Need help finding the fundamental set of solutions to the differential equation: $y^{(6)}+y=cos(t)$ I factor it to $(r^2+1)(r^4-r^2+1)$. Then I sub $x=r^2$ to get $(x^2-x+1)$. The roots I am getting aren't correct and I'm not sure if I should be subbing the second polynomial like how I did or not.
Wolfy says
y(t) = c_5 sin(t) + c_4 e^(-(sqrt(3) t)/2) sin(t/2) + c_6 e^((sqrt(3) t)/2) sin(t/2) + c_1 e^((sqrt(3) t)/2) cos(t/2) + c_2 cos(t) + c_3 e^(-(sqrt(3) t)/2) cos(t/2) + 1/6 t sin(t).

I entered it as
y''''''+y=cos(t).
Wolfy says
y(t) = c_5 sin(t) + c_4 e^(-(sqrt(3) t)/2) sin(t/2) + c_6 e^((sqrt(3) t)/2) sin(t/2) + c_1 e^((sqrt(3) t)/2) cos(t/2) + c_2 cos(t) + c_3 e^(-(sqrt(3) t)/2) cos(t/2) + 1/6 t sin(t).

I entered it as
y''''''+y=cos(t).