Solve (x_1)' = x_1 -5 x_2
(x_2)' =  x_1 + 3 x_2

To solve the system of differential equations:

\[
\begin{cases}
(x_1)' = x_1 - 5x_2 \\
(x_2)' = x_1 + 3x_2
\end{cases}
\]

we can express the system in matrix form:

\[
\frac{d}{dt} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & -5 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}
\]

Let \( \mathbf{x}(t) = \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix} \), and let \( A = \begin{pmatrix} 1 & -5 \\ 1 & 3 \end{pmatrix} \). The system can be written as:

\[
\frac{d\mathbf{x}}{dt} = A\mathbf{x}
\]

### Step 1: Find the Eigenvalues of the Matrix \( A \)

The eigenvalues \( \lambda \) of the matrix \( A \) are found by solving the characteristic equation:

\[
\det(A - \lambda I) = 0
\]

where \( I \) is the identity matrix. Thus,

\[
\det \begin{pmatrix} 1-\lambda & -5 \\ 1 & 3-\lambda \end{pmatrix} = (1-\lambda)(3-\lambda) - (-5)(1) = 0
\]

Expanding the determinant:

\[
(1-\lambda)(3-\lambda) + 5 = \lambda^2 - 4\lambda + 3 + 5 = \lambda^2 - 4\lambda + 8 = 0
\]

The eigenvalues are the solutions to the quadratic equation:

\[
\lambda^2 - 4\lambda + 8 = 0
\]

Using the quadratic formula:

\[
\lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times 8}}{2 \times 1} = \frac{4 \pm \sqrt{16 - 32}}{2} = \frac{4 \pm \sqrt{-16}}{2} = \frac{4 \pm 4i}{2} = 2 \pm 2i
\]

So the eigenvalues are \( \lambda_1 = 2 + 2i \) and \( \lambda_2 = 2 - 2i \).

### Step 2: Find the Eigenvectors

For \( \lambda_1 = 2 + 2i \), we solve:

\[
(A - \lambda_1 I)\mathbf{v} = 0
\]

This leads to:

\[
\begin{pmatrix} 1-(2+2i) & -5 \\ 1 & 3-(2+2i) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} -1-2i & -5 \\ 1 & 1-2i \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0
\]

This simplifies to:

\[
\begin{cases}
(-1 - 2i)v_1 - 5v_2 = 0 \\
v_1 + (1 - 2i)v_2 = 0
\end{cases}
\]

This gives the eigenvector \( \mathbf{v}_1 \). We follow similar steps for \( \lambda_2 = 2 - 2i \) to find \( \mathbf{v}_2 \).

### Step 3: Form the General Solution

The general solution is:

\[
\mathbf{x}(t) = c_1 e^{(2+2i)t} \mathbf{v}_1 + c_2 e^{(2-2i)t} \mathbf{v}_2
\]

### Step 4: Simplify the Solution

Using Euler's formula \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), the general solution can be expressed in real terms as:

\[
x_1(t) = e^{2t} \left[ c_1 \cos(2t) - c_2 \sin(2t) \right]
\]
\[
x_2(t) = e^{2t} \left[ c_1 \cos(2t) - c_2 \sin(2t) \right]
\]

where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you determine the eigenvectors of a matrix for complex eigenvalues?
2. What is the physical interpretation of the solution to this system of differential equations?
3. How do initial conditions affect the constants in the general solution?
4. Can we solve this system using another method, such as diagonalization?
5. What does the presence of complex eigenvalues indicate about the behavior of the system over time?

**Tip:** Always check if a matrix is diagonalizable before attempting to solve using eigenvalues and eigenvectors. This can significantly simplify the problem.

Learn how to solve the system of differential equations (x1)' = x1 - 5x2, (x2)' = x1 + 3x2 using eigenvalues and eigenvectors. This problem covers key concepts in differential equations, matrix operations, and complex numbers, with a detailed step-by-step solution. Suitable for advanced undergraduate students and researchers.

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