Find a particular solution of the indicated linear system that satisfies the initial conditions x₁ (0) = 5, x₂ (0)=2, and x3 (0) = 6. -13-17 -2 3 3t x' = 10 14 2 x X₁5 -2.x₂ 04 - 10 - 10 2 2 The particular solution is x₁ (t) = x₂(t)=, and x3 (t) =

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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**Problem Statement:**

Find a particular solution of the indicated linear system that satisfies the initial conditions \(x_1(0) = 5\), \(x_2(0) = 2\), and \(x_3(0) = 6\).

\[ \mathbf{x'} = \begin{bmatrix} -13 & -17 & -2 \\ 10 & 14 & 2 \\ -10 & -10 & 2 \end{bmatrix} \mathbf{x}; \quad \mathbf{x_1} = e^{-3t} \begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix}, \quad \mathbf{x_2} = e^{2t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad \mathbf{x_3} = e^{4t} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \]

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**Diagram Explanation:**

1. **Initial Conditions and Given Matrices:**
   - A 3x3 coefficient matrix in the differential equation \( \mathbf{x'} \).
   - Three linearly independent solutions:
     - \( \mathbf{x_1} = e^{-3t} \begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix} \)
     - \( \mathbf{x_2} = e^{2t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \)
     - \( \mathbf{x_3} = e^{4t} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \)
   
2. **Particular Solution to Find:**
   - The particular solution must satisfy the given initial conditions: \(x_1(0) = 5\), \(x_2(0) = 2\), and \(x_3(0) = 6\).

---

**Solution Format:**

The particular solution is:
\[ x_1(t) = \_\_\_\_ \]
\[ x_2(t) = \_\_\_\_ \]
\[ x_3(t) = \_\_\_\_ \]

Please enter the specific solutions for \(x_1(t)\), \(x_2(t)\), and \(x
Transcribed Image Text:**Problem Statement:** Find a particular solution of the indicated linear system that satisfies the initial conditions \(x_1(0) = 5\), \(x_2(0) = 2\), and \(x_3(0) = 6\). \[ \mathbf{x'} = \begin{bmatrix} -13 & -17 & -2 \\ 10 & 14 & 2 \\ -10 & -10 & 2 \end{bmatrix} \mathbf{x}; \quad \mathbf{x_1} = e^{-3t} \begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix}, \quad \mathbf{x_2} = e^{2t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad \mathbf{x_3} = e^{4t} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \] --- **Diagram Explanation:** 1. **Initial Conditions and Given Matrices:** - A 3x3 coefficient matrix in the differential equation \( \mathbf{x'} \). - Three linearly independent solutions: - \( \mathbf{x_1} = e^{-3t} \begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix} \) - \( \mathbf{x_2} = e^{2t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \) - \( \mathbf{x_3} = e^{4t} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \) 2. **Particular Solution to Find:** - The particular solution must satisfy the given initial conditions: \(x_1(0) = 5\), \(x_2(0) = 2\), and \(x_3(0) = 6\). --- **Solution Format:** The particular solution is: \[ x_1(t) = \_\_\_\_ \] \[ x_2(t) = \_\_\_\_ \] \[ x_3(t) = \_\_\_\_ \] Please enter the specific solutions for \(x_1(t)\), \(x_2(t)\), and \(x
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