Finding particular integral to $ (D^3 + 4D)y = \sin2x$ The question is to find particular integral of this differential equation 

$$ (D^3 + 4D)y = \sin2x.$$

where $$ D = \frac{d}{dx} $$
Please looking for your quick tip guys.
the first step is finding the complementary solution 
$$D(D^2+4)=0$$
$$D=0$$
$$D=\pm2i$$
so the complementary solution is
$$y_c=c_1+c_2\cos2x+c_3\sin 2x$$
now assume the particular solution as follow
$$y_p=A\cos 2x+B\sin 2x$$
because the similarity with complementary solution, we should multiply the particular solution by $x$, so
$$y_p=x(A\cos 2x+B\sin 2x)$$

then you can complete the solution
Given:- (D³+4D)y=sin2x
Condition : 1/f(D), if after equal to sinax or cosax given means you have to replace D² into -(a²)
Ans-  (1/D³+4D)*sin2x
[1/D(D²+4)]*sin2x    Here, D² is replaced by -(a²) that is
D²= -(2²)
[1/D(-4+4)]*sin2x  (important first square the number and put (-) symbol.)
[1/D(0)]*sin2x [1/0]*sinx ( condition if denomerator  is zero multiply  numerator by x and differentiate f(D)).
[x/3D²+4]*sin2x ( substitute D² value)
[x/3(-4)+4]*sin2x [x/(-12+4)]*sin2x
Therefore [-x/8]*sin2x.