D-operator method

Question

Evaluate a particular value of $$\frac{1}{D^2+4D}\sin 2x$$ I know the complementary function(C.F.) & particular integral(P.I.)....but I can't understand what is particular value, please explain

Answer 1

$$y=\frac{\sin 2x}{D^2+4D}=\frac{\sin 2x}{-2^2+4D}=\frac{\sin 2x}{4(D-1)}=\frac{(D+1)\sin 2x}{4(D^2-1)}=\frac{2 \cos 2x+\sin 2x}{4(-2^2-1)}$$ $$\implies y=-\frac{1}{20}(\sin 2x+2 \cos 2x)$$

One can check that $$(D^2+4D)y=\sin 2x$$