Find a particular solution of the indicated linear system that satisfies the initial conditions x₁ (0) = 7, x₂ (0) = -5. 3 - 1 1 2t 1 X'= ~~ = e x2 = e D 5 - 3 1 5 The particular solution is x₁ (t)= and x₂ (t) =
Find a particular solution of the indicated linear system that satisfies the initial conditions x₁ (0) = 7, x₂ (0) = -5. 3 - 1 1 2t 1 X'= ~~ = e x2 = e D 5 - 3 1 5 The particular solution is x₁ (t)= and x₂ (t) =
Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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![**Finding a Particular Solution for a Linear System with Given Initial Conditions**
**Problem Statement:**
Find a particular solution of the indicated linear system that satisfies the initial conditions \( x_1(0) = 7 \), \( x_2(0) = -5 \).
\[ x' = \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} x \]
\[ x_1 = e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \, x_2 = e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \]
*Note:* The problem specifies the solution vectors \( x_1 \) and \( x_2 \) for the system.
**Solution:**
To find the particular solution that meets the initial conditions, we combine the given solutions and constants (generally denoted as constants \( c_1 \) and \( c_2 \)) to form the general solution:
\[ x(t) = c_1 e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \]
We then use the initial conditions to solve for the constants \( c_1 \) and \( c_2 \).
*Given Initial Conditions:*
At \( t = 0 \),
\[ x(0) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 5 \end{bmatrix} = \begin{bmatrix} 7 \\ -5 \end{bmatrix} \]
This gives us the following equations:
\[ c_1 + c_2 = 7 \]
\[ c_1 + 5c_2 = -5 \]
Solve these equations simultaneously to find \( c_1 \) and \( c_2 \).
1. Subtract the first equation from the second:
\[ 4c_2 = -12 \]
\[ c_2 = -3 \]
2. Substitute \( c_2 \) back into the first equation:
\[ c_1 - 3 = 7 \]
\[ c_1 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06bbd155-c9d1-4bdc-8881-0cd893ac016a%2F180cfa7c-6750-4831-8bca-1c0e8d334ff3%2Fzs9tb6k_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding a Particular Solution for a Linear System with Given Initial Conditions**
**Problem Statement:**
Find a particular solution of the indicated linear system that satisfies the initial conditions \( x_1(0) = 7 \), \( x_2(0) = -5 \).
\[ x' = \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} x \]
\[ x_1 = e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \, x_2 = e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \]
*Note:* The problem specifies the solution vectors \( x_1 \) and \( x_2 \) for the system.
**Solution:**
To find the particular solution that meets the initial conditions, we combine the given solutions and constants (generally denoted as constants \( c_1 \) and \( c_2 \)) to form the general solution:
\[ x(t) = c_1 e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \]
We then use the initial conditions to solve for the constants \( c_1 \) and \( c_2 \).
*Given Initial Conditions:*
At \( t = 0 \),
\[ x(0) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 5 \end{bmatrix} = \begin{bmatrix} 7 \\ -5 \end{bmatrix} \]
This gives us the following equations:
\[ c_1 + c_2 = 7 \]
\[ c_1 + 5c_2 = -5 \]
Solve these equations simultaneously to find \( c_1 \) and \( c_2 \).
1. Subtract the first equation from the second:
\[ 4c_2 = -12 \]
\[ c_2 = -3 \]
2. Substitute \( c_2 \) back into the first equation:
\[ c_1 - 3 = 7 \]
\[ c_1 =
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