91Ó°ÊÓ

To solve a system of differential equations, we first need to convert the system into a format that allows for finding solutions easily. This involves the **characteristic equation**.
A characteristic equation helps us find the eigenvalues of a matrix formed from the system.
For our example, we have the system:
\(\begin{cases} x' = y\ y' = -x - 2y \ \).
When written as a matrix it becomes: \[ \mathbf{x}' = A \mathbf{x}, \] where \[A = \begin{pmatrix} 0 & 1 \ -1 & -2 \ \end{pmatrix}. \]
The next step is to find the characteristic equation by solving for \(|A - \lambda I| = 0\). This will help us find the eigenvalues needed for the solution.

Eigenvalues are found from the characteristic equation. They are crucial for solving differential equations as they provide the foundation for finding eigenvectors and the general solution.
From our characteristic equation: \[ \begin{vmatrix} 0 - \lambda & 1\ -1 & -2 - \lambda\ \end{vmatrix} = 0, \]
we simplify to: \[ \lambda^2 + 2\lambda + 1 = 0. \]
Solving the quadratic equation, we get a repeated root: \[ \lambda = -1. \]
This eigenvalue is used next to find eigenvectors.

Initial conditions are crucial since they help determine specific solutions from the general solution.
For our example, we have the initial conditions: \[ x(0) = 2, y(0) = 0. \]
These provide the values at time \(t = 0\), helping us find the constants in our general solution and ensure our answer is precise and fits the given problem.

Eigenvectors, found using eigenvalues, are essential for forming the general solution.
With eigenvalue \(\begin{cases} \lambda = -1 \end{cases}\), we solve:
\[ (A - \lambda I)V = 0. \]
For \(\begin{cases} \lambda = -1 \end{cases}\), we solve: \[ \begin{pmatrix} 0 + 1 & 1 \ -1 & -2 + 1 \ \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \ \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \end{pmatrix}. \]
which simplifies to \[ \begin{pmatrix} 1 & 1 \ -1 & -1 \ \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \ \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \end{pmatrix}. \]
This leads to the eigenvector: \[ \begin{pmatrix} 1 \ -1 \ \end{pmatrix}. \]

Finally, let’s wrap it all together to solve the system.
Using the eigenvalues and eigenvectors, we create the general solution: \[ \begin{pmatrix} x(t) \ y(t) \ \end{pmatrix} = c_1 e^{-t} \begin{pmatrix} 1 \ -1 \ \end{pmatrix} + c_2 t e^{-t} \begin{pmatrix} 1 \ -1 \ \end{pmatrix}. \]
Next, apply the initial conditions. Substituting \( t = 0\):
\[ \begin{pmatrix} 2 \ 0 \ \end{pmatrix} = c_1 \begin{pmatrix} 1 \ -1 \ \end{pmatrix}, \]
solving, \(c_1 = 2 \) and \(c_2 = 0\).
Hence, the specific solution is: \[ \begin{pmatrix} x(t) \ y(t) \ \end{pmatrix} = 2 e^{-t} \begin{pmatrix} 1 \ -1 \ \end{pmatrix}. \]
Separating components:
\[ x(t) = 2 e^{-t}, \] \[ y(t) = -2 e^{-t}. \]
And that's how we solve systems of differential equations!