Answer
See below
Work Step by Step
We can notice that: $y^{vi}+3y^{iv}+3y''+y=2\sin x$
Obtain: $(D^6+3D^4+3D^2+1)y=2\sin x$
Solve the auxiliary equation for the differential equation.
$r^6+3r^4+3r^2+1=0\\
(r^2+1)^3=0$
Roots are: $r_1=-i$ as a multiplicity of 3 and $r_2=i$ as a multiplicity of 3.
Therefore, the general solution for the given differential equation is:
$y(x)=c_1\cos x+c_2x\cos x+c_3x^2\cos x+c_4\sin x+c_5x\sin x+c_6x^2\sin x$
We can see that $D^2+4$ is the annihilator of $G(x)=2\sin x$
Multiply $(D^2+4)$ to both sides of the given equation we get
$(D^2+4)(D^2+1)^3=0$
Solve the auxiliary equation for the differential equation.
$(r^2+4)^2(r^2+1)^3=0$
Roots are: $r_1=-i$ as a multiplicity of 3, $r_2=-i$ as a multiplicity of 3 and $r_3=-2i, r_4=2i$ as a multiplicity of 1.
The general solution to $(D^2+4)(D^2+1)^3=0$ is
$y(x)=c_1\cos x+c_2x\cos x+c_3x^2\cos x+c_4\sin x+c_5x \sin x+c_6x^2\sin x+c_7\cos 2x+c_8\sin 2x$
The trial solution to initial solution is $y_p=c_7\cos 2x+c_8\sin 2x$.
We obtain the same trial solution with the use of annihilators.
