To determine The general solution of the given ODE. Explanation Given: The given ODE is y ′ ′ ′ + 3 y ″ + 3 y ′ + y = e x − x − 1 . Procedure used: Let y ( n ) + a n − 1 ( x ) y ( n − 1 ) + ⋯ + a 1 ( x ) y ′ + a 0 ( x ) y = r ( x ) be the n th order homogeneous ODE with constant coefficients, where y ( n ) = d n y d x n is the first term and r ( x ) ≠ 0 . A general solution of the above equation on an open interval I of the x -axis is of the form r ( x ) ≠ 0 Obtain the characteristic equation by substituting y = e λ x in the ODE as follows. y ( x ) = y h ( x ) + y p ( x ) Here y h ( x ) is a general solution of the corresponding homogeneous ODE. And y p ( x ) is any solution of the given equation on I containing no arbitrary constants. Calculation: Consider the given ODE, y ′ ′ ′ + 3 y ″ + 3 y ′ + y = e x − x − 1 Consider the corresponding homogeneous equation of the given ODE, y ′ ′ ′ + 3 y ″ + 3 y ′ + y = 0 Obtain the characteristic equation as follows. λ 3 + 3 λ 2 + 3 λ + 1 = 0 Compute the roots of the characteristic equation as follows. λ 3 + 3 λ 2 + 3 λ + 1 = 0 ( λ + 1 ) ( λ 2 + 2 λ + 1 ) = 0 ( λ + 1 ) ( λ + 1 ) 2 = 0 ( λ + 1 ) 3 = 0 λ 1 = λ 2 = λ 3 = − 1 The homogeneous solution of the given ODE is given as follows. y h = ( c 1 + c 2 x + c 3 x 2 ) e − x Let the particular solution is of the form y p = A e x + B x + C

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