Explanation Given: d y d x + 2 x y 1 + x 2 = sin x , with y = 1 when x = 0 Concept used: Solution to linear differential equation of the form, d y d x + P y = Q , where P are Q are constants or functions of x only is, y ⋅ ( IF ) = ∫ Q × IF d x + C , where IF = e ∫ P d x & ∫ f ( x ) g ( x ) d x = f ( x ) ∫ g ( x ) d x − ∫ [ f ′ ( x ) ⋅ ∫ g ( x ) d x ] d x + C Calculation: Comparing the differential equation with d y d x + P y = Q , we get P = 2 x 1 + x 2 and Q = sin x Therefore integrating factor is, IF = e ∫ P d x = e ∫ 2 x 1 + x 2 d x = e ln ( 1 + x 2 ) [ ∵ ∫ f ′ ( x ) f ( x ) d x = ln [ f ( x ) ] + C ] = 1 + x 2 Hence the solution is, y ⋅ ( IF ) = ∫ Q × IF d x + C ⇒ y ⋅ ( 1 + x 2 ) = ∫ ( ( 1 + x 2 ) ⋅ sin x ) d x + C ⇒ ( 1 + x 2 ) y

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J

Publisher: Kendall Hunt Publishing

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Differential Equations

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Chapter 7.6, Problem 23PS

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To find the particular solution of the differential equation dydx+2xy1+x2=sinx

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Chapter 7.6, Problem 22PS

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To find the particular solution of the differential equation d y d x + 2 x y 1 + x 2 = sin x

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To find the particular solution of the differential equation dydx+2xy1+x2=sinx

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