Explanation Theorem used: Separation of Variables: 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 ′ = x ( x − 1 ) 1 2 ( y − 1 ) 1 2 1 ( y − 1 ) 1 2 y ′ = x ( x − 1 ) 1 2 ( y − 1 ) − 1 2 y ′ = x ( x − 1 ) 1 2 That is. ( y − 1 ) − 1 2 y ′ = x ( x − 1 ) 1 2 (1) The given differential equation ( y − 1 ) − 1 2 y ′ = x ( x − 1 ) 1 2 is in the form of f ( y ) y ′ = g ( x ) where, f ( y ) = ( y − 1 ) − 1 2 and g ( x ) = x ( x − 1 ) 1 2 . By the above mentioned theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( x ) d x ∫ ( y − 1 ) − 1 2 d y = ∫ ( x ( x − 1 ) 1 2 ) d x That is, ∫ ( y − 1 ) − 1 2 d y = ∫ ( x ( x − 1 ) 1 2 ) d x (2) Obtain the integral value of ∫ ( y − 1 ) − 1 2 d y . ∫ ( y − 1 ) − 1 2 d y = ( y − 1 ) − 1 2 + 1 − 1 2 + 1 = ( y − 1 ) 1 2 1 2 = 2 ( y − 1 ) 1 2 Thus, the value of the integral ∫ ( y − 1 ) − 1 2 d y is 2 ( y − 1 ) 1 2 . Use the method of substitution to obtain the integral ∫ ( x ( x − 1 ) 1 2 ) d x . Substitute u = x − 1 , x = u + 1 and d u = d x in ∫ ( x ( x − 1 ) 1 2 ) d x , ∫ ( x ( x − 1 ) 1 2 ) d x = ∫ ( u + 1 ) u 1 2 d u = ∫ ( u 3 2 + u 1 2 ) d u = u 3 2 + 1 3 2 + 1 + u 1 2 + 1 1 2 + 1 = u 5 2 5 2 + u 3 2 3 2 Simplify further as, ∫ ( x ( x − 1 ) 1 2 ) d x = 2 5 u 5 2 + 2 3 u 3 2 = 6 15 u 5 2 + 10 15 u 3 2 = 6 u 5 2 + 10 u 3 2 15 Thus, the value of the integral ∫ ( x ( x − 1 ) 1 2 ) d x is 6 u 5 2 + 10 u 3 2 15

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 50E

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To find: The particular solution for the differential equation y′=x(x−1)12(y−1)12 satisfying the initial condition y(1)=1.

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Chapter 9.2, Problem 49E

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

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The particular solution for the differential equation y ′ = x ( x − 1 ) 1 2 ( y − 1 ) 1 2 satisfying the initial condition y ( 1 ) = 1 .

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To find: The particular solution for the differential equation y′=x(x−1)12(y−1)12 satisfying the initial condition y(1)=1.

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