https://r2.erweima.ai/mathbot/image/1726359826417gm52t5b2.jpg

The system of differential equations provided in the image is:

\[
\begin{aligned}
x' - 3x + 4y &= 0 \\
y' - 2x + 3y &= 0
\end{aligned}
\]

We are tasked with solving this system step by step.

### Step 1: Write the system in matrix form

We can rewrite the system of equations in matrix form:

\[
\begin{pmatrix}
x' \\
y'
\end{pmatrix}
=
\begin{pmatrix}
3 & -4 \\
2 & -3
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
\]

This is a linear system of first-order differential equations, where the matrix \(A = \begin{pmatrix} 3 & -4 \\ 2 & -3 \end{pmatrix}\) represents the coefficients of the system.

### Step 2: Solve the characteristic equation

To solve the system, we need to find the eigenvalues and eigenvectors of the matrix \(A\). First, we find the characteristic equation by solving \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues.

\[
\det \begin{pmatrix} 3 - \lambda & -4 \\ 2 & -3 - \lambda \end{pmatrix} = 0
\]

The determinant is:

\[
(3 - \lambda)(-3 - \lambda) - (-4)(2) = 0
\]

Expanding:

\[
(3 - \lambda)(-3 - \lambda) + 8 = 0
\]

\[
\lambda^2 - 9 + 8 = 0
\]

\[
\lambda^2 - 1 = 0
\]

Solving for \(\lambda\), we get:

\[
\lambda = \pm 1
\]

Thus, the eigenvalues of the system are \(\lambda_1 = 1\) and \(\lambda_2 = -1\).

### Step 3: Find the eigenvectors

Now we find the eigenvectors corresponding to each eigenvalue.

#### For \(\lambda_1 = 1\):

We solve \((A - I)v = 0\):

\[
\begin{pmatrix} 3 - 1 & -4 \\ 2 & -3 - 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

\[
\begin{pmatrix} 2 & -4 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

From the first equation: \(2v_1 - 4v_2 = 0\), which gives \(v_1 = 2v_2\). Let \(v_2 = 1\), so \(v_1 = 2\). The eigenvector corresponding to \(\lambda_1 = 1\) is:

\[
v_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}
\]

#### For \(\lambda_2 = -1\):

We solve \((A + I)v = 0\):

\[
\begin{pmatrix} 3 + 1 & -4 \\ 2 & -3 + 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

\[
\begin{pmatrix} 4 & -4 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

From the first equation: \(4v_1 - 4v_2 = 0\), which gives \(v_1 = v_2\). Let \(v_2 = 1\), so \(v_1 = 1\). The eigenvector corresponding to \(\lambda_2 = -1\) is:

\[
v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\]

### Step 4: General solution

The general solution to the system is given by:

\[
\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 e^{\lambda_1 t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{\lambda_2 t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\]

Substituting the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = -1\):

\[
\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 e^t \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\]

Thus, the solution is:

\[
x(t) = 2c_1 e^t + c_2 e^{-t}
\]
\[
y(t) = c_1 e^t + c_2 e^{-t}
\]

### Summary:

The general solution to the system of differential equations is:

\[
x(t) = 2c_1 e^t + c_2 e^{-t}
\]
\[
y(t) = c_1 e^t + c_2 e^{-t}
\]

Let me know if you'd like me to save this solution in a specific format such as PDF, and feel free to ask for further clarifications! 

---

Here are 5 related questions you may find helpful:
1. How do we determine the initial conditions for this system of equations?
2. What are some methods to solve systems of differential equations besides eigenvalue methods?
3. How would the solution change if the system had complex eigenvalues?
4. Can this system be solved using Laplace transforms? How would the approach differ?
5. What is the physical significance of finding eigenvalues and eigenvectors in such systems?

**Tip**: Always check the stability of the solutions by analyzing the signs of the real parts of the eigenvalues.

Resuelva paso a paso el siguiente sistema de ecuaciones diferenciales: x' - 3x + 4y = 0, y' - 2x + 3y = 0. Siendo x(t) y y(t) funciones que dependen de una variable t. Guarde su solución en un archivo (.jpg, .docx, .pdf) y envíelo por este medio para su calificación.

Learn how to solve the system of differential equations x' - 3x + 4y = 0 and y' - 2x + 3y = 0 with a step-by-step breakdown. This solution uses the eigenvalue and eigenvector method, detailing how to find the general solution to this homogeneous linear system. Suitable for undergraduate-level students.

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