Explanation Formula used: The solution of a linear differential equation of first order and is of the form d y + P y d x = Q d x , where P and Q are functions of x only is given by y e ∫ P d x = ∫ Q e ∫ P d x + c . Calculation: The differential equation is y ' = 4 x − 2 y . Rewrite the given differential equation as follows. d y d x = 4 x − 2 y d y d x + 2 y = 4 x d y + 2 y d x = 4 x d x Note that the differential equation has the form d y + P y d x = Q d x where P = 2 y and Q = 4 x . Now, find the integrating factor e ∫ P d x as follows. e ∫ P d x = e ∫ 2 d x = e 2 x Then by the above mentioned formula the solution can be calculated as follows. y e ∫ P d x = ∫ Q e ∫ P d x d x + c ︸ ⇓ y e 2 x = ∫ 4 x e 2 x d x + c Now, integrate on both sides as follows

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 31, Problem 43RE

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The indicated particular solution of the differential equation y'=4x−2y; x=0 when y=−2.

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Chapter 31, Problem 42RE

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Chapter 31 Solutions

Basic Technical Mathematics

Ch. 31.1 - Show that is a solution of . Is it the general...Ch. 31.1 - Prob. 1E Ch. 31.1 - In Exercises 1 and 2, show that the indicated...Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 4E Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 6E Ch. 31.1 - In Exercises 7–10, show that each function y =...Ch. 31.1 - Prob. 8E Ch. 31.1 - In Exercises 7–10, show that each function y =...

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The indicated particular solution of the differential equation y ' = 4 x − 2 y ; x = 0 when y = − 2 .

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The indicated particular solution of the differential equation y'=4x−2y; x=0 when y=−2.

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