$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$'s General Solution I was looking for the general solution of the differential equation $$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$$
My work
We need to get the auxillary equation of the given differential equation above...
The auxillary equation would be:
$$r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$$
We need to get the roots of the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ using the rational zero theorem...
The rational zero theorem states that if $P(x)$ is a polynomial with integer coefficients and if is a zero of $P(x) ( P(\frac{p}{q} ) = 0 )$, 
then $p$ is a factor of the constant term of $P(x)$ and $q$ is a factor of the leading coefficient of $P(x)$.
With that in mind....
The factors of $p$ are:
$$-1,12,1,-12$$
$$-2,6,2,-6$$
$$-3,4,3,-4$$
The factors of $q$ are:
$$1,1,-1,-1$$
Then the values of $\frac{p}{q}$ would be:
$$-1,1,2,-3,4,-4,12,-12,6,-6,-2,3$$
It was found out that only $r= -2$ and $r= 3$ could make the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ true, instead of $5$
Now....if I only got $2$ real and distinct roots out of possible $5$, then the solution of the differential equation $(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$ 
would be:
$$ y = c_1e^{-2x} + c_2e^{3x} + (unknown \space term) + (unknown \space term) + (unknown \space term) $$
I'm stuck....How do you get the solution of the given differential equation above?
UPDATE
After some feedback from intrepid answerers....I was able to factor out the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ into $(r+2)(r+2)(r-3)(r^2 +1)=0$, so I think the solution to the differential equation above would be:
$$ y = c_1e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + e^{x}(c_4\cos (-x) + c_5\sin (-x)) $$
or 
$$ y = c_1e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + e^{x}(c_4\cos x - c_5\sin x) $$
Is my solution now correct?
 A: Hint: Since you know that two roots are $-2$ and $3$, you can exploit factoring by grouping or polynomial long division:
\begin{align}\begin{aligned}
&r^5+r^4-7r^3-11r^2-8r-12 \\&\quad= r^5+2r^4-r^4-2r^3-5r^3-10r^2-r^2-2r-6r-12\\
&\quad= (r+2)(r^4-r^3-5r^2-r-6)\\
&\quad= (r+2)(r^4-3r^3+2r^3-6r^2+r^2-3r+2r-6)\\
&\quad= (r+2)(r-3)(r^3+2r^2+r+2)
\end{aligned}\end{align}
That last cubic factors in a particularly nice way. (You can also try applying the rational roots theorem to the cubic $r^3+2r^2+r+2$ to try to pick out a root, and then use factoring by grouping again; since the last term is a +2, there's only 4 values to try, namely $\pm 1$ and $\pm 2$.)

EDIT:
It is true that this becomes
$$(r+2)^2(r-3)(r^2+1) = 0$$
which yields roots of $r = -2,-2,3,\pm i$
However, it is not true that your solution becomes
$$y = c_1 e^{-2x} + c_2 xe^{-2x} + c_3 e^{3x} + e^x(c_4 \cos(x)+c_5\sin(x)$$
The problem is with those last two terms. You can check that the function $$f(x) = e^x(c_4\cos(x) + c_5\sin(x))$$ is a solution to 
$$((D-1)^2+1)f = 0$$
i.e., it corresponds to complex roots of $r=1\pm i$, not $r=\pm i$. The piece you're missing is Euler's equation:
$$e^{i\alpha} = \cos(\alpha) + i\sin(\alpha)$$
Using this you can conclude your solution is
$$y = c_1 e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + c_4\cos(x) + c_5\sin(x)$$

Really, it all hinges on the solution to $(D^2+1)y = 0$, which one can derive in the following manner. We know that a candidate solution is (with $A$ and $B$ arbitrary real constants)
\begin{align}\begin{aligned}
y &= Ae^{ix} + Be^{-ix}\\
&= A(\cos(x)+i\sin(x)) + B(\cos(x)-i\sin(x))\\
&= (A+B)\cos(x) + (A-B)i\sin(x)
\end{aligned}\end{align}
and now we notice that if this is a solution, then so is the real part (i.e., $\cos(x)$) and the imaginary part (i.e., $\sin(x)$), hence arriving at the desired solution
$$y = c_1\cos(x) + c_2\sin(x)$$
A: $(r+2)^2(r-3)(r^2+1)$
We have one double root and one pair of complex roots.
With complex roots:
$Ae^{it} + Be^{-it} = C_1 \cos t+ C_2\sin t$
are solutions
With roots of multiplicity.
$C_3^{-2t}+ C_4te^{-2t}$ will be solutions. 
