$ \text { Put x =} e^z, \text {z= logx} \\ $

$ D(D-1)y -Dy + 4y = cos z + e^z sinz \\ $

$ (D^2 - 2D + 4) y = cos z + e^z sinz \\ $

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

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

$ D = 1\pm i \sqrt3 \\ $

$ \text{ C.F. is } y = e^z(c_1cos \sqrt3z + c_2sin \sqrt3z) \\ $

$ \text{P.I = } \frac{1}{D^2 - 2D + 4}cosz + \frac{1}{D^2 - 2D + 4}e^zsinz \\ $

$ =\frac{1}{-1 - 2D + 4}cosz + e^z\frac{1}{(D+1)^2 - 2(D+1) + 4}sinz \\ $

$ =\frac{1}{3 - 2D}cosz + e^z\frac{1}{D^2 + 2D + 1 -2D-2 + 4}sinz \\ $

$ =\frac{3+2D}{9 - 4D^2}cosz + e^z\frac{1}{D^2 +3}sinz \\ $

$ =\frac{3+2D}{13}cosz + e^z\frac{1}{-1 +3}sinz \\ $

$ =\frac{1}{13}[3cosz - 2sinz) + \frac{1}{2}e^zsinz \\ $

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

$ \therefore y = e^z(c_1cos \sqrt3z + c_2sin \sqrt3z) + \frac{1}{13}[3cosz - 2sinz) + \frac{1}{2}e^zsinz \\ $

$ \therefore y = x(c_1cos \sqrt3logx + c_2sin \sqrt3logx) + \frac{1}{13}[3coslogx - 2sinlogx) + \frac{1}{2}xsinlogx \\ $