A homogeneous second-order linear differential equation, two functions y, and y₂, and a pair of initial conditions are given. First verify that y, and y₂ are solutions of the differential equation. Then find a particular solution of the form y=c₁y₁ + C₂Y₂ that satisfies the given initial conditions. Primes denote derivatives with respect to x y" + 2y'+y=0; y₁=ex, y₂ = xe*; y(0) = 10, y'(0) = -2 Why is the function y₁ = e a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice. A. The function y₁ = ex is a solution because when the function, its first derivative y₁= and its second derivative, y₁", are substituted into the equation, the result is a true statement. B. The function y₁=e* is a solution because when the function and its indefinite integral, are substituted into the equation, the result is a true statement. Why is the function y₂ = xe a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice. OA. The function y₂ = xex is a solution because when the function and its indefinite integral, are substituted into the equation, the result is a true statement. OB. The function y₂ = xex is a solution because when the function, its derivative, y,' =, and its second derivative, y," = The particular solution of the form y=C₁Y₁ + C₂y₂ that satisfies the initial conditions y(0) = 10 and y'(0) = -2 is y= are substituted into
A homogeneous second-order linear differential equation, two functions y, and y₂, and a pair of initial conditions are given. First verify that y, and y₂ are solutions of the differential equation. Then find a particular solution of the form y=c₁y₁ + C₂Y₂ that satisfies the given initial conditions. Primes denote derivatives with respect to x y" + 2y'+y=0; y₁=ex, y₂ = xe*; y(0) = 10, y'(0) = -2 Why is the function y₁ = e a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice. A. The function y₁ = ex is a solution because when the function, its first derivative y₁= and its second derivative, y₁", are substituted into the equation, the result is a true statement. B. The function y₁=e* is a solution because when the function and its indefinite integral, are substituted into the equation, the result is a true statement. Why is the function y₂ = xe a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice. OA. The function y₂ = xex is a solution because when the function and its indefinite integral, are substituted into the equation, the result is a true statement. OB. The function y₂ = xex is a solution because when the function, its derivative, y,' =, and its second derivative, y," = The particular solution of the form y=C₁Y₁ + C₂y₂ that satisfies the initial conditions y(0) = 10 and y'(0) = -2 is y= are substituted into
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
![A homogeneous second-order linear differential equation, two functions \( y_1 \) and \( y_2 \), and a pair of initial conditions are given. First verify that \( y_1 \) and \( y_2 \) are solutions of the differential equation. Then find a particular solution of the form \( y = c_1 y_1 + c_2 y_2 \) that satisfies the given initial conditions. Primes denote derivatives with respect to \( x \).
\[ y'' + 2y' + y = 0; \, y_1 = e^{-x}, \, y_2 = xe^{-x}; \, y(0) = 10, \, y'(0) = -2 \]
**Question 1:**
Why is the function \( y_1 = e^{-x} \) a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice.
- \( \circ \) A. The function \( y_1 = e^{-x} \) is a solution because when the function, its first derivative, \( y_1' = \) \[\boxed{-e^{-x}}\] and its second derivative, \( y_1'' = \) \[\boxed{e^{-x}}\], are substituted into the equation, the result is a true statement.
- \( \bullet \) B. The function \( y_1 = e^{-x} \) is a solution because when the function and its indefinite integral, \[\boxed{-e^{-x}}\], are substituted into the equation, the result is a true statement.
**Question 2:**
Why is the function \( y_2 = xe^{-x} \) a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice.
- \( \circ \) A. The function \( y_2 = xe^{-x} \) is a solution because when the function and its indefinite integral, \[\boxed{-(x+1)e^{-x}}\], are substituted into the equation, the result is a true statement.
- \( \bullet \) B. The function \( y_2 = xe^{-x} \) is a solution because when the function, its derivative, \( y_2' = \) \[\boxed{-xe^{-x}+e^{-x}}\], and its second derivative, \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F848bc833-9fb3-4eda-a7de-b585ddd0b1dc%2F52301abb-def9-4947-b768-c289e58b235e%2F87cg999_processed.png&w=3840&q=75)
Transcribed Image Text:A homogeneous second-order linear differential equation, two functions \( y_1 \) and \( y_2 \), and a pair of initial conditions are given. First verify that \( y_1 \) and \( y_2 \) are solutions of the differential equation. Then find a particular solution of the form \( y = c_1 y_1 + c_2 y_2 \) that satisfies the given initial conditions. Primes denote derivatives with respect to \( x \).
\[ y'' + 2y' + y = 0; \, y_1 = e^{-x}, \, y_2 = xe^{-x}; \, y(0) = 10, \, y'(0) = -2 \]
**Question 1:**
Why is the function \( y_1 = e^{-x} \) a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice.
- \( \circ \) A. The function \( y_1 = e^{-x} \) is a solution because when the function, its first derivative, \( y_1' = \) \[\boxed{-e^{-x}}\] and its second derivative, \( y_1'' = \) \[\boxed{e^{-x}}\], are substituted into the equation, the result is a true statement.
- \( \bullet \) B. The function \( y_1 = e^{-x} \) is a solution because when the function and its indefinite integral, \[\boxed{-e^{-x}}\], are substituted into the equation, the result is a true statement.
**Question 2:**
Why is the function \( y_2 = xe^{-x} \) a solution to the differential equation? Select the correct choice below and fill in the answer box to complete your choice.
- \( \circ \) A. The function \( y_2 = xe^{-x} \) is a solution because when the function and its indefinite integral, \[\boxed{-(x+1)e^{-x}}\], are substituted into the equation, the result is a true statement.
- \( \bullet \) B. The function \( y_2 = xe^{-x} \) is a solution because when the function, its derivative, \( y_2' = \) \[\boxed{-xe^{-x}+e^{-x}}\], and its second derivative, \(
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 4 steps with 3 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