Explanation Given: The Cauchy-Euler equation is x 3 y ‴ − 3 x 2 y ″ + 6 x y ′ − 6 y = x − 1 , x > 0 and { x , x 2 , x 3 } is a fundamental solution set for the corresponding homogeneous equation. Approach: a) First write down 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 is y = y c + y p . Calculation: The complimentary function of the homogeneous equation is- C F = c 1 x + c 2 x 2 + c 3 x 3 y c = c 1 x + c 2 x 2 + c 3 x 3 Compare with C F = c 1 y 1 ( x ) + c 2 y 2 ( x ) + c 3 y 3 ( x ) . Then, y 1 ( x ) = x , y 2 ( x ) = x 2 and y 3 ( x ) = x 3 . 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 | = | x x 2 x 3 1 2 x 3 x 2 0 2 6 x | After solve it becomes, W [ y 1 , y 2 , y 3 ] ( x ) = x ( 12 x 2 − 6 x 2 ) − x 2 ( 6 x ) + x 3 ( 2 ) = 12 x 3 − 6 x 3 − 6 x 3 + 2 x 3 = 2 x 3 . 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 2 x 3 2 x 3 x 2 | , W 1 ( x ) = x 4 Similarly for W 2 follow same procedure, W 2 ( x ) = ( − 1 ) 3 − 2 | x x 3 1 3 x 2 | = − 2 x 3 . Similarly W 3 is written as, W 3 ( x ) = ( − 1 ) 3 − 3 | x x 2 1 2 x | = x 2

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ISBN: 9780135902738

Author: Nagle

Publisher: PEARSON

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