Explanation Given: The Cauchy-Euler equation x 3 y ‴ − 2 x 2 y ″ + 3 x y ′ − 3 y = x 2 , x > 0 and { x , l n x , x 3 } is a fundamental solution set for the corresponding homogeneous equation. Approach: a) Write the complimentary function of the given equation. b) Find out four Wronskian of determinants. c) Formula for finding the particular solution of the equation is y p ( x ) = ∑ k = 1 n y k ( x ) ∫ g ( x ) W K ( x ) W [ y 1 , ... , y n ] ( x ) d x . d) The general solution of the equation is y = y c + y p Calculation: The complimentary function of the homogeneous equation is, C F = c 1 x + c 2 x ln x + c 3 x 3 y c = c 1 x + c 2 x ln x + c 3 x 3 Compare with C F = c 1 y 1 ( x ) + c 2 y 2 ( x ) + c 3 y 3 ( x ) . Now y 1 ( x ) = x , y 2 ( x ) = x ln x and y 3 ( x ) = x 3 To find particular solution divide the given equation by x 3 . The given equation become, y ‴ x − 2 y ″ + 3 y ′ x 2 − 3 y x 3 = 1 x Now g ( x ) = 1 x Write four Wronskian of determinant, W [ y 1 , y 2 , y 3 ] ( x ) = | y 1 y 2 y 3 y ′ 1 y ′ 2 y ′ 3 y ″ 1 y ″ 2 y ″ 3 | W [ y 1 , y 2 , y 3 ] ( x ) = | x x ln x x 3 1 1 + ln x 3 x 2 0 1 x 6 x | This implies, W [ y 1 , y 2 , y 3 ] ( x ) = x ( 6 x + 6 ln x − 3 x ) − x ln x ( 6 x − 0 ) + x 3 ( 1 x − 0 ) = 3 x 2 + 6 x 2 ln x − 6 x 2 ln x + x 2 = 4 x 2 Find the value of W 1 , W 2 and W 3 , W 1 ( x ) = ( − 1 ) 3 − 1 W [ y 2 , y 3 , ] ( x ) = ( − 1 ) 2 | x ln x x 3 1 + x ln x 3 x 2 | W 1 ( x ) = 3 x 3 ln x − x 3 − x 3 ln x W 1 ( x ) = 2 x 3 ln x − x 3 Similarly for W 2 follow same procedure. W 2 ( x ) = ( − 1 ) 3 − 2 | x x ln x 1 1 + ln x | = x + x ln x − x ln x = x , Similarly W 3 is written as,

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ISBN: 9780321977175

Author: Nagle

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The general solution to the Cauchy-Euler equation.

Solution Summary: The author explains how the Cauchy-Euler equation is a fundamental solution set for the corresponding homogeneous equation.

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