Explanation Given: The differential equation is y ( 1 + x 2 ) y ′ − x ( 1 + y 2 ) = 0 with the initial condition given as y ( 0 ) = 3 . Formula used: The differential equation is in the variable separable form, f ( y ) d y = g ( x ) d x where f ( y ) and g ( x ) are functions of y and x respectively. The integration formula is ∫ x n d x = x n + 1 n + 1 + C . Calculation: Consider the differential equation given as, y ( 1 + x 2 ) y ′ − x ( 1 + y 2 ) = 0 This implies, y ( 1 + x 2 ) y ′ = x ( 1 + y 2 ) y ′ = x ( 1 + y 2 ) y ( 1 + x 2 ) d y d x = x ( 1 + y 2 ) y ( 1 + x 2 ) y ( 1 + y 2 ) d y = x ( 1 + x 2 ) d x This differential equation is of the variable separable form, f ( y ) d y = g ( x ) d x Now, integrate both sides, ∫ y ( 1 + y 2 ) d y = ∫ x ( 1 + x 2 ) d x 1 2 ∫ 2 y ( 1 + y 2 ) d y = 1 2 ∫ 2 x ( 1 + x 2 ) d x Use ∫ 1 x d x = ln x + c Put 1 <

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ISBN: 9781285057095

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 6.3, Problem 19E

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To calculate: The particular solution of the differential equation given as, y(1+x2)y′−x(1+y2)=0 with the initial condition y(0)=3.

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Chapter 6.3, Problem 18E

Chapter 6.3, Problem 20E

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The particular solution of the differential equation given as, y ( 1 + x 2 ) y ′ − x ( 1 + y 2 ) = 0 with the initial condition y ( 0 ) = 3 .

Solution Summary: The author explains the differential equation in the variable separable form, y(1+x2) and the integration formula.

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To calculate: The particular solution of the differential equation given as, y(1+x2)y′−x(1+y2)=0 with the initial condition y(0)=3.

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