Determining a Solution In Exercises 23-30, determine whether the function is a solution of the differential equation xy' - 2y = x³ex, x > 0. 25. y = x²ex
Determining a Solution In Exercises 23-30, determine whether the function is a solution of the differential equation xy' - 2y = x³ex, x > 0. 25. y = x²ex
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...
Related questions
Question
![### Determining a Solution
In Exercises 23–30, determine whether the function is a solution of the differential equation:
\[ xy' - 2y = x^3e^x, \quad x > 0. \]
**25.** \( y = x^2e^x \)
**Explanation:**
To determine if \( y = x^2e^x \) is a solution to the differential equation, follow these steps:
1. **Differentiate \( y \):**
Find \( y' \) by using the product rule. The product rule states that if you have a function \( y = u(x)v(x) \), then \( y' = u'v + uv' \).
For \( y = x^2e^x \):
- \( u = x^2 \) and \( v = e^x \)
- \( u' = 2x \) and \( v' = e^x \)
Applying the product rule, we get:
\[
y' = (2x)e^x + (x^2)(e^x) = e^x(2x + x^2)
\]
2. **Substitute into the differential equation:**
Substitute \( y \) and \( y' \) into the equation \( xy' - 2y = x^3e^x \).
Left-hand side:
\[
xy' - 2y = x(e^x(2x + x^2)) - 2(x^2e^x)
\]
Simplifying:
\[
= xe^x(2x + x^2) - 2x^2e^x
\]
\[
= x(e^x(2x) + e^xx^2) - 2x^2e^x
\]
\[
= e^x(2x^2) + e^xx^3 - 2x^2e^x
\]
\[
= x^3e^x
\]
3. **Compare both sides:**
The simplified left-hand side is \( x^3e^x \), which matches with the right-hand side of the original differential equation. Therefore, \( y = x^2e^x \) is indeed a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa1490c8-ab05-4172-8498-f7edf81cd3ae%2Fc7009134-4d47-49fd-aa07-5d61576c24b5%2F890yss1_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining a Solution
In Exercises 23–30, determine whether the function is a solution of the differential equation:
\[ xy' - 2y = x^3e^x, \quad x > 0. \]
**25.** \( y = x^2e^x \)
**Explanation:**
To determine if \( y = x^2e^x \) is a solution to the differential equation, follow these steps:
1. **Differentiate \( y \):**
Find \( y' \) by using the product rule. The product rule states that if you have a function \( y = u(x)v(x) \), then \( y' = u'v + uv' \).
For \( y = x^2e^x \):
- \( u = x^2 \) and \( v = e^x \)
- \( u' = 2x \) and \( v' = e^x \)
Applying the product rule, we get:
\[
y' = (2x)e^x + (x^2)(e^x) = e^x(2x + x^2)
\]
2. **Substitute into the differential equation:**
Substitute \( y \) and \( y' \) into the equation \( xy' - 2y = x^3e^x \).
Left-hand side:
\[
xy' - 2y = x(e^x(2x + x^2)) - 2(x^2e^x)
\]
Simplifying:
\[
= xe^x(2x + x^2) - 2x^2e^x
\]
\[
= x(e^x(2x) + e^xx^2) - 2x^2e^x
\]
\[
= e^x(2x^2) + e^xx^3 - 2x^2e^x
\]
\[
= x^3e^x
\]
3. **Compare both sides:**
The simplified left-hand side is \( x^3e^x \), which matches with the right-hand side of the original differential equation. Therefore, \( y = x^2e^x \) is indeed a
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning