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

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

$D = 1,2 \\ $

$ \text{ C.F. is } y_c = c_1 e^x + c_2 e^{2x} \\ $

$ \text{P.I = } \frac{1}{D^2 - 3D +2}2e^x cos\frac{x}{2} \\ $

$ = 2e^x \frac{1}{(D+1)^2 - 3(D+1) +2} cos\frac{x}{2} \\ $

$ = 2e^x \frac{1}{D^2 -D} cos\frac{x}{2} \\ $

$ = 2e^x \frac{1}{\frac{-1}{4} -D} cos\frac{x}{2} \\ $

$ = 2e^x \frac{4}{-1-4D} cos\frac{x}{2} \\ $

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

$ = -8e^x \frac{4D-1}{(4D + 1) (4D-1)} cos\frac{x}{2} \\ $

$ = -8e^x \frac{4D-1}{16D^2 -1} cos\frac{x}{2} \\ $

$ = -8e^x \frac{4D-1}{16(\frac{-1}{4}) -1} cos\frac{x}{2} \\ $

$ = -8e^x \frac{4D-1}{-5} cos\frac{x}{2} \\ $

$ = \frac{8}{5} e^x [4( - sin\frac{x}{2}) \frac{1}{2} - cos\frac{x}{2} ] \\ $

$ = \frac{8}{5} e^x [ -2 sin\frac{x}{2} - cos\frac{x}{2} ] \\ $

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

$ \therefore y =c_1 e^x + c_2 e^{2x} - \frac{8}{5} e^x [ 2 sin\frac{x}{2} + cos\frac{x}{2} ] \\ $