Transform the given system of differential equations into an equivalent system of first-order differential equations. x" - 4x' + 6x - 5y = 0 y"' + 2y' + 3x-2y = cos t Let x₁ = x, x₂ = x', Y₁ = y, and y₂ =y'. Complete the system below. 8 X X₂' X 11 II Y2: 11
Transform the given system of differential equations into an equivalent system of first-order differential equations. x" - 4x' + 6x - 5y = 0 y"' + 2y' + 3x-2y = cos t Let x₁ = x, x₂ = x', Y₁ = y, and y₂ =y'. Complete the system below. 8 X X₂' X 11 II Y2: 11
Calculus: Early Transcendentals
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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![### Transforming Second-Order Differential Equations to First-Order Systems
Given the system of second-order differential equations:
\[ x'' - 4x' + 6x - 5y = 0 \]
\[ y'' + 2y' + 3x - 2y = \cos t \]
We aim to transform this system into an equivalent system of first-order differential equations.
#### Step-by-Step Process
1. Define new variables to represent the first-order derivatives:
Let \( x_1 = x \), \( x_2 = x' \), \( y_1 = y \), and \( y_2 = y' \).
2. Rewrite the original equations using these new variables:
From \( x_1' = x_2 \)
Given \( x'' - 4x' + 6x - 5y = 0 \), substitute \( x'' = x_2' \), \( x' = x_2 \), \( x = x_1 \), and \( y = y_1 \) to get:
\[ x_2' - 4x_2 + 6x_1 - 5y_1 = 0 \]
From \( y_1' = y_2 \)
Given \( y'' + 2y' + 3x - 2y = \cos t \), substitute \( y'' = y_2' \), \( y' = y_2 \), \( y = y_1 \), and \( x = x_1 \) to get:
\[ y_2' + 2y_2 + 3x_1 - 2y_1 = \cos t \]
#### Completed System:
\[
\begin{aligned}
& x_1' = x_2 \\
& x_2' = 4x_2 - 6x_1 + 5y_1 \\
& y_1' = y_2 \\
& y_2' = -2y_2 - 3x_1 + 2y_1 + \cos t
\end{aligned}
\]
This is the equivalent system of first-order differential equations.
#### Summary
Transforming a given system of higher-order differential equations into a first-order system involves:
- Introducing new variables to represent the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F896deee6-4ebc-4afd-8502-502eb7aa6712%2Fad92a54e-6202-4019-a323-730544d36581%2F123dto_processed.png&w=3840&q=75)
Transcribed Image Text:### Transforming Second-Order Differential Equations to First-Order Systems
Given the system of second-order differential equations:
\[ x'' - 4x' + 6x - 5y = 0 \]
\[ y'' + 2y' + 3x - 2y = \cos t \]
We aim to transform this system into an equivalent system of first-order differential equations.
#### Step-by-Step Process
1. Define new variables to represent the first-order derivatives:
Let \( x_1 = x \), \( x_2 = x' \), \( y_1 = y \), and \( y_2 = y' \).
2. Rewrite the original equations using these new variables:
From \( x_1' = x_2 \)
Given \( x'' - 4x' + 6x - 5y = 0 \), substitute \( x'' = x_2' \), \( x' = x_2 \), \( x = x_1 \), and \( y = y_1 \) to get:
\[ x_2' - 4x_2 + 6x_1 - 5y_1 = 0 \]
From \( y_1' = y_2 \)
Given \( y'' + 2y' + 3x - 2y = \cos t \), substitute \( y'' = y_2' \), \( y' = y_2 \), \( y = y_1 \), and \( x = x_1 \) to get:
\[ y_2' + 2y_2 + 3x_1 - 2y_1 = \cos t \]
#### Completed System:
\[
\begin{aligned}
& x_1' = x_2 \\
& x_2' = 4x_2 - 6x_1 + 5y_1 \\
& y_1' = y_2 \\
& y_2' = -2y_2 - 3x_1 + 2y_1 + \cos t
\end{aligned}
\]
This is the equivalent system of first-order differential equations.
#### Summary
Transforming a given system of higher-order differential equations into a first-order system involves:
- Introducing new variables to represent the
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