1 36-x² Let y'= 2xy²; y= Then by considering y=(x) as a 11 solution of the differential equation, give at least one interval I of definition. " a. (-6,∞0), (-6,6), [6,∞) b. (-6,6), (-∞, -6] c. (-6,6) d. (-6,∞0), (6,∞0), e. (-∞, -6) y(5)=
1 36-x² Let y'= 2xy²; y= Then by considering y=(x) as a 11 solution of the differential equation, give at least one interval I of definition. " a. (-6,∞0), (-6,6), [6,∞) b. (-6,6), (-∞, -6] c. (-6,6) d. (-6,∞0), (6,∞0), e. (-∞, -6) y(5)=
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Given the differential equation \( y' = 2xy^2 \) and the function \( y = \frac{1}{36 - x^2} \) with the condition \( y(5) = \frac{1}{11} \), determine at least one interval \( I \) of definition for the solution \( y = \phi(x) \) of the differential equation.
**Possible Intervals:**
a. \((-6, \infty), (-6, 6), [6, \infty)\)
b. \((-6, 6), (-\infty, -6] \)
c. \((-6, 6)\)
d. \((-6, \infty), (6, \infty)\)
e. \((- \infty, -6)\)
**Explanation:**
Examine and determine appropriate intervals where the given solution function, \( y = \frac{1}{36 - x^2} \), is defined and continuous. This function has singularities where the denominator becomes zero, i.e., at \( x = \pm 6 \). Therefore, the intervals must avoid these points to ensure the function remains valid and meaningful.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92c05ba6-e775-4ea8-9e55-a6761b98c61f%2F4ad21f46-68a3-4dd0-b0ee-ff4b92d3fc45%2Fd8dnv9_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Given the differential equation \( y' = 2xy^2 \) and the function \( y = \frac{1}{36 - x^2} \) with the condition \( y(5) = \frac{1}{11} \), determine at least one interval \( I \) of definition for the solution \( y = \phi(x) \) of the differential equation.
**Possible Intervals:**
a. \((-6, \infty), (-6, 6), [6, \infty)\)
b. \((-6, 6), (-\infty, -6] \)
c. \((-6, 6)\)
d. \((-6, \infty), (6, \infty)\)
e. \((- \infty, -6)\)
**Explanation:**
Examine and determine appropriate intervals where the given solution function, \( y = \frac{1}{36 - x^2} \), is defined and continuous. This function has singularities where the denominator becomes zero, i.e., at \( x = \pm 6 \). Therefore, the intervals must avoid these points to ensure the function remains valid and meaningful.
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