Find the general solution of $y^{(6)}+2y^{(4)}+y'' = 0$.
$r^6+2r^4+r^2=0$
$r^2(r^4+2r^2+1)=0$
$r^2[(r^2+1)(r^2+1)]$
So we have the roots:
$0$: Multiplicity 2
$+i$: Multiplicity 2
$-i$: Multiplicity 2
Now I'm not sure. I'm supposed to arrive at:
$y(x) = c_1+c_2x+c_3\cos x+c_4\sin x+c_5x\cos x+c_6x\sin x$
EDIT: I'm particularly curious as to what the multiplicity does to the general solution.