We have an inhomogeneous difference equation:
$$ y(n+2) - 5y(n+1) + 6y(n) = 1 + n $$
So to find the particular solution, we must find an annihilator $N(E)$, such that $N(E)g(n) = 0 $.
What is the process of finding it. What if there is something else instead of $1+n$.
In the end we have to solve it. For that answer, I am particularly interested how you go about finding the right Annihalators.
So I know that we look at how the inhomogeneous term looks:
So if it is linear, quadratic, or trigonometric.
But how do we go about if the term is a mix of those.
Lets say what do we do if the inhomogeneous term is $$g(n) = 2n - 2 + 10 \sin{\frac{n \pi}{2}}$$
I am studying from the textbook http://bmm.univ-lyon1.fr/bmm/data/references/Elaydi-2005.pdf and it really is not explained how to do this there.
To find a particular solution for $g(n) = 2n-2$ is trivial. Considering instead
$$ y(n+2) - 5y(n+1) + 6y(n) = 10\sin\left(\frac{n\pi}{2}\right) $$
making
$$ y_p(n) = a\sin\left(\frac{n\pi}{2}\right) $$
after substitution we have
$$ 6 a \sin \left(\frac{\pi n}{2}\right)-5 a \sin \left(\frac{1}{2} \pi (n+1)\right)+a \sin \left(\frac{1}{2} \pi (n+2)\right)-10 \sin \left(\frac{\pi n}{2}\right)=0 $$
Solving for $a$
$$ a = \frac{2 \sin \left(\frac{\pi n}{2}\right)}{\sin \left(\frac{\pi n}{2}\right)-\cos \left(\frac{\pi n}{2}\right)} $$
then
$$ y_p(n) = \frac{2 \left(\sin \left(\frac{\pi n}{2}\right)\right)^2}{\sin \left(\frac{\pi n}{2}\right)-\cos \left(\frac{\pi n}{2}\right)} $$