Explanation Given: { e x , e − x , e 2 x } is a fundamental solution set for the homogeneous equation corresponding to the equation y ‴ − 2 y ″ − y ′ + 2 y = g ( x ) . Approach: a) First write down the complimentary function of the given equation. b) Then find out four Wronskian of determinants. c) Formula for 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 Calculation: To determine the particular solution of the equation first write down the complimentary function by considering fundamental set of solution, C F = c 1 e x + c 2 e − x + c 3 x e 2 x y c = c 1 e x + c 2 e − x + c 3 x e 2 x Compare with C F = c 1 y 1 ( x ) + c 2 y 2 ( x ) + c 3 y 3 ( x ) . Now y 1 ( x ) = e x , y 2 ( x ) = e − x and y 3 ( x ) = x e 2 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 | = | e x e − x e 2 x e x − e − x 2 e 2 x e x e − x 4 e 2 x | Now solve and apply operation R 3 = R 3 − R 1 , W [ y 1 , y 2 , y 3 ] ( x ) = | e x e − x e 2 x e x − e − x 2 e 2 x 0 0 3 e 2 x | = 3 e 2 x ( − e x e − x − e − x e x ) W [ y 1 , y 2 , y 3 ] ( x ) = − 6 e 2 x Now 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 | e x x 2 e x ( x + 1 ) e x ( x 2 + 2 x ) e x | Write common factor from each of the column. W 1 ( x ) = ( − 1 ) 3 − 1 W [ y 2 , y 3 , ] ( x ) = | e − x e 2 x − e − x 2 e 2 x | = ( 2 e x + e x ) W 1 ( x ) = 3 e x Similarly for W 2 follow the same procedure.

Pearson eText Fundamentals of Differential Equations with Boundary Value Problems -- Instant Access (Pearson+)

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