Explanation Theorem used: The solution of the differential equation, f ( y ) y ′ = g ( x ) is given implicitly by the equation ∫ f ( y ) d y = ∫ g ( x ) d x . Calculation: The given differential equation can be rewritten as, y ′ ( y − 3 ) = 4 x 3 ( 2 + x 4 ) . Comparing the above equation with an equation ∫ f ( y ) d y = ∫ g ( x ) d x , it is observed that, f ( y ) = 1 y − 3 and g ( x ) = 4 x 3 ( 2 + x 4 ) . By the above theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( x ) d x ∫ d y y − 3 = ∫ 4 x 3 d x ( 2 + x 4 ) Thus, ∫ d y y − 3 = ∫ 4 x 3 d x ( 2 + x 4 ) (1) Evaluate the integrals separately as follows. ∫ d y y − 3 = ln | y − 3 | + C 1 Substitute u = 2 + x 4 and d u = 4 x 3 d x in ∫ 4 x 3 d x ( 2 + x 4 ) . ∫ 4 x 3 d x ( 2 + x 4 ) = ∫ d u u = ln | u | + C 2 = ln | 2 + x 4 | + C 2 Thus, the equation (1) becomes, ∫ d y y − 3 = ∫ 4 x 3 d x ( 2 + x 4 )

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

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Chapter 9.2, Problem 2MP

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To find: The general solution of the differential equation (2+x4)y′=4x3(y−3) and its particular solution that satisfies the initial condition y(0)=5.

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Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

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The general solution of the differential equation ( 2 + x 4 ) y ′ = 4 x 3 ( y − 3 ) and its particular solution that satisfies the initial condition y ( 0 ) = 5 .

Solution Summary: The author explains the general solution of the differential equation (2+x4)yprime and its particular solution that satisfies the initial condition.

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To find: The general solution of the differential equation (2+x4)y′=4x3(y−3) and its particular solution that satisfies the initial condition y(0)=5.

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Chapter 9 Solutions