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Theory and Applications(Continuation of Exercise 3.)

Thomas' Calculus: Early Transcendentals | 13th Edition | ISBN: 9780321884077 | Authors: George B. Thomas Jr., Maurice D. Weir, Joel R. Hass ISBN: 9780321884077 57

Solution for problem 4AAE Chapter 9

Thomas' Calculus: Early Transcendentals | 13th Edition

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Thomas' Calculus: Early Transcendentals | 13th Edition | ISBN: 9780321884077 | Authors: George B. Thomas Jr., Maurice D. Weir, Joel R. Hass

Thomas' Calculus: Early Transcendentals | 13th Edition

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Problem 4AAE

Theory and Applications

(Continuation of Exercise 3.) Assume the hypotheses of Exercise 3, and assume that y1(x) and y2(x) are both solutions to the first order linear equation satisfying the initial condition y(x0) = y0 .

a. Verify that y(x) = y1(x) - y2(x) satisfies the initial value problem

b. For the integrating factor show that

Conclude that v(x)[ y1(x) - y2(x)] ≡ constant

c. From part (a), we have y1(x0) - y2(x0) = 0. Since v(x) > 0 for a < x < b use part (b) to establish that y1(x) - y2(x) ≡ 0 on the interval (a, b). Thus y1(x) = y2(x) for all a < x < b

Reference: Exercise 3

a. Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Fundamental Theorem of Calculus, Part 1 to show that any function y satisfying the equation.

for  is a solution to the first-order linear equation

b. If then show that any solution y in part (a) satisfies the initial condition y(x0) = y0

Step-by-Step Solution:
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Chapter 9, Problem 4AAE is Solved
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Textbook: Thomas' Calculus: Early Transcendentals
Edition: 13
Author: George B. Thomas Jr., Maurice D. Weir, Joel R. Hass
ISBN: 9780321884077

Thomas' Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321884077. This full solution covers the following key subjects: part, initial, equation, assume, show. This expansive textbook survival guide covers 138 chapters, and 9198 solutions. The answer to “Theory and Applications(Continuation of Exercise 3.) Assume the hypotheses of Exercise 3, and assume that y1(x) and y2(x) are both solutions to the first order linear equation satisfying the initial condition y(x0) = y0 .a. Verify that y(x) = y1(x) - y2(x) satisfies the initial value problem b. For the integrating factor show thatConclude that v(x)[ y1(x) - y2(x)] ? constant c. From part (a), we have y1(x0) - y2(x0) = 0. Since v(x) > 0 for a < x < b use part (b) to establish that y1(x) - y2(x) ? 0 on the interval (a, b). Thus y1(x) = y2(x) for all a < x < bReference: Exercise 3a. Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Fundamental Theorem of Calculus, Part 1 to show that any function y satisfying the equation. for is a solution to the first-order linear equation b. If then show that any solution y in part (a) satisfies the initial condition y(x0) = y0” is broken down into a number of easy to follow steps, and 167 words. Since the solution to 4AAE from 9 chapter was answered, more than 227 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 4AAE from chapter: 9 was answered by , our top Calculus solution expert on 08/01/17, 02:37PM. This textbook survival guide was created for the textbook: Thomas' Calculus: Early Transcendentals , edition: 13.

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