To solve the given system of linear ordinary differential equations:

$y_1' = -2y_2$ $y_2' = -y_1 + y_2$

we can rewrite this system in matrix form. Let $\mathbf{Y} = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$. Then, we can represent the system as:

$\mathbf{Y}' = \begin{pmatrix} 0 & -2 \\ -1 & 1 \end{pmatrix} \mathbf{Y}$

The matrix form is:

$\begin{pmatrix} y_1' \\ y_2' \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$

To solve this system, we need to find the eigenvalues and eigenvectors of the coefficient matrix:

$A = \begin{pmatrix} 0 & -2 \\ -1 & 1 \end{pmatrix}$

1. Find the eigenvalues of $A$ by solving the characteristic equation $\det(A - \lambda I) = 0$:

   $\det \begin{pmatrix} 0 - \lambda & -2 \\ -1 & 1 - \lambda \end{pmatrix} = \lambda^2 - \lambda + 2 = 0$

   Solving this quadratic equation for $\lambda$, we find the eigenvalues.

2. Find the eigenvectors corresponding to each eigenvalue.

3. Construct the general solution by forming a linear combination of the solutions corresponding to each eigenpair.

Let me go through each step and present the complete solution.import numpy as np from sympy import symbols, Matrix, solve

Define the matrix and lambda symbol

lambda_ = symbols('lambda') A = Matrix([[0, -2], [-1, 1]])

Compute the characteristic polynomial

char_poly = A.charpoly(lambda_).as_expr()

Solve for eigenvalues

eigenvalues = solve(char_poly, lambda_) char_poly, eigenvalues(lambda2 - lambda - 2, [-1, 2])(lambda2 - lambda - 2, [-1, 2])successThe characteristic polynomial of the matrix $A$ is:

$\lambda^2 - \lambda - 2 = 0$

Solving this equation, we find the eigenvalues:

$\lambda_1 = -1 \quad \text{and} \quad \lambda_2 = 2$

Next, we will determine the eigenvectors corresponding to each eigenvalue.# Calculate eigenvectors for each eigenvalue eigenvectors = {} for eigenvalue in eigenvalues: eigenvectors[eigenvalue] = A.eigenvects()[eigenvalues.index(eigenvalue)][2][0]

eigenvectors{-1: Matrix([ [2], [1]]), 2: Matrix([ [-1], [ 1]])}{-1: Matrix([ [2], [1]]), 2: Matrix([ [-1], [ 1]])}successThe eigenvectors corresponding to each eigenvalue are as follows:

• For $\lambda = -1$, the eigenvector is $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.
• For $\lambda = 2$, the eigenvector is $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$.

Using these eigenvalues and eigenvectors, we can write the general solution to the system as:

$\mathbf{Y}(t) = c_1 e^{-t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} -1 \\ 1 \end{pmatrix}$

Expanding this solution, we get:

$y_1(t) = 2c_1 e^{-t} - c_2 e^{2t}$ $y_2(t) = c_1 e^{-t} + c_2 e^{2t}$

where $c_1$ and $c_2$ are constants determined by initial conditions.

Would you like further details on any part of this solution, or do you have any specific questions?

Here are five related questions you might find helpful:

1. How do we determine initial conditions for a system of differential equations?
2. What if the eigenvalues were complex—how would that affect the solution?
3. Can we solve non-homogeneous systems using a similar approach?
4. How does matrix diagonalization simplify solving systems of differential equations?
5. What methods can be used if the matrix is not diagonalizable?

Tip: For systems with repeated eigenvalues, the solution may involve generalized eigenvectors to account for missing independent solutions.