<s>[INST] $(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$'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?
 [/INST] $(r+2)^2(r-3)(r^2+1)$
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. 
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