Explanation Given: The Cauchy-Euler equation, x 3 y ‴ − 3 x y ′ + 3 y = x 4 cos x ... ( 1 ) x > 0 . Approach: To determine the general solution of the equation first find the complimentary function of the given equation. Then find out four Wronskian of determinants. Use the formula to find the particular solution of the equation. y p ( x ) = ∑ k = 1 n y k ( x ) ∫ g ( x ) W K ( x ) W [ y 1 , ... , y n ] ( x ) d x The general solution of the equation is, y = y c + y p . The general equation of complementary function is, C F = c 1 y 1 ( x ) + c 2 y 2 ( x ) + c 3 y 3 ( x ) ... ( 2 ) Calculation: Consider, y = x r y ′ = r ( x r − 1 ) y ″ = r ( r − 1 ) x r − 2 y ‴ = r ( r − 1 ) ( r − 2 ) x r − 3 Substitute the above value in equation ( 1 ) . x 3 ( r ( r − 1 ) ( r − 2 ) ) x r − 3 − 3 x ( r x r − 1 ) + 3 x r = 0 ( r ( r − 1 ) ( r − 2 ) ) x r − 3 r x r + 3 x r = 0 [ ( r ( r − 1 ) ( r − 2 ) ) − 3 r + 3 ] x r = 0 ( r ( r − 1 ) ( r − 2 ) ) − 3 r + 3 = 0 ( r 2 − r ) ( r − 2 ) − 3 r + 3 = 0 r 2 ( r − 2 ) − r ( r − 2 ) − 3 r + 3 = 0 r 3 − 2 r 2 − r 2 + 2 r − 3 r + 3 = 0 r 3 − 3 r 2 − r + 3 = 0 ( r − 1 ) ( r 2 − 2 r − 3 ) = 0 ( r − 1 ) ( r − 3 ) ( r + 1 ) = 0 r = 1 , − 1 , 3 So, the fundamental set of the homogeneous equation is, { x r ( r = 1 , − 1 , 3 ) } = { x , x − 1 , x 3 } . The complimentary function is, C F = c 1 x + c 2 x − 1 + c 3 x 3 . Compare the above Equation with Equation ( 2 ) . y 1 ( x ) = x y 2 ( x ) = x − 1 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 − 1 x 3 1 − x − 2 3 x 2 0 2 x − 3 6 x | = x ( − 6 x − 1 − 6 x − 1 ) − x − 1 ( 6 x − 0 ) + x 3 ( 2 x − 3 − 0 ) = − 12 − 6 + 2 And, W [ y 1 , y 2 , y 3 ] ( x ) = − 16 Multiply the equation ( 1 ) by 1 x 3 . ( x 3 y ‴ − 3 x y ′ + 3 y ) 1 x 3 = ( x 4 cos x ) 1 x 3 y ‴ − 3 x 2 y ′ + 3 x 3 y = x cos x Consider, g ( x ) = x cos x . Now, the value of W 1 , W 2 and W 3 are, W 1 ( x ) = ( − 1 ) 3 − 1 W [ y 2 , y 3 , ] ( x ) = ( − 1 ) 2 | x − 1 x 3 − x − 2 3 x 2 | = ( 1 ) [ ( x − 1 ) ( 3 x 2 ) − ( − x − 2 ) ( x 3 ) ] = 3 x + x And, W 1 ( x ) = 4 x Similarly, W 2 follow same procedure. W 2 ( x ) = ( − 1 ) 3 − 2 | x x 3 1 3 x 2 | W 2 ( x ) = − ( 3 x 3 − x 3 ) W 2 ( x ) = − 2 x 3 Similarly W 3 is written as, W 3 ( x ) = ( − 1 ) 3 − 3 | x x − 1 1 − x − 2 | = x − 1 − x − 1 = 2 x − 1 Now particular solution is written as with the help of formula. y p ( x ) = ∑ k = 1 n y k ( x ) ⋅ v k ( x ) , k = 1 , ... , n y p ( x ) = y 1 ( x ) ⋅ v 1 ( x ) + y 2 ( x ) ⋅ v 2 ( x ) + y 3 ( x ) ⋅ v 3 ( x )

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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The general solution to the given Cauchy-Euler equation x3y‴−3xy′+3y=x4cosx, x>0.

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The general solution to the given Cauchy-Euler equation x 3 y ‴ − 3 x y ′ + 3 y = x 4 cos x , x > 0 .

Solution Summary: The author explains how to determine the general solution of the Cauchy-Euler equation.

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The general solution to the given Cauchy-Euler equation x3y‴−3xy′+3y=x4cosx, x>0.

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The general solution to the given Cauchy-Euler equation x 3 y ‴ − 3 x y ′ + 3 y = x 4 cos x , x &gt; 0 .

Solution Summary: The author explains how to determine the general solution of the Cauchy-Euler equation.

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The general solution to the given Cauchy-Euler equation x3y‴−3xy′+3y=x4cosx, x>0.

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The general solution to the given Cauchy-Euler equation x 3 y ‴ − 3 x y ′ + 3 y = x 4 cos x , x > 0 .