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.
thanks for your help

$(D-\lambda)$ 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
$$
       y(t) = Re^{(3+\sqrt{8})t}+Se^{(3-\sqrt{8})t}+\frac{1}{28}e^{3t}+\frac{1}{6}\cos(t).
$$
