by-step solution

Write down the given matrix

We are given the following matrix: $$ A = \left[\begin{array}{rrr} 0 & 2 & -2 \\ -1 & 5 & -3 \\ 1 & 1 & 1 \\ \end{array}\right] $$

Find the eigenvalues of the matrix A

To find the eigenvalues, we will need to solve the characteristic equation, which is given by \(|A - \lambda I|=0\). This gives us: $$ \left| \begin{array}{ccc} 0-\lambda & 2 & -2 \\ -1 & 5-\lambda & -3 \\ 1 & 1 & 1-\lambda \end{array} \right|=0 $$ Calculate the determinant and solve for \(\lambda\): $$ (-\lambda)((5-\lambda)(1-\lambda)-(-3)(1)) - 2((-1)(1-\lambda)-(-3)(1)) - 2((-1)(1)+(-1)(1)) = 0 $$ Simplifying and solving for \(\lambda\), we get the eigenvalues: $$ \lambda = -1,~2,~4 $$

Find the corresponding eigenvectors for each eigenvalue

Now, for each eigenvalue, we will find the corresponding eigenvector by solving the homogeneous system \((A-\lambda I)\mathbf{v} =\mathbf{0}\): 1. For \(\lambda_1 = -1\), we have the following system: $$ \left[\begin{array}{ccc} 1 & 2 & -2 \\ -1 & 6 & -3 \\ 1 & 1 & 2 \end{array}\right] \left[\begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] $$ Solving this system of equations, we get the eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 6 \\ 3 \\ 5 \end{bmatrix}\). 2. For \(\lambda_2 = 2\), we have the following system: $$ \left[\begin{array}{ccc} -2 & 2 & -2 \\ -1 & 3 & -3 \\ 1 & 1 & -1 \end{array}\right] \left[\begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] $$ Solving this system of equations, we get the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\). 3. For \(\lambda_3 = 4\), we have the following system: $$ \left[\begin{array}{ccc} -4 & 2 & -2 \\ -1 & 1 & -3 \\ 1 & 1 & -3 \end{array}\right] \left[\begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] $$ Solving this system of equations, we get the eigenvector \(\mathbf{v}_3 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}\).

Write the general solution

Using the eigenvalues and eigenvectors, we can write the general solution for the system of first-order linear differential equations in the following form: $$ \mathbf{y}(t) = c_1e^{-t}\begin{bmatrix} 6 \\ 3 \\ 5 \end{bmatrix} + c_2e^{2t}\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} + c_3e^{4t}\begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix} $$ Where \(c_1, c_2,\) and \(c_3\) are constants determined by the initial conditions.

Key concepts

Eigenvalues and Eigenvectors

Understanding the concept of eigenvalues and eigenvectors is crucial in various fields like quantum mechanics, vibration analysis, and in solving differential equations. An eigenvalue is a scalar that determines the factor by which an eigenvector is scaled during a linear transformation. Mathematically, if you have a matrix A and a vector v, and if Av is parallel to v, then v is an eigenvector of A, and the scale factor is the corresponding eigenvalue λ.
To put it simply, an eigenvector doesn't change its direction under the associated linear transformation represented by A. In our exercise example, once we find the eigenvalues, we use them to compute the corresponding eigenvectors, which are crucial in expressing the general solution to the system of differential equations.

Characteristic Equation of a Matrix

The characteristic equation is a fundamental aspect of linear algebra that connects eigenvalues to a matrix. It is obtained by subtracting λI, where I is the identity matrix and λ is a scalar, from our matrix A, and setting the determinant of the resulting matrix to zero, symbolically shown as |A - λI| = 0. The roots of the resulting polynomial give us the eigenvalues of the matrix A.

In simpler terms, it's a special equation derived from a matrix that allows us to find those values of λ for which we can find non-trivial solutions (eigenvectors) that satisfy Ax = λx. This is an essential step in the process as seen in the exercise, where we compute the characteristic equation to find the necessary eigenvalues.

Homogeneous Systems

A homogeneous system of linear equations is a set where all the constant terms are zero. In our matrix form, it is typically represented as Ax = 0. Such a system always has at least one solution, the trivial solution where all variables equal zero.

When dealing with eigenvalues and eigenvectors, we often set up a homogeneous system because we're looking for non-zero solutions that satisfy the equation (A - λI)v = 0. Here, the matrix (A - λI) is derived from our given matrix A by subtracting the eigenvalue times the identity matrix. The non-trivial solutions to this system are the eigenvectors. This is demonstrated in the step-by-step solution where we find the eigenvectors corresponding to each eigenvalue by solving a homogeneous system.

First-Order Linear Differential Equations

First-order linear differential equations are fundamental in modeling rates of change. These equations can often be written as dy/dt = ay + b, where dy/dt is the rate of change, y is the variable of interest, and a and b are constants with b being zero for a homogeneous equation.

When we extend this idea to systems of equations, they can be expressed in matrix form as y' = Ay, which is exactly what we have in our exercise. By finding the eigenvalues and eigenvectors of the matrix A, we can solve these systems and express the general solution as a combination of exponential functions multiplied by the eigenvectors, where the exponents involve the eigenvalues, as seen in the general solution of our differential system.