Well, I'm not sure what to say. You say "using superposition, already found Yc, so please don't thread derail" but it is precisely yh that is important! You got yh= C1cos(x)+ C2sin(x), didn't you? So your "Bcos(x)" and "Dsin(x)" gives you nothing new. You also say "When i put Yp in the equation (y''+y= 2x sin(x)), it worked fine as in i didn't get 0=2*x*sin(x)" and I don't see how that is at all relevant.
If y= (Ax+ B)cos(x)+ (Cx+ D)sin(x), then y'= Acos(x)- (Ax+ B)sin(x)+ Csin(x)+ (Cx+ D)cos(x)= (Cx+A+D)cos(x)- (Ax+B- C)sin(x) and y''= Ccos(x)- (Cx+A+D)sin(x)- Asin(x)- (Ax+B-C)cos(x)=
-(Ax+ B- 2C)cos(x)- (Cx+ 2A+ D)sin(x) adding y= (Ax+B)cos(x)+ (Cx+D)sin(x) cancels those terms giving 2C cos(x)+ 2A sin(x) which has no "x" multiplying the sine or cosine and so cannot be equal to 2x sin(x) for any choice of A and B.
And, of course, those terms cancelled out precisely because cos(x) and sin(x) satified the associated homogenous equation.