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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![### Problem 47
Show that \( y = x^3 + 3x + 1 \) satisfies \( y''' + xy'' - 2y' = 0 \).
---
To solve this, we need to take the necessary derivatives of \( y \) and then substitute them into the given differential equation to verify the solution.
1. **First Derivative \( y' \)**:
\[
y = x^3 + 3x + 1
\]
\[
y' = \frac{d}{dx}(x^3 + 3x + 1) = 3x^2 + 3
\]
2. **Second Derivative \( y'' \)**:
\[
y'' = \frac{d}{dx}(3x^2 + 3) = 6x
\]
3. **Third Derivative \( y''' \)**:
\[
y''' = \frac{d}{dx}(6x) = 6
\]
Now we substitute \( y''' \), \( y'' \), and \( y' \) into the differential equation \( y''' + xy'' - 2y' \):
\[
y''' + xy'' - 2y' = 6 + x(6x) - 2(3x^2 + 3)
\]
Simplify the expression:
\[
6 + 6x^2 - 6x^2 - 6
\]
Combine like terms:
\[
6 - 6 + 6x^2 - 6x^2 = 0
\]
Therefore, \( y''' + xy'' - 2y' = 0 \) is verified, confirming that \( y = x^3 + 3x + 1 \) is a solution to the differential equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7f605619-bbba-4f43-942e-b505f4944796%2F756c25a0-e72e-4c00-951b-030d5420e0c6%2Fe2ih95r_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 47
Show that \( y = x^3 + 3x + 1 \) satisfies \( y''' + xy'' - 2y' = 0 \).
---
To solve this, we need to take the necessary derivatives of \( y \) and then substitute them into the given differential equation to verify the solution.
1. **First Derivative \( y' \)**:
\[
y = x^3 + 3x + 1
\]
\[
y' = \frac{d}{dx}(x^3 + 3x + 1) = 3x^2 + 3
\]
2. **Second Derivative \( y'' \)**:
\[
y'' = \frac{d}{dx}(3x^2 + 3) = 6x
\]
3. **Third Derivative \( y''' \)**:
\[
y''' = \frac{d}{dx}(6x) = 6
\]
Now we substitute \( y''' \), \( y'' \), and \( y' \) into the differential equation \( y''' + xy'' - 2y' \):
\[
y''' + xy'' - 2y' = 6 + x(6x) - 2(3x^2 + 3)
\]
Simplify the expression:
\[
6 + 6x^2 - 6x^2 - 6
\]
Combine like terms:
\[
6 - 6 + 6x^2 - 6x^2 = 0
\]
Therefore, \( y''' + xy'' - 2y' = 0 \) is verified, confirming that \( y = x^3 + 3x + 1 \) is a solution to the differential equation.
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