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.\(\mathbf{y}^{r}=\left[\begin{array}{ll}6 & 0 \\ 2 & 6\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}-2 \\ 0\end{array}\right]\)

Short answer

Answer: The solution to the initial value problem is the zero vector, \(\mathbf{y}(t)=\left[\begin{array}{l}0\\0\end{array}\right]\).

Step-by-step solution

Find the eigenvalues and eigenvectors of matrix A

Let's start by finding the eigenvalues of the given matrix \(A = \left[\begin{array}{ll}6&0\\2&6\end{array}\right]\). To do this, we need to solve the characteristic equation: $$\det(A - \lambda I) = 0$$ Where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. For our matrix, the characteristic equation becomes: $$\det\left(\left[\begin{array}{ll}6-\lambda & 0 \\ 2 & 6-\lambda\end{array}\right]\right)=(6-\lambda)^{2}-0(2)=\lambda^{2}-12\lambda+36$$ Next, we will solve the quadratic equation for \(\lambda\): $$\lambda^{2}-12\lambda+36=(\lambda-6)^{2}=0$$ This equation has only one distinct eigenvalue: $$\lambda=6$$ Now, let's find the eigenvectors corresponding to this eigenvalue. We will solve the following equation: $$(A-\lambda I)\mathbf{x}=\mathbf{0}$$ Substitute the eigenvalue and form the matrix vector equation: $$\left[\begin{array}{ll}-0&0\\2&0\end{array}\right]\left[\begin{array}{r}x_{1}\\x_{2}\end{array}\right]=\left[\begin{array}{r}0\\0\end{array}\right]$$ This equation is equivalent to \(2x_{1}=0\). The eigenvector for this single eigenvalue is: $$\mathbf{x}=\left[\begin{array}{l}0\\1\end{array}\right]$$

Construct a fundamental set of solutions

The general solutions for a homogeneous linear system of ODEs can be written as: $$y(t)=c_{1}e^{\lambda_{1} t}\mathbf{v}_{1}+c_{2}e^{\lambda_{2} t}\mathbf{v}_{2}$$ Since we have only one distinct eigenvalue (\(\lambda = 6\)) and the corresponding eigenvector (\(\mathbf{v}=\left[\begin{array}{l}0\\1\end{array}\right]\)), the general solution can be written as: $$\mathbf{y}(t)=c_{1}e^{6t}\mathbf{v}$$ Where \(c_{1}\) is an arbitrary constant.

Solve the initial value problem

Now, we will use the initial condition (\(\mathbf{y}(0)=\left[\begin{array}{r}-2\\0\end{array}\right]\)) to solve for the constant \(c_{1}\): $$\mathbf{y}(0)=c_{1}e^{6(0)}\mathbf{v}=c_{1}\mathbf{v}$$ Substitute the initial value and the eigenvector: $$\left[\begin{array}{r}-2\\0\end{array}\right]=c_{1}\left[\begin{array}{l}0\\1\end{array}\right]$$ Since \(x_{1}=0\) for the eigenvector, we find \(c_{1}\) from the second component: $$0=c_{1}$$ In this case, the constant \(c_{1}\) is zero, and the initial value problem does not have a unique solution. The corresponding homogeneous solution space is spanned by the eigenvector: $$\mathbf{y}(t)=0 \cdot e^{6t}\left[\begin{array}{l}0\\1\end{array}\right]=\left[\begin{array}{l}0\\0\end{array}\right]$$ The solution to the initial value problem is simply the zero vector: $$\mathbf{y}(t)=\left[\begin{array}{l}0\\0\end{array}\right]$$

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are essential in understanding matrix behavior. An eigenvalue is a scalar that helps characterize a transformation, while an eigenvector gives the direction in which the transformation acts.
To find an eigenvalue, you solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( A \) is the matrix in question. This involves finding the determinant of \( A - \lambda I \), leading to a polynomial equation. Solving it gives the eigenvalues.
Next, eigenvectors are found by solving \((A - \lambda I)\mathbf{x} = \mathbf{0}\). This linear system gives the eigenvectors for each eigenvalue. For our matrix, we found that there was one eigenvalue \( \lambda = 6 \), and the eigenvector associated was \( \mathbf{x} = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Fundamental Set of Solutions

In solving differential equations, particularly linear ones, we look for a fundamental set of solutions. This set forms the basis for all solutions of a homogeneous linear system.
For a system with distinct eigenvalues, each solution is a combination of exponential terms involving eigenvalues, \( \lambda_i \), and eigenvectors, \( \mathbf{v}_i \). If there are repeated eigenvalues, the solution might include polynomial terms.
In our case, the solution to the system \( \mathbf{y}' = A\mathbf{y} \) involved only one eigenvalue \( \lambda = 6 \). Due to this repetition, using \( \mathbf{v} = \begin{bmatrix} 0 \ 1 \end{bmatrix} \), the solution is \( \mathbf{y}(t) = c_1 e^{6t} \mathbf{v} \). This represents a simple exponential growth along the direction of the eigenvector.

Homogeneous Linear System

A homogeneous linear system of differential equations is one where all terms are proportional to the unknown function or its derivatives, with no constant term present. They often appear in the form \( \mathbf{y}' = A\mathbf{y} \).
The goal is to determine the behavior of solutions, which are typically combinations of exponentials. Each exponential term corresponds to an eigenvalue of the matrix \( A \).
With distinct eigenvalues, each solution is unique for each eigenvector. However, when eigenvalues are repeated, the solution might not be as straightforward, sometimes requiring additional linearly independent solutions, potentially involving polynomials.
For the initial value problem presented, we found a simple solution due to the presence of repeated eigenvalues, leading to the trivial solution \( \mathbf{y}(t) = \begin{bmatrix} 0 \ 0 \end{bmatrix} \).

Matrix Algebra

Matrix algebra is the mathematical structure used in dealing with linear transformations and systems of linear equations. It includes operations such as addition, multiplication, and finding determinants and inverses.
For solving differential equations, the manipulation of matrices, like finding eigenvalues and eigenvectors, is crucial. These concepts help reformulate differential equations into simpler, solvable forms.

• Eigenvalues aid in finding characteristic polynomials, which indicate system behavior.
• Eigenvectors indicate the direction of influence, helping to frame solutions.

Understanding this algebra forms the backbone of linear system analysis, crucial for modeling complex systems accurately.