# annihilator method confusion

I have a final in the morning and I am extremely confused on the annihilator method. I have been googling different explanations all night and I just dont get it at all. I am looking at an example: $$\ddot{y}+6\dot{y}+y=e^{(3x)}-\sin(x)$$

now I get that the annihilator of the $$e$$ term is $$(D-3)$$ but the answer is $$(D-3)(D+1)(D^2 +6D +8)$$ can someone explain the second part and if you are feeling generous how to do other annihilators maybe with examples in really simple language. I get so lost with these explanations that use "math language"

also is there a list or something I can study for what annihilates what? i have found one that I understand but it's really limited. A lot of them are written in extremely complicated language.

The annihilator method is a systematic way to find the particular solutions to a nonhomogeneous linear ODE. Initially, on the left hand side of your equation, the differential operator $f_1(D)$ "generates" the term $y''+6y'+y$. you can see this as $$f_1(D)y=\left(D^2+6D+1\right)y=\left(\frac{d^2}{dx^2}+6\frac{d}{dx}+1\right)y=\frac{d^2}{dx^2}y+6\frac{d}{dx}y+1y=y''+6y'+y$$
Now that we see what a differential operator does, we can investigate the annihilator method. It is a systematic way to generate the guesses that show up in the method of undetermined coefficients. You look for differential operators such that when they act on the terms on the right hand side they become zero. In your example you have $e^{3x}$ and $-\sin x$. So we see: $$f_2(D)e^{3x} = \left(D-3\right)e^{3x} = \left(\frac{d}{dx}-3\right)e^{3x}=\frac{d}{dx}e^{3x}-3e^{3x}=0$$ and $$f_3(D)(-\sin x) = \left(D^2+1\right)(-\sin x) = \left(\frac{d^2}{dx^2}+1\right)(-\sin x)=\sin x - \sin x=0$$
and now we have the full differential operator for this problem: $$f_1f_2f_3 = (D^2+6D+1)(D^2+1)(D-3)$$ The roots of this polynomial become the coefficients in the arguments of exponentials for the general solution. So we have $$y_1(x)=e^{(-3+2\sqrt{2})x}, y_2(x)=e^{(-3-2\sqrt{2})x},y_3(x)=e^{ix},y_4(x)=e^{-ix},y_5(x)=e^{3x}$$ and from this "basis" set we form the general solution: $$y(x)=c_1y_1+c_2y_2+c_3y_3+c_4y_4+c_5y_5$$ $c_1$ and $c_2$ are arbitrary constants and $c_3,c_4,c_5$ are fixed by the ODE.
$(D-\lambda)$ annihilates $e^{\lambda t}$, whether $\lambda$ is real or complex. Normally, if $\lambda$ is complex, you want to use $(D-\lambda)(D-\overline{\lambda})$ so that the resulting operator will annihilate $e^{\Re\lambda t}\cos(\Im\lambda t)$ and $e^{\Re\lambda t}\sin(\Im\lambda t)$, where $\overline{\lambda}$ is the complex conjugate of $\lambda$, $\Re\lambda$ is the real part of $\lambda$, and $\Im\lambda$ is the imaginary part of $\lambda$. You need both to annihilate the sin and cos terms. So, for example, $(D-i)(D+i)=D^{2}+1$ annihilates $\sin t$ and $\cos t$, and $(D-(3+i))(D-(3-i))=(D-3)^{2}+1$ annihilates both $e^{3t}\sin t$ and $e^{3t}\cos t$.
Taking the (n+1)-st power of such operators annihilates any polynomial $p(t)=a_{n}t^{n}+a_{n-1}t^{n-1}+\cdots +a_{1}t + a_{0}$ times what is annihilated by the first power of the operator. So, for example, $(D-3)^{4}$ annihilates $(a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0})e^{3t}$. And $(D^{2}+1)^{2}$ annihilates $\sin t, t\sin t,\cos t,t\cos t$.
So, in your case of $(D^{2}+6D+1)y=e^{3x}-\sin(x)$, you need $(D-3)(D^{2}+1)$ to annihilate the right side: $(D-3)$ annihilates $e^{3x}$ and $(D^{2}+1)$ annihilates $\sin t$ and $\cos t$. So your $y$ is also a solution of $(D^{2}+6D+1)(D-3)(D^{2}+1)y=0$. Applying annihilators introduces additional solutions, which must be weeded out later, but at least it gives you the general form of solution. In this case, $$(D^{2}+6D+1)=(D-3)^{2}-8=(D-3-\sqrt{8})(D-3+\sqrt{8}).$$ So the general solution $y$ must have the form $$y=Re^{(3+\sqrt{8})t}+Se^{(3-\sqrt{8})t}+Te^{3t}+U\cos(t)+V\sin(t),$$ where $R,S,T,U,V$ are constants to be determined. The constants $R$ and $S$ can be anything because $(D^{2}+6D+1)y$ annihilates the terms corresponding to $R$ and $S$. However, the other contants are determined by $$(D^{2}+6D+1)y = e^{3t}-\sin t.$$ Applying $(D^{2}+6D+1)$ to $Te^{3t}$ gives $28Te^{3t}$, which must equal the $e^{3t}$ term on the right. So $T=1/28$. Applying $(D^{2}+6D+1)=(D^{2}+1)+6D$ to $U\cos(t)+V\sin(t)$ is the same as applying $6D$ to $U\cos(t)+V\sin(t)$ because $D^{2}+1$ annihilates $U\cos(t)+V\sin(t)$. This gives $$-6U\sin(t)+6V\cos(t) = -\sin t.$$ So, $U=1/6$ and $V=0$. The solution of the ODE is $$y(t) = Re^{(3+\sqrt{8})t}+Se^{(3-\sqrt{8})t}+\frac{1}{28}e^{3t}+\frac{1}{6}\cos(t).$$