Question

Solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -7 & 3 \\ -3 & -1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 0 \\ 2 \end{array}\right] $$

Short answer

Based on the above step by step solution, write a short answer to the question: The solution to the initial value problem for the given system of linear differential equations is: $$ \mathbf{y}(t) = -\frac{2}{3} e^{-6t} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + \frac{4}{3} e^{-2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Step-by-step solution

Find the eigenvalues and eigenvectors of the matrix

First, let's find the eigenvalues (\(\lambda\)) of the given matrix: $$ \begin{vmatrix} A - \lambda I \end{vmatrix} = \begin{vmatrix} -7 - \lambda & 3 \\ -3 & -1 - \lambda \end{vmatrix} = \lambda^2 + 8 \lambda + 10 = (\lambda + 6)(\lambda + 2) $$ This gives us two eigenvalues: \(\lambda_1 = -6\) and \(\lambda_2 = -2\). Now we find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = -6\): $$ (A - (-6)I)\mathbf{v}_1 = \begin{bmatrix} 1 & 3 \\ -3 & 5 \end{bmatrix}\mathbf{v}_1 = 0 $$ Row reducing the augmented matrix, we get: $$ \begin{bmatrix} 1 & 3 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ Thus, \(\mathbf{v}_1 = \begin{bmatrix} 3 \\ -1 \end{bmatrix}\) is the eigenvector corresponding to the eigenvalue \(\lambda_1 = -6\). For \(\lambda_2 = -2\): $$ (A - (-2)I)\mathbf{v}_2 = \begin{bmatrix} -5 & 3 \\ -3 & -3 \end{bmatrix}\mathbf{v}_2 = 0 $$ Row reducing the augmented matrix, we get: $$ \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ Thus, \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is the eigenvector corresponding to the eigenvalue \(\lambda_2 = -2\).

Find the general solution

The general solution of the system of linear differential equations can be written as: $$ \mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{v}_1 + C_2 e^{\lambda_2 t} \mathbf{v}_2 = C_1 e^{-6t} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + C_2 e^{-2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Apply the initial conditions

Now we use the initial conditions \(\mathbf{y}(0) = \begin{bmatrix} 0 \\ 2 \end{bmatrix}\) to find the constants \(C_1\) and \(C_2\): $$ \begin{bmatrix} 0 \\ 2 \end{bmatrix} = C_1 e^{0} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + C_2 e^{0} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = C_1 \begin{bmatrix} 3 \\ -1 \end{bmatrix} + C_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$ Solving the system of linear equations, we find that \(C_1 = -\frac{2}{3}\) and \(C_2 = \frac{4}{3}\).

Write the particular solution

Now that we have found the constants \(C_1\) and \(C_2\), we can write the particular solution of the initial value problem: $$ \mathbf{y}(t) = -\frac{2}{3} e^{-6t} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + \frac{4}{3} e^{-2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$ This is the solution to the initial value problem.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial when solving systems of linear differential equations. They reveal a lot about the nature and behavior of such systems. In a given square matrix A, eigenvalues are scalars (/lambda) that satisfy the characteristic equation det(A - (/lambda)I) = 0, where I is the identity matrix of the same dimension as A. Each (/lambda) has an associated eigenvector (/mathbf{v}), a nonzero vector that satisfies (A - (/lambda)I)(/mathbf{v}) = 0.

In our exercise, we calculate these by setting up and solving determinant equations after subtracting (/lambda) times the identity matrix from the given matrix. Row reduction helps us find a simplified system where an eigenvector can be easily identified. These eigenvectors, paired with their respective eigenvalues, provide a foundational base for constructing the general solution to differential equations.
In a teaching environment, it would be beneficial to demonstrate additional examples of finding eigenvalues and eigenvectors for a variety of matrices, with students solving these step by step to ensure strong conceptual understanding.

Systems of Linear Differential Equations

A system of linear differential equations consists of multiple equations that describe the rate of change of several interdependent variables. In matrix terms, this is often represented as (/mathbf{y}^{(/prime)}) = A(/mathbf{y}), where A is a matrix of coefficients, and (/mathbf{y}) is a vector of functions. A general solution to this system involves linear combinations of eigenvalues and corresponding eigenvectors found from the matrix A.

(/mathbf{y}(t)) = C_1 e^{(/lambda_1 t)} (/mathbf{v}_1) + C_2 e^{(/lambda_2 t)} (/mathbf{v}_2) represents the general solution where C_1 and C_2 are constants determined by the initial conditions of the problem. This form highlights how solutions to linear systems will have components that either grow or decay exponentially based on the real parts of the eigenvalues. Clear visualization through graphing software or sketches showing exponential growth or decay informed by eigenvalues can reinforce students’ understanding of the concepts.

Particular Solution of Differential Equations

A particular solution to a differential equation is a solution that satisfies the initial conditions of the problem. In contrast to a general solution, which includes constants representing a family of solutions, a particular solution speaks to one specific instance. To obtain it, we substitute the initial values into the general solution and solve for these constants.

(/mathbf{y}(0)) = (/mathbf{C}) provides the necessary information to determine the unique values of (/mathbf{C}). After solving for C_1 and C_2, we substitute them back into the general solution to get a formula for (/mathbf{y}(t)) that gives us the expected value of the variables for any given time t.
Students often struggle with the transition from the general to the particular solution, as this step involves solving a system of equations. Real-life contextualization of initial conditions, like populations at time zero or initial concentrations of a substance, can solidify the relevance and importance of finding particular solutions.