Question

The particular integral of (x2D2-xD+4)Y=COS(LOGX)is

Answer 1

$(x^2D^2-xD+4)y=cos(logx)\\ \text{By euler equation.}\\ (D'(D'-1)-D'+4)y=cosz\\ (D'^2-2D'+4)y=cosz\\ \text{homogeneous differential equation is given by}\\ (D'^2-2D'+4)y=0\\ \text{Auxiliary equation,}\\ m^2-m+4=0\\ m=1\pm\sqrt3i\\ \text{Therefore, general solution of homogeneous differential equationis}\\y=e^z(c_1cos(\sqrt3z)+c_2sin(\sqrt3z))\\ y=x(c_1cos(\sqrt3logx)+c_2sin(\sqrt3logx))\\ \text{Now,finding the particular solution }\\ y_p=\frac{1}{(D'^2-2D'+4)}cosz\\ =\frac{1}{(-(1)^2-2D'+4)}cosz\\ =\frac{1}{(3-2D')}cosz\\ =\frac{3+2D'}{(9-4D'^2)}cosz\space (\text{Rationalise})\\ =\frac{3+2D'}{(9-4(-(1)^2)}cosz\\ =\frac{3+2D'}{13}cosz\\ =\frac{1}{13}(3cosz-2sinz)\\ =\frac{1}{13}(3cos(logx)-2sin(logx))\\ \text{Therefore, general solution of the given differential equation is}\\ y=x(c_1cos(\sqrt3logx)+c_2sin(\sqrt3logx))+\frac{1}{13}(3cos(logx)-2sin(logx))$