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.\(\mathbf{y}^{\prime}=\left[\begin{array}{lll}2 & 1 & 0 \\ 0 & 2 & 0 \\\ 0 & 0 & 1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}1 \\ 3 \\ -2\end{array}\right]\)

Short answer

Answer: \(\mathbf{y}(t) = \left[\begin{array}{r}-1 \\ 1 \\ 0\end{array}\right]e^{t} + 2\left[\begin{array}{r}1 \\ 0 \\ 0\end{array}\right]e^{2t} + \left[\begin{array}{r}t \\ 1 \\ 0\end{array}\right]e^{2t}\)

Step-by-step solution

Find the Eigenvalues and Eigenvectors of Matrix A

To find the eigenvalues (\(\lambda\)), we need to solve the characteristic equation \(|A - \lambda I| = 0\), where \(I\) is the identity matrix of the same order as matrix \(A\). \(|A - \lambda I| = \left|\begin{array}{ccc} 2-\lambda & 1 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 1-\lambda \end{array}\right| = (2-\lambda)((2-\lambda)(1-\lambda)) = (\lambda-2)^2(\lambda-1)\) Solving this equation, we get the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = \lambda_3 = 2\) (repeated eigenvalue). Now, we'll find the eigenvectors for each eigenvalue: For \(\lambda_1 = 1:\) \((A-\lambda_1 I)\mathbf{v}_1=0 \Rightarrow \left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right]\mathbf{v}_1=0\) We can solve this system of linear equations to find the eigenvector \(\mathbf{v}_1=\left[\begin{array}{r}-1 \\ 1 \\ 0\end{array}\right]\). For \(\lambda_2 = 2:\) \((A-\lambda_2 I)\mathbf{v}_2=0 \Rightarrow \left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]\mathbf{v}_2=0\) We can solve this system of linear equations to find the eigenvector \(\mathbf{v}_2=\left[\begin{array}{r}1 \\ 0 \\ 0\end{array}\right]\). Since we have repeated eigenvalue \(\lambda_3 = 2\), instead of finding another eigenvector, we need to find the generalized eigenvector, by solving \((A-\lambda_3 I)(\mathbf{v}_3-\mathbf{v}_2)=\mathbf{v}_2 \Rightarrow \left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]\mathbf{v}_3=\left[\begin{array}{r}1 \\ 0 \\ 0\end{array}\right]\). Solving this system, we find the generalized eigenvector: \(\mathbf{v}_3=\left[\begin{array}{r}0 \\ 1 \\ 0\end{array}\right]\).

Construct a Fundamental Set of Solutions

Using the eigenvalues and eigenvectors (including the generalized eigenvector), a fundamental set of solutions can be obtained: \(\mathbf{y}_1(t) = \mathbf{v}_1 e^{\lambda_1 t} = \left[\begin{array}{r}-1 \\ 1 \\ 0\end{array}\right]e^{t},\) \(\mathbf{y}_2(t) = \mathbf{v}_2 e^{\lambda_2 t} = \left[\begin{array}{r}1 \\ 0 \\ 0\end{array}\right]e^{2t},\) \(\mathbf{y}_3(t) = \left(\mathbf{v}_3 + \mathbf{v}_2 t\right) e^{\lambda_3 t} = \left[\begin{array}{r}t \\ 1 \\ 0\end{array}\right]e^{2t}\).

Solve the Initial Value Problem

Now, we'll solve the initial value problem \(\mathbf{y}(0) = \left[\begin{array}{r}1 \\ 3 \\ -2\end{array}\right]\). We can express the general solution as a linear combination of the fundamental set of solutions: \(\mathbf{y}(t) = c_1\mathbf{y}_1(t) + c_2\mathbf{y}_2(t) + c_3\mathbf{y}_3(t)\). Plugging in the initial condition, we need to find the constants \(c_1\), \(c_2\), and \(c_3\): \(\mathbf{y}(0) = c_1\mathbf{y}_1(0) + c_2\mathbf{y}_2(0) + c_3\mathbf{y}_3(0) = \left[\begin{array}{r}1 \\ 3 \\ -2\end{array}\right]\). This gives us the system of linear equations: \(-c_1 + c_2 = 1\) \(c_1 + c_3 = 3\) \(-2c_3 = -2\) Solving this system, we find \(c_1=1\), \(c_2=2\), and \(c_3=1\). Thus, the solution of the initial value problem is: \(\mathbf{y}(t) = \left[\begin{array}{r}-1 \\ 1 \\ 0\end{array}\right]e^{t} + 2\left[\begin{array}{r}1 \\ 0 \\ 0\end{array}\right]e^{2t} + \left[\begin{array}{r}t \\ 1 \\ 0\end{array}\right]e^{2t}\).

Key concepts

Eigenvalues and Eigenvectors

When exploring linear differential equations with constant coefficients, eigenvalues and eigenvectors are essential. They help you understand the system's behavior. Eigenvalues are numbers that signify how the system evolves over time. For a matrix \(A\), the eigenvalues are found by solving the characteristic equation \(|A - \lambda I| = 0\). Here, \(\lambda\) represents the eigenvalues, and \(I\) is the identity matrix.
In the example provided, the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = \lambda_3 = 2\). Notice \(\lambda_2\) and \(\lambda_3\) are repeated, having implications for the eigenvectors.
Eigenvectors, linked to each eigenvalue, show directions in which the transformation only stretches. These are derived by solving \((A - \lambda I)\mathbf{v} = 0\). If an eigenvalue is repeated, you might need to find generalized eigenvectors as well. This involves additional computations to ensure enough linearly independent solutions.

Initial Value Problem

The initial value problem (IVP) involves finding a specific solution to a differential equation that satisfies given initial conditions. In this case, you start with a given state \(\mathbf{y}(0)\). Solving an IVP means finding a solution that passes through this initial state.
In the exercise, the initial value condition is \(\mathbf{y}(0) = \left[\begin{array}{r}1 \ 3 \ -2\end{array}\right]\). By using this information, you can determine constants in the general solution to make sure the solution exactly matches the initial condition at \(t = 0\). This ensures that your solution isn't just any solution, but the one that starts at your specified initial point.

Fundamental Set of Solutions

The fundamental set of solutions is vital in solving linear differential equations. It consists of linearly independent solutions that capture all possible behaviors of the system. Each solution is associated with different eigenvalues and eigenvectors, including generalized ones if eigenvalues are repeated.
In this case, solutions derived from eigenvectors \(\mathbf{y}_1(t)\), \(\mathbf{y}_2(t)\), and \(\mathbf{y}_3(t)\) form this set. They are:

• \(\mathbf{y}_1(t) = \left[\begin{array}{r}-1 \ 1 \ 0\end{array}\right]e^{t}\)
• \(\mathbf{y}_2(t) = \left[\begin{array}{r}1 \ 0 \ 0\end{array}\right]e^{2t}\)
• \(\mathbf{y}_3(t) = \left[\begin{array}{r}t \ 1 \ 0\end{array}\right]e^{2t}\)

These solutions are fundamental because any other solution can be expressed as a combination of them.

General Solution of Differential Equations

The general solution encompasses all possible solutions to a differential equation. It's often expressed as a linear combination of the fundamental set of solutions. This means you use the solutions derived from the eigenvectors, adjusting with constants to match conditions like an initial value problem.
The formula \( \mathbf{y}(t) = c_1\mathbf{y}_1(t) + c_2\mathbf{y}_2(t) + c_3\mathbf{y}_3(t) \) is a perfect example. The constants \(c_1, c_2,\) and \(c_3\) are determined by initial conditions or boundary values.
In the solution, these constants were computed using the given initial condition \(\mathbf{y}(0) = \left[\begin{array}{r}1 \ 3 \ -2\end{array}\right]\). By setting time \(t = 0\), equations involving \(c_1, c_2, \) and \(c_3\) were solved to fit the starting point exactly. The outcome is a specific solution that both matches the fundamental solutions and satisfies the initial condition.