#(D^2+4)y=cos2x+cos4x#
Characteristic equation of differential equation #r^2+4=0#
Roots of it are #r_1=-2i# and #r_2=2i#
Hence homogeneous part of it #y_h=c_1*sin2x+c_2*cos2x#
Due to #cos2x# is root of homogeneous part of it, particular solution of it is #y_p=Axsin2x+Bxcos2x+Csin4x+Dcos4x#
Consequently,
#D^2(Axsin2x+Bxcos2x+Csin4x+Dcos4x)+4(Axsin2x+Bxcos2x+Csin4x+Dcos4x)=cos2x+cos4x#
#4Acos2x-4Axsin2x-4Bsin2x-4Bxcos2x-16Csin4x-16Dcos4x+4Axsin2x+4Bxcos2x+4Csin4x+4Dcos4x=cos2x+cos4x#
#4Acos2x-4Bsin2x-12Csin4x-12Dcos4x=cos2x+cos4x#
After equating coefficients,
#4A=1#, #-4B=0#, #-12C=0# and #-12D=1#
So, #B=C=0#, #A=1/4# and #D=-1/12#
Consequently, #y_p=1/4xsin2x-1/12cos4x#
Thus, #y=y_h+y_p=c_1*sin2x+c_2*cos2x+1/4xsin2x-1/12cos4x#
