To determine To solve: The differential equation x d y d x + 2 y = 1 − 1 x , x > 0 . Explanation Procedure used: To solve the linear equation y ′ + P ( x ) y = Q ( x ) , multiply both sides by the integrating factor v ( x ) = e ∫ P ( x ) d x and integrate on both sides. The equation can also be solved directly by using the formula y = 1 v ( x ) ∫ v ( x ) Q ( x ) d x . Calculation: Rewrite the differential equation x d y d x + 2 y = 1 − 1 x , x > 0 in the standard form as follows. x d y d x + 2 y = 1 − 1 x d y d x + ( 2 x ) y = 1 x − 1 x 2 Compare the above equation with the standard form d y d x + P ( x ) y = Q ( x ) . Here, P ( x ) = 2 x and Q ( x ) = 1 x − 1 x 2 . Evaluate the value of ∫ P ( x ) d x as follows. ∫ P ( x ) d x = ∫ 2 x d x = 2 ln | x | = 2 ln x ( ∵ x > 0 ) = ln x 2 Evaluate the integrating factor as follows

University Calculus: Early Transcendentals, Loose-leaf Edition (4th Edition)

4th Edition

ISBN: 9780135164860

Author: Joel R. Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 16.2, Problem 5E

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To solve: The differential equation xdydx+2y=1−1x,x>0.

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Chapter 16.2, Problem 4E

Chapter 16.2, Problem 6E

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Chapter 16 Solutions

University Calculus: Early Transcendentals, Loose-leaf Edition (4th Edition)

Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 5 and 6, copy the slope fields, and...Ch. 16.1 - In Exercises 5 and 6, copy the slope fields, and...Ch. 16.1 - Prob. 7E Ch. 16.1 - Prob. 8E Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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The differential equation x d y d x + 2 y = 1 − 1 x , x &gt; 0 .

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To solve: The differential equation xdydx+2y=1−1x,x>0.

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Chapter 16 Solutions

The differential equation x d y d x + 2 y = 1 − 1 x , x > 0 .