Solve (D2-3D+2)y =x(x+4) and show that its general solution is given by y=Aex + Be2x +(x2/2) + (7x/2) + (19/4)

My work :

auxiliary eqn , m2 - 3m + 2= 0 m=1 , m=2

Reduced eqn : yn = Aex + Be2x

P.I. = 1/f(D)F(x) = (1/(D2-3D+2))(x(x+4))

P.I. = [(x2+4x) + ( (1/2)( 2- 3(2x) - 3(4) ) ) + (1/4)(92) )

P.I. = x2 + (4x/2) - (6x/4) - (5/2) + (18/4)

P.I. =(x2/2) + x + 8/4

General solution : y= Aex + Be2x + (x2/2) + x + 8/4

Could anyone point out my mistakes?

Answers
while solving for P.I, 1/f(D) becomes 1/f(D) = (1/2)*{1+(D^2 -3D)/2}^(-1)  , now expanding it as (1+ x)^(-1) = 1 - x + x^2 - ...this will give u answeryou expanded it in wrong way ..i also solved the answer but unable to share it with you, don't know why.
gopalsharma
25 December 2019
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