The annihilator of $n(2^n)\sin({n\pi \over 2})$ I have to solve this problem:
$y(n+2)-y(n)=n(2^n)\sin({n\pi \over 2})$
And I know the annihilator of $n(2^n) = (E-2)^2$, but I don't know how I should find the other part of the annihilator.
$y_p(n)=c_0+nc_1$

I'm not so familiar with applying annihilators to recurrence relations, but I found the idea interesting and did some reading. I'm fairly sure I have a solution, but it may not be the simplest.
Use Euler's formula ($ e^{i x} = \cos{x} + i \sin{x} $) to derive a more useful expression for $\sin(x)$ as follows:
$ e^{-i x} = \cos{-x} + i \sin{-x} = \cos{x} - i \sin{x} $
Hence $ e^{i x} - e^{-i x} = (\cos{x} + i \sin{x}) - (\cos{x} - i \sin{x}) = 2i \sin(x) $
$ \therefore \sin{x} = \frac{e^{i x} - e^{-i x}}{2i} $
Now, applying the new expression, we have:
$$ \sin{\frac{n \pi}{2}} = \frac{e^{i \frac{n \pi}{2}} - e^{-i \frac{n \pi}{2}}}{2 i} = \frac{(e^{i \frac{\pi}{2}})^n - (e^{-i \frac{\pi}{2}})^n}{2 i} = \frac{i^n - (-i)^n}{2 i} $$
And thus that $ n 2^n \sin{\frac{n \pi}{2}} = n 2^n \frac{i^n - (-i)^n}{2 i} = \frac{n}{2i}(2^n i^n - 2^n (-i)^n) = \frac{n}{2i}(2i)^n - \frac{n}{2i}(-2i)^n$
Which has annihilator $ (E - 2i)^2 (E + 2i)^2  = (E^2 + 4)^2$
