VIPSolutionsgive answer in 3 step with explanation at the end of each step and final answer at the end:Problems Find general solutions of the linear systems in Problems | tion field and typical solution curves for the given system. through 20. If initial conditions are given, find the particu- I lar solution that satisfies them. In Problems I through 6, use a. computer system or graphing calculator 0 construct a direc: 2 RR a i wilt RE
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give answer in 3 step with explanation at the end of each step and final answer at the end:
Problems
Find general solutions of the linear systems in Problems | tion field and typical solution curves for the given system.
through 20. If initial conditions are given, find the particu- I
lar solution that satisfies them. In Problems I through 6, use a.
computer system or graphing calculator 0 construct a direc: 2 RR a
i wilt RE
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Asked by: Test User 100161
Created at: 2025-05-07 07:08:51
Asked by: Test User 100161
Created at: 2025-05-07 07:08:51
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Let's solve the given linear system step by step:
Given system:
\[
\begin{cases}
x' = -2x + y \\
y' = 2x - 3y
\end{cases}
\]
---
### **Step 1: Write the system in matrix form**
Rewrite the system as a matrix differential equation:
\[
\begin{bmatrix}
x' \\
y'
\end{bmatrix}
=
\begin{bmatrix}
-2 & 1 \\
2 & -3
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
\]
**Explanation:**
This step makes it easier to analyze the system and find eigenvalues/eigenvectors for the general solution.
---
### **Step 2: Find the eigenvalues of the coefficient matrix**
Let \( A = \begin{bmatrix} -2 & 1 \\ 2 & -3 \end{bmatrix} \).
Find the eigenvalues \(\lambda\) from the characteristic equation:
\[
\det(A - \lambda I) = 0
\]
\[
\begin{vmatrix}
-2 - \lambda & 1 \\
2 & -3 - \lambda
\end{vmatrix}
= 0
\]
\[
(-2-\lambda)(-3-\lambda) - 2 \cdot 1 = 0
\]
\[
(\lambda + 2)(\lambda + 3) - 2 = 0
\]
\[
\lambda^2 + 5\lambda + 6 - 2 = 0
\]
\[
\lambda^2 + 5\lambda + 4 = 0
\]
Solve this quadratic:
\[
\lambda = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2}
\]
So, \(\lambda_1 = -1\), \(\lambda_2 = -4\).
**Explanation:**
Eigenvalues are crucial for constructing the general solution.
---
### **Step 3: Find the eigenvectors and write the general solution**
For \(\lambda_1 = -1\):
\[
(A + I)\vec{v} = 0
\]
\[
\begin{bmatrix}
-1 & 1 \\
2 & -2
\end{bmatrix}
\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
\]
So, \(-v_1 + v_2 = 0 \implies v_2 = v_1\).
Let \(v_1 = 1\), then \(v_2 = 1\). So, eigenvector: \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\).
For \(\lambda_2 = -4\):
\[
(A + 4I)\vec{v} = 0
\]
\[
\begin{bmatrix}
2 & 1 \\
2 & 1
\end{bmatrix}
\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
\]
So, \(2v_1 + v_2 = 0 \implies v_2 = -2v_1\).
Let \(v_1 = 1\), then \(v_2 = -2\). So, eigenvector: \(\begin{bmatrix} 1 \\ -2 \end{bmatrix}\).
---
### **Final Answer: General Solution**
\[
\boxed{
\begin{aligned}
x(t) &= C_1 e^{-t} + C_2 e^{-4t} \\
y(t) &= C_1 e^{-t} - 2C_2 e^{-4t}
\end{aligned}
}
\]
Where \(C_1\) and \(C_2\) are constants determined by initial conditions.
---
**Summary:**
1. Wrote the system in matrix form for easier manipulation.
2. Found eigenvalues to understand the solution's form.
3. Found eigenvectors and constructed the general solution using them.
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