Question

Find the general solution. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}-2 & 2 & -6 \\ 2 & 6 & 2 \\ -2 & -2 & 2\end{array}\right] \mathbf{y}\)

Short answer

Question: Find the general solution of the given system of linear differential equations: \(\mathbf{y}'=\boldsymbol{A}\mathbf{y}\), where \(\boldsymbol{A} = \begin{bmatrix}-2 & 2 & -6 \\ 2 & 6 & 2 \\ -2 & -2 & 2\end{bmatrix}\) Answer: The general solution of the given system of linear differential equations is: \[\mathbf{y}(t)=c_1e^{-2t}\begin{bmatrix}6 \\ 0 \\ 2\end{bmatrix}+c_2e^{2t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}+c_3e^{4t}\begin{bmatrix}-3 \\ 0 \\ 1\end{bmatrix}\] where \(c_1\), \(c_2\), and \(c_3\) are constants determined by initial conditions.

Step-by-step solution

Find the eigenvalues of the matrix

To find the eigenvalues, first we need to find the determinant of the following matrix: \[|\boldsymbol{A}-\lambda \boldsymbol{I}|=\begin{vmatrix} -2-\lambda & 2 & -6 \\ 2 & 6-\lambda & 2 \\ -2 & -2 & 2-\lambda \end{vmatrix}\] We will compute the determinant: \begin{align*} |\boldsymbol{A}-\lambda \boldsymbol{I}| &= (-2-\lambda)((6-\lambda)(2-\lambda)-(-2\cdot2))-2(2(2-\lambda)-(-2\cdot 2))+(-6)(-2(6-\lambda)-(-2\cdot 2)) \\ &= (-2-\lambda)(\lambda^2 - 8\lambda + 4) - 2(-4 + 2\lambda + 4) - 6(-12 + 2\lambda + 4) \\ &= (-2-\lambda)(\lambda^2 - 8\lambda + 4) - 24\lambda +12 + 72\lambda \\ \end{align*} Now we can find the eigenvalues by setting the determinant equal to zero: \begin{align*} & (-2-\lambda)(\lambda^2 - 8\lambda + 4) = 0 \\ \end{align*} This equation leads us to the following eigenvalues: \(\lambda_1 = -2, \lambda_2 = 2, \lambda_3 = 4\)

Find the eigenvectors

Now, for each eigenvalue, we will find the eigenvectors by solving the following system of linear equations: \((\boldsymbol{A} - \lambda \boldsymbol{I}) \boldsymbol{x}=0\) For \(\lambda_1 = -2\): \[\begin{bmatrix}0 & 2 & -6 \\ 2 & 8 & 4 \\ -2 & -2 & 4\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}=0\] This matrix can be simplified by dividing the first row by 2: \[\begin{bmatrix}0 & 1 & -3 \\ 2 & 8 & 4 \\ -2 & -2 & 4\end{bmatrix}\] The third row can be obtained by adding the first row to the second row: \[\begin{bmatrix}0 & 1 & -3 \\ 2 & 8 & 4 \\ -2 & -1 & 1\end{bmatrix}\] From these three equations: \\ If (2): \(2x_1 + 8 x_2 + 4x_3 = 0\) \\ If (3): \(-2x_1 - x_2 + x_3 = 0\) Solving these equations, the eigenvector corresponding to \(\lambda_1 = -2\) is \[\boldsymbol{x}_1= \begin{bmatrix}6 \\ 0 \\ 2\end{bmatrix}\] Similarly, eigenvectors corresponding to \(\lambda_2 = 2\) and \(\lambda_3 = 4\) can be found: \(\lambda_2 = 2\): \[ \boldsymbol{x}_2=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}\] \(\lambda_3 = 4\): \[ \boldsymbol{x}_3=\begin{bmatrix}-3 \\ 0 \\ 1\end{bmatrix}\]

Construct the general solution

Finally, we will combine the eigenvalues and eigenvectors to construct the general solution: \[\mathbf{y}(t)=c_1e^{-2t}\begin{bmatrix}6 \\ 0 \\ 2\end{bmatrix}+c_2e^{2t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}+c_3e^{4t}\begin{bmatrix}-3 \\ 0 \\ 1\end{bmatrix}\] where \(c_1\), \(c_2\), and \(c_3\) are constants determined by initial conditions.

Key concepts

Eigenvalues

Eigenvalues are crucial when working with linear algebra and differential equations. They provide valuable insight into the properties of a matrix in a system of linear equations.
For a given square matrix \( \mathbf{A} \), an eigenvalue is a scalar \( \lambda \) that satisfies the equation:
\( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{x} = 0 \),
where \( \mathbf{I} \) is the identity matrix and \( \mathbf{x} \) is the eigenvector. This equation indicates that the matrix \( \mathbf{A} \) transforms the vector \( \mathbf{x} \) by scaling it by \( \lambda \).
The process of finding eigenvalues involves computing the determinant of \( (\mathbf{A} - \lambda \mathbf{I}) \) and setting it equal to zero to solve for \( \lambda \). This, in turn, results in a characteristic polynomial, where solving it gives the possible eigenvalues for the matrix. In the exercise solution, the eigenvalues \( -2, 2, \) and \( 4 \) are determined precisely through this method, which are then used to explore the behavior of the system.

Eigenvectors

With each eigenvalue, there corresponds an eigenvector, which is a non-zero vector that, when multiplied by the matrix, only scales by the eigenvalue and does not change direction. Eigenvectors are key in understanding how matrices operate.
To find an eigenvector for a given eigenvalue \( \lambda \), we solve the linear system:
\( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{x} = 0 \).
This involves substituting each eigenvalue into the matrix and solving the resulting system of equations for vector \( \mathbf{x} \). In matrix terms, this results in a system of linear equations where the solution provides the direction vectors or eigenvectors.
In our case, the solution process provided the eigenvectors \[ \begin{bmatrix}6 \ 0 \ 2\end{bmatrix}, \begin{bmatrix}1 \ 1 \ 1\end{bmatrix}, \begin{bmatrix}-3 \ 0 \ 1\end{bmatrix} \] corresponding to the eigenvalues \( -2, 2, \) and \( 4 \) respectively. By solving these equations, you can understand how the system evolves over time.

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. The system in the context of our problem is represented in the form of a matrix equation where the goal is to determine the unknowns or vectors that satisfy the equation.
For example, by transforming the differential equation into a system using matrix operations, you can use linear algebra techniques to solve for eigenvalues and eigenvectors.
Each row in a matrix corresponds to a linear equation, and the solution to the system is found when all the equations are satisfied simultaneously. When working with eigenvalues and eigenvectors, you convert a complex differential equation problem into a solvable system of linear equations, making it easier to determine the system's behavior at different states.

General Solution

The general solution in differential equations provides a formula that describes all possible solutions of a differential equation. It incorporates constants that can be adjusted or determined based on initial conditions.
In the case of a system represented by \( \mathbf{y}' = \mathbf{Ay} \), where \( \mathbf{A} \) is a matrix, the solution is constructed from the eigenvalues and eigenvectors of \( \mathbf{A} \). The general solution formula is:
\[ \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{x}_1 + c_2 e^{\lambda_2 t} \mathbf{x}_2 + c_3 e^{\lambda_3 t} \mathbf{x}_3 \]
Here, \( \lambda_1, \lambda_2, \lambda_3 \) are eigenvalues, and \( \mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3 \) are corresponding eigenvectors.
Combining these, the general solution captures all trajectories of the system's state. In the provided exercise, it forms a pattern that includes exponential growth or decay (depending on the eigenvalues' nature), wrangled with the span of the eigenvectors to provide valuable insights into the movement and stability of the system.