Question

Find the general solution. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}3 & 2 & -2 \\ -2 & 7 & -2 \\\ -10 & 10 & -5\end{array}\right] \mathbf{y}\)

Short answer

The general solution to the given system of linear differential equations is given by: \(\mathbf{y}(t) = C_1 e^{3t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} + C_2 e^{5t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} + C_3 e^{-3t}\begin{bmatrix}1 \\ -3 \\ -1\end{bmatrix}\)

Step-by-step solution

Find the eigenvalues and eigenvectors of the coefficient matrix

First, we have to find the eigenvalues of the given matrix. To do this, we'll find the determinant of the matrix subtracted by the identity matrix multiplied by the eigenvalue (\(\lambda\)) and set it equal to zero (i.e., \(|A - \lambda I|=0\)), where \(A = \begin{bmatrix}3 & 2 & -2 \\ -2 & 7 & -2 \\ -10 & 10 & -5\end{bmatrix}\): \(|A - \lambda I| = \left|\begin{array}{ccc}3-\lambda & 2 & -2 \\ -2 & 7-\lambda & -2 \\ -10 & 10 & -5-\lambda\end{array}\right| = 0\) Now, calculate the determinant and solve for lambda: \((3-\lambda)((7-\lambda)(-5-\lambda)-(-2)(-2))-2(-2(-2)(-5-\lambda)-(-2)(10))+-2(-10(-2)-(-2)(10)) = 0\) After expanding and simplifying, we get: \(\lambda^3 - 5\lambda^2 - 63\lambda + 315 = 0\) The eigenvalues for this matrix are \(\lambda_1 = 3\), \(\lambda_2 = 5\), and \(\lambda_3 = -3\). Next, we have to find the eigenvectors for each eigenvalue. For \(\lambda_1 = 3\): \((A-\lambda_1 I)\mathbf{v_1} = \left[\begin{array}{ccc}0 & 2 & -2 \\ -2 & 4 & -2 \\ -10 & 10 & -8\end{array}\right]\mathbf{v_1} = \mathbf{0}\) From the first row, we have \(2v_2 - 2v_3 = 0\). Let \(v_2 = 1\), then \(v_3 = 1\). The eigenvector is \(\mathbf{v_1} = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}\). For \(\lambda_2 = 5\): \((A-\lambda_2 I)\mathbf{v_2} = \left[\begin{array}{ccc}-2 & 2 & -2 \\ -2 & 2 & -2 \\ -10 & 10 & -10\end{array}\right]\mathbf{v_2} = \mathbf{0}\) From the first row, we have \(-2v_1+2v_2-2v_3 = 0\). Let \(v_1 = 1\), then \(v_2=1\), and \(v_3 = 1\). The eigenvector is \(\mathbf{v_2} = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}\). For \(\lambda_3 = -3\): \((A-\lambda_3 I)\mathbf{v_3} = \left[\begin{array}{ccc}6 & 2 & -2 \\ -2 & 10 & -2 \\ -10 & 10 & -2\end{array}\right]\mathbf{v_3} = \mathbf{0}\) From the first row, we have \(6v_1 + 2v_2 - 2v_3 = 0\). Let \(v_1 = 1\), then \(v_2 = -3\), and \(v_3 = -1\). The eigenvector is \(\mathbf{v_3} = \begin{bmatrix}1 \\ -3 \\ -1\end{bmatrix}\).

Form the solution in terms of eigenvalues and eigenvectors

For each eigenvalue and eigenvector pair, form a solution using the form: \(\mathbf{y} = C_i e^{\lambda_i t}\mathbf{v_i}\) We will have: \(\mathbf{y_1} = C_1 e^{3t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}\) \(\mathbf{y_2} = C_2 e^{5t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}\) \(\mathbf{y_3} = C_3 e^{-3t}\begin{bmatrix}1 \\ -3 \\ -1\end{bmatrix}\)

Combine the solutions to form the general solution

Now, combine the solutions obtained in Step 2 to get the general solution of the given system of linear differential equations: \(\mathbf{y}(t) = \mathbf{y_1} + \mathbf{y_2} + \mathbf{y_3} = C_1 e^{3t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} + C_2 e^{5t}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} + C_3 e^{-3t}\begin{bmatrix}1 \\ -3 \\ -1\end{bmatrix}\) We have found the general solution to the given linear differential equation system.

Key concepts

Eigenvalues and Eigenvectors

When working with systems of linear differential equations, eigenvalues and eigenvectors are incredibly valuable tools. The eigenvalues of a matrix reveal much about the behavior of the system over time. To find them, apply the characteristic equation, which involves setting the determinant of the matrix \( A - \lambda I \) to zero. Here, \( A \) is the coefficient matrix and \( I \) is the identity matrix. In our exercise, this leads to the cubic equation \( \lambda^3 - 5\lambda^2 - 63\lambda + 315 = 0 \) resulting in the eigenvalues: \( 3, 5, \text{and} -3 \).

• Eigenvectors offer direction; they point the way the system behaves near eigenvalues.
• Find them by solving \( (A - \lambda I) \mathbf{v} = \mathbf{0} \).

For each eigenvalue, an eigenvector emerges that complements the understanding of the system's dynamics. This combination of eigenvalues and eigenvectors aids in building solutions that mirror the system's behaviors.

System of Linear Equations

A system of linear equations in the context of differential equations involves a matrix of coefficients multiplying a vector of functions. The task is to find the functions that satisfy all the given equations simultaneously.

• Each equation in the system reflects a particular dynamic aspect of the entire process being modeled.
• In our problem, the system is represented by \( \mathbf{y}^{\prime}=A\mathbf{y} \), where \( A \) is a \( 3 \times 3 \) matrix.

Solving such systems requires finding values for the vector \( \mathbf{y} \) such that, when multiplied by \( A \), they satisfy the differential equations for any given point in time. This approach translates physical dynamics into a mathematical framework, providing insight and predictions about the behavior of complex systems over time.

General Solution

The general solution of a system of linear differential equations combines individual solutions for each eigenvalue/eigenvector pair. This solution accounts for all possible behaviors of the system. It offers an overarching formula that describes the system dynamics for any possible state.

• It leverages exponential functions of time multiplied by constant vectors.
• These solutions are encapsulated by expressions of the form \( \mathbf{y} = C_i e^{\lambda_i t}\mathbf{v_i} \).

In the exercise solved, the general solution is the sum of all individual solutions derived from the eigenvalues \( 3, 5, \) and \(-3\). When added together, \( \mathbf{y}(t) = C_1 e^{3t}\begin{bmatrix}1 \ 1 \ 1\end{bmatrix} + C_2 e^{5t}\begin{bmatrix}1 \ 1 \ 1\end{bmatrix} + C_3 e^{-3t}\begin{bmatrix}1 \ -3 \ -1\end{bmatrix} \), it shows all possible trajectories the original system might follow under different conditions.

Matrix Algebra

Matrix algebra is a critical mathematical tool for manipulating and solving systems of equations, especially those involved in linear systems of differential equations. It simplifies the way we handle large sets of linear equations by encapsulating them within matrix operations.

• Matrices can represent entire systems, reducing complex calculations to more manageable forms.
• Operations such as matrix addition, multiplication, and finding determinants are fundamental to solving these equations.

In the provided exercise, matrix algebra helps derive the coefficients for each eigenvalue by handling operations efficiently. The expression \( A - \lambda I \), essential in finding eigenvalues and eigenvectors, is based on matrix subtraction and scalar multiplication. Thus, matrix algebra forms the backbone of turning abstract concepts into calculable and solvable problems, facilitating solutions to complex differential equation systems.