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

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

$ (D-2)^2 = 0 \\ $

$D= 2,2 \\ $

$ \text{ C.F. is } y = (c_1 + c_2x) e^{2x}\\ $

$ \text{P.I = } y_p = \frac{1}{ D^2 - 4D + 4 } 8(x^2 + sin2x + e^{2x} ) \\ $

$= 8[ \frac{1}{D^2 - 4D + 4} x^2 + \frac{1}{D^2 - 4D + 4} sin2x + \frac{1}{D^2 - 4D + 4}e^{2x} ]$

$ = 8 [ y_{p_1} + y_{p_2} + y_{p_3} ] \\ $

$ y_{p_1} = \frac{1}{D^2 - 4D + 4} x^2 \\ $

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

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

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

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

$= \frac{1}{4} (x^2 - \frac{2}{4} + 2x+ 2 ) \\ $

$= \frac{1}{4} (x^2 + 2x + \frac{3}{2} ) \\ $

$ y_{p_2} = \frac{1}{D^2 - 4D + 4} sin2x\\ $

$= \frac{1}{-4 - 4D + 4} sin2x \\ $

$= \frac{1}{- 4D} sin2x \\ $

$= \frac{1}{-4} \int sin2x dx \\ $

$= \frac{1}{-4} ( \frac{- cos2x}{2} ) \\ $

$ = \frac{cos2x}{8} \\ $

$ y_{p_3} = \frac{1}{D^2 - 4D + 4} e^{2x} \\ $

$= \frac{x}{2D - 4} e^{2x} \\ $

$= \frac{x^2}{2} e^{2x} \\ $

$\therefore \text{ P.I. is }y_p = 8 [ \frac{1}{4} (x^2 + 2x + \frac{3}{2}) +\frac{cos2x}{8}+\frac{x^2}{2} e^{2x} ]$

$ \therefore y = y_c + y_p $

$ = (c_1 + c_2 x)e^{2x} + (2x^2 + 4x + 3 ) +cos2x + {4x^2} e^{2x} $