Explanation Formula used: The formula for integrating factor is I ( x ) = e ∫ P ( x ) d x . Calculation: Obtain the solution of the differential equation x y ′ + 6 x 2 y = 10 − x 3 . Divide the differential equation x y ′ + 6 x 2 y = 10 − x 3 by x . x y ′ + 6 x 2 y x = 10 − x 3 x y ′ + ( 6 x ) y = 10 x − x 2 x Thus, the redefined differential equation is y ′ + ( 6 x ) y = 10 x − x 2 x . This is a first order linear differential equation of the form y ′ + P ( x ) y = Q ( x ) . Comparing the coefficients of y and constant terms with y ′ + P ( x ) y = Q ( x ) . Thus, P ( x ) = 6 x and Q ( x ) = 10 x − x 2 x

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

8th Edition

ISBN: 9781305279148

Author: Stewart, James, St. Andre, Richard

Publisher: Cengage Learning

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Differential Equations

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Chapter 9.5, Problem 2PT

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To choose: The appropriate option for the integer factor of xy′+6x2y=10−x3.

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Chapter 9 Solutions

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

Ch. 9.1 - Prob. 1PT Ch. 9.1 - Prob. 2PT Ch. 9.1 - Prob. 3PT Ch. 9.1 - Prob. 4PT Ch. 9.2 - Prob. 1PT Ch. 9.2 - Prob. 2PT Ch. 9.2 - Prob. 3PT Ch. 9.3 - Prob. 1PT Ch. 9.3 - True or False: y′ + xey = ex+y is separable. Ch. 9.3 - Prob. 3PT

Ch. 9.3 - Prob. 4PT Ch. 9.4 - Prob. 1PT Ch. 9.4 - Prob. 2PT Ch. 9.4 - Prob. 3PT Ch. 9.4 - Prob. 4PT Ch. 9.5 - Prob. 1PT Ch. 9.5 - Prob. 2PT Ch. 9.5 - Prob. 3PT Ch. 9.6 - Prob. 1PT Ch. 9.6 - Prob. 2PT Ch. 9.6 - Prob. 3PT

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aIf HP=4, PJ=5, and PM=2, find LP. _ bIf HP=x+1, PJ=x1, LP=8, and PM=3, find x. _

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To choose: The appropriate option for the integer factor of x y ′ + 6 x 2 y = 10 − x 3 .

Solution Summary: The author explains the formula for integrating factor, which is underset_a) e3x2.

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To choose: The appropriate option for the integer factor of xy′+6x2y=10−x3.

Chapter 9 Solutions