$ \text{ Put x-1 =} e^z ,z=log(x-1)\\ $

$ D(D-1)(D-2)y + 2D(D-1)y -4Dy + 4y = 4z \\ $

$ [(D^2-D)(D-2) + 2D^2- 2D -4D + 4]y = 4z \\ $

$ [D^3-2D^2-D^2 + 2D + 2D^2- 2D -4D + 4]y = 4z \\ $

$ [D^3-D^2 -4D + 4]y = 4z \\ $

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

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

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

$ \text{ C.F. is } y_c = c_1e^{-2z} + c_2e^{2z} + c_3 e^z \\ $

$ \text{P.I = } \frac{1}{D^3-D^2 -4D + 4}4z \\ $

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

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

$ = z -1 \\ $

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

$ \therefore y = c_1e^{2z} + c_2e^{-2z} + c_3 e^z + z -1 \\ $

$ = c_1(x-1)^{-2} + c_2(x-1)^{2z} + c_3 (x-1) + log(x-1) -1 \\ $