Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Consider the system
\[ \frac{x'}{x} = xy \]
\[ \frac{y'}{y} = y \]
#### a) Show, by direct substitution, that:
\[ x_1 = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ e^t \end{bmatrix} \quad \text{and} \quad x_2 = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} y \\ 0 \end{bmatrix} \]
are solutions of the system.
#### b) Show that \(\mathbf{x_{NOT}} = x_1(t) + x_2(t)\) *is NOT a solution of the system*.
#### c) For linear systems, linear combinations of solutions to systems are also solutions. Why doesn't this principle apply for this example?
---
### Diagram Explanation:
There are no graphs or diagrams included in the provided image.
---
### Detailed Explanation:
**Part (a)**
To show that \(\mathbf{x_{1}}\) and \(\mathbf{x_{2}}\) are solutions to the given system by direct substitution:
1. **Verify \( \mathbf{x_{1}} \)**
- Substitute \( x \) and \( y \) from \( \mathbf{x_{1}} = \begin{bmatrix} 0 \\ e^t \end{bmatrix} \) into the system:
- For \( \frac{x'}{x} = xy \):
\[ x = 0 \]
\[ y = e^t \]
Since \( x = 0 \), the left-hand side \(\frac{x'}{x}\) is undefined, but \( x' = 0 \) as \( x \) is a constant zero.
- For \( \frac{y'}{y} = y \):
\[ y' = e^t \]
\[ y = e^t \]
\[ \frac{y'}{y} = \frac{e^t}{e^t} = 1 \]
Since \( y' = e^t \) and \( y = e^t \), the equation holds true.
2. **Verify \( \mathbf{x_{2}} \)**
- Substitute \( x \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05bd8196-d1b6-458c-8115-79c4148218b1%2F15009059-a55f-4027-9fcd-d66491d1d194%2Fcsgeyz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Consider the system
\[ \frac{x'}{x} = xy \]
\[ \frac{y'}{y} = y \]
#### a) Show, by direct substitution, that:
\[ x_1 = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ e^t \end{bmatrix} \quad \text{and} \quad x_2 = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} y \\ 0 \end{bmatrix} \]
are solutions of the system.
#### b) Show that \(\mathbf{x_{NOT}} = x_1(t) + x_2(t)\) *is NOT a solution of the system*.
#### c) For linear systems, linear combinations of solutions to systems are also solutions. Why doesn't this principle apply for this example?
---
### Diagram Explanation:
There are no graphs or diagrams included in the provided image.
---
### Detailed Explanation:
**Part (a)**
To show that \(\mathbf{x_{1}}\) and \(\mathbf{x_{2}}\) are solutions to the given system by direct substitution:
1. **Verify \( \mathbf{x_{1}} \)**
- Substitute \( x \) and \( y \) from \( \mathbf{x_{1}} = \begin{bmatrix} 0 \\ e^t \end{bmatrix} \) into the system:
- For \( \frac{x'}{x} = xy \):
\[ x = 0 \]
\[ y = e^t \]
Since \( x = 0 \), the left-hand side \(\frac{x'}{x}\) is undefined, but \( x' = 0 \) as \( x \) is a constant zero.
- For \( \frac{y'}{y} = y \):
\[ y' = e^t \]
\[ y = e^t \]
\[ \frac{y'}{y} = \frac{e^t}{e^t} = 1 \]
Since \( y' = e^t \) and \( y = e^t \), the equation holds true.
2. **Verify \( \mathbf{x_{2}} \)**
- Substitute \( x \)
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