Question

Consider the given initial value problem \(\mathbf{y}^{\prime}=A \mathbf{y}, \mathbf{y}\left(t_{0}\right)=\mathbf{y}_{0}\). (a) Find the eigenvalues and eigenvectors of the coefficient matrix \(A\). (b) Construct a fundamental set of solutions. (c) Solve the initial value problem.\(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}-3 & 0 & -36 \\ 0 & 1 & 0 \\ 1 & 0 & 9\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}0 \\ -1 \\ 2\end{array}\right]\)

Short answer

Question: Find the solution to the given initial value problem involving the system of first-order linear ordinary differential equations with matrix A and initial condition 𝐲(0) = [0, -1, 2]: $A = \left[\begin{array}{rrr} -3 & 0 & -36 \\ 0 & 1 & 0 \\ 1 & 0 & 9 \end{array}\right]$ Answer: The solution to the initial value problem is: \(\mathbf{y}(t) = -\frac{2}{3}e^{-3t}\left[\begin{array}{r}12 \\ 0 \\ 1\end{array}\right] - \frac{1}{3} e^{t}\left[\begin{array}{r}9 \\ 0 \\ 1\end{array}\right] + \frac{5}{3} e^{9t}\left[\begin{array}{r}3 \\ 0 \\ 1\end{array}\right]\)

Step-by-step solution

Find Eigenvalues

First, we need to find the eigenvalues of the coefficient matrix \(A\). To do this, we need to compute the determinant of (\(A - \lambda I\)), set it equal to zero, and solve for \(\lambda\). Here, \(I\) is the identity matrix of the same dimensions as \(A\). $A - \lambda I = \left[\begin{array}{rrr} -3 - \lambda & 0 & -36 \\ 0 & 1 - \lambda & 0 \\ 1 & 0 & 9 - \lambda \end{array}\right]$ Now we compute the determinant: $\det(A-\lambda I) = \left|\begin{array}{ccc} -3-\lambda & 0 & -36 \\ 0 & 1-\lambda & 0 \\ 1 & 0 & 9-\lambda \end{array}\right| = (-3-\lambda)((1-\lambda)(9-\lambda))$ Setting this determinant equal to zero and solving for \(\lambda\), we get: \(\lambda_1 = -3, \lambda_2 = 1, \lambda_3 = 9\)

Find Eigenvectors

Now, we will find the eigenvectors corresponding to each eigenvalue by solving the system \((A-\lambda_i I)\mathbf{x}_i=\mathbf{0}\) for each eigenvalue \(\lambda_i\). For \(\lambda_1 = -3\): $(A + 3I)\mathbf{x}_1 = \left[\begin{array}{rrr} 0 & 0 & -36\\ 0 & 4 & 0\\ 1 & 0 & 12 \end{array}\right] \mathbf{x}_1 = \mathbf{0}$ The eigenvector corresponding to \(\lambda_1\) is \(\mathbf{x}_1 = \left[\begin{array}{r}12 \\ 0 \\ 1\end{array}\right]\). For \(\lambda_2 = 1\): $(A - I)\mathbf{x}_2 = \left[\begin{array}{rrr} -4 & 0 & -36\\ 0 & 0 & 0 \\ 1 & 0 & 8 \\ \end{array}\right] \mathbf{x}_2 = \mathbf{0}$ The eigenvector corresponding to \(\lambda_2\) is \(\mathbf{x}_2 = \left[\begin{array}{r}9 \\ 0 \\ 1\end{array}\right]\). For \(\lambda_3 = 9\): $(A - 9I)\mathbf{x}_3 = \left[\begin{array}{rrr} -12 & 0 & -36 \\ 0 & -8 & 0 \\ 1 & 0 & 0 \end{array}\right] \mathbf{x}_3 = \mathbf{0}$ The eigenvector corresponding to \(\lambda_3\) is \(\mathbf{x}_3 = \left[\begin{array}{r}3 \\ 0 \\ 1\end{array}\right]\).

Construct a Fundamental Set of Solutions

Now that we have the eigenvalues and eigenvectors, we can construct a fundamental set of solutions. The Eigenvalue Method tells us that a fundamental set of solutions is given by \(\mathbf{y}_i(t) = e^{\lambda_i t}\mathbf{x}_i\), where \(i=1,2,3\). Therefore, the fundamental set of solutions is: \(\mathbf{y}_1(t) = e^{-3t}\left[\begin{array}{r}12 \\ 0 \\ 1\end{array}\right]\), \(\mathbf{y}_2(t) = e^{t}\left[\begin{array}{r}9 \\ 0 \\ 1\end{array}\right]\), \(\mathbf{y}_3(t) = e^{9t}\left[\begin{array}{r}3 \\ 0 \\ 1\end{array}\right]\)

Solve the Initial Value Problem

The general solution to the given initial value problem can be written as a linear combination of the fundamental set of solutions: \(\mathbf{y}(t)=c_1\mathbf{y}_1(t)+c_2\mathbf{y}_2(t)+c_3\mathbf{y}_3(t)\) We are given the initial condition \(\mathbf{y}(0) = \left[\begin{array}{r}0 \\ -1 \\ 2\end{array}\right]\). Applying the initial condition to the general solution: \(\mathbf{y}(0) = c_1\mathbf{y}_1(0) + c_2\mathbf{y}_2(0) + c_3\mathbf{y}_3(0) = \left[\begin{array}{r}0 \\ -1 \\ 2\end{array}\right]\) This gives us a system of equations that we can solve for \(c_1\), \(c_2\), and \(c_3\): $ \left[\begin{array}{ccc} 12 & 9 & 3\\ 0 & 0 & 0\\ 1 & 1 & 1 \end{array}\right] \left[\begin{array}{r} c_1\\ c_2\\ c_3 \end{array}\right] = \left[\begin{array}{r} 0\\ -1\\ 2 \end{array}\right] $ Solving this system of equations, we find that \(c_1=-\frac{2}{3}\), \(c_2 = -\frac{1}{3}\), and \(c_3 = \frac{5}{3}\). Hence, the solution to the initial value problem is: \(\mathbf{y}(t) = -\frac{2}{3}e^{-3t}\left[\begin{array}{r}12 \\ 0 \\ 1\end{array}\right] - \frac{1}{3} e^{t}\left[\begin{array}{r}9 \\ 0 \\ 1\end{array}\right] + \frac{5}{3} e^{9t}\left[\begin{array}{r}3 \\ 0 \\ 1\end{array}\right]\)

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are key concepts in linear algebra and play a significant role in solving matrix differential equations. They provide insights into the system's behavior and stability. Let's understand these concepts better.

• Eigenvalues (\( \lambda \)). These are scalar values that help identify the scaling factor when a matrix is multiplied by an eigenvector. They are found by solving the characteristic equation, which involves setting the determinant of the matrix \( (A - \lambda I) \) to zero. Here, \( I \) is the identity matrix. The solutions to this equation are the eigenvalues.


• Eigenvectors (\( \mathbf{x} \)). Once you have the eigenvalues, eigenvectors are non-zero vectors \( \mathbf{x} \) that satisfy the equation \( (A - \lambda I) \mathbf{x} = \mathbf{0} \). For each eigenvalue, there is a corresponding eigenvector that defines the direction in which the transformation operates.

In the problem, the matrix's eigenvalues were \( \lambda_1 = -3, \lambda_2 = 1, \lambda_3 = 9 \), with corresponding eigenvectors determined through system equations.

Fundamental Set of Solutions

A fundamental set of solutions is an essential component when solving systems of differential equations, especially those described by matrices. When you have the eigenvalues and corresponding eigenvectors, constructing a fundamental set of solutions becomes straightforward.

The general approach involves associating each eigenvalue \( \lambda_i \) with its respective eigenvector \( \mathbf{x}_i \) and exponential solution function \( e^{\lambda_i t} \). Thus, each solution takes the form:

• Expression: \( \mathbf{y}_i(t) = e^{\lambda_i t} \mathbf{x}_i \)
• Explanation: This method ensures that each component solution effectively captures the system's behavior across time.

In the exercise, a set of three fundamental solutions was constructed, which forms the basis for the complete solution. The fundamentals: \( \mathbf{y}_1(t) = e^{-3t}[12, 0, 1]^T \), \( \mathbf{y}_2(t) = e^{t}[9, 0, 1]^T \), and \( \mathbf{y}_3(t) = e^{9t}[3, 0, 1]^T \), provide a comprehensive solution set for the initial value problem.

Matrix Differential Equations

Matrix differential equations form a cornerstone in the study of linear algebra applied to dynamics and systems. These equations involve derivatives of vector-valued functions, and they can often be expressed using matrices to simplify or frame problems involving multiple functions or variables.

• Structure: The typical form is \( \mathbf{y}^{\prime} = A \mathbf{y} \), where \( A \) is a matrix and \( \mathbf{y} \) a vector function. This format sets up a system of first-order linear differential equations.


• Solving such Equations: The process generally starts with identifying the eigenvalues and eigenvectors of matrix \( A \). These eigen-components are crucial, as they help establish the solution's general form via the fundamental set of solutions.

When dealing with matrix differential equations, it's vital to also apply any given initial conditions. In our problem, using the initial condition \( \mathbf{y}(0) = [0, -1, 2]^T \), you can solve for coefficients that fine-tune the solution to fit the specific problem at hand, leading to a precise answer that meets all conditions imposed by the initial setup.