$ \text{ The Auxiliary equation is } \\ $

$ D^3 - 2D + 4 = 0 \\ $

$D= -2, 1 \pm i\\ $

$ \text{ C.F. is } y = c_1 e^{-2x} + e^x[c_2cosx + c_3sinx] \\ $

$ \text{P.I = } \frac{1}{D^3 - 2D + 4} (3x^2 - 5x + 2) \\ $

$= \frac{1}{4} \frac{1}{[1 + \frac{D^3 - 2D}{4}]} (3x^2 - 5x + 2)\\ $

$= \frac{1}{4} [1 + \frac{D^3 - 2D}{4}]^{-1} (3x^2 - 5x + 2)\\ $

$= \frac{1}{4} [1 - \frac{(D^3 - 2D)}{4} + (\frac{D^3 - 2D}{4})^2] (3x^2 - 5x + 2)\\ $

$= \frac{1}{4} (1 - \frac{D^3}{4} + \frac{2D}{4} + \frac{4D^2}{16}) (3x^2 - 5x + 2)\\ $

$= \frac{1}{4} [3x^2 - 5x + 2 + \frac{1}{2} (6x-5) + \frac{1}{4} (6)]\\ $

$= \frac{1}{4} [3x^2 - 2x + 1] \\ $

$\therefore \text{ The complete solution is y = C.F. + P.I. } \\ $

$ \therefore y = c_1 e^{-2x} + e^x[c_2cosx + c_3sinx]+ \frac{1}{4} [3x^2 - 2x + 1] $