Solve non-homogeneous recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$ I'm trying to solve this recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$, but the solution of the homogeneous is $a_n=b+cn$, and I don't know how to change it to find the non-homogeneous solution.
 A: Try the Ansatz $x_n=kn^2$ so$$8=k[n^2+4n+4-2n^2-4n-2+n^2]=2k\implies k=4.$$
A: $$x_{n+2} - 2x_{n+1} + x_n = 8\tag{1}$$
Set $n=n+1$
$$x_{n+3}-2x_{n+2}+x_{n+1}=8$$
Subtract the two equations
$$x_{n+3}-2x_{n+2}+x_{n+1}-x_{n+2} + 2x_{n+1} - x_n=0$$
$$x_{n+3}-3x_{n+2}+3x_{n+1}-x_n=0$$
Characteristic equation is
$$\lambda^3-3\lambda^2+3\lambda-1=0\to (\lambda-1)^3=0\to \lambda=1$$
and finally
$$x_n=an^2+bn+c\tag{2}$$

edit
from $(1)$ we get $x_0=8+2x_1-x_2$
from $(2)$ we have $x_0=c,\;x_1=a+b+c;\;x_2=4a+2b+c$
putting all together
$$\begin{cases}
c=8+2x_1-x_2\\
a+b+c=x_1\\
4a+2b+c=x_2\\
\end{cases}
$$
$$\begin{cases}
a+b+8+2x_1-x_2=x_1\to a+b=x_2-x_1-8\\
4a+2b+8+2x_1-x_2=x_2\to 4a+2b=2x_2-2x_1-8\\
\end{cases}
$$
$$\begin{cases}
a+b=x_2-x_1-8\\
2a+b=x_2-x_1-4\\
\end{cases}
$$
subtracting the two equations, we get $a=4$ so the $(2)$ becomes
$$x_n=4n^2+bn+c$$
A: If the non-homogeneous part is constant you can always take the difference and make it homogeneous.  In your case,
$$(\mathbb E-1)^2 x_n = 8 \implies (\mathbb E-1)^3 x_n = (\mathbb E - 1)8=0$$
And you get your characteristic function immediately which is $(\lambda-1)^3=1$.
Here $\mathbb E$ is the forward shift operator, $\mathbb E^i x_n = x_{n+i}, \forall i \in \mathbb N$.
For a non-constant $f(x)$ example, check out this answer.
A: Another way is applying unilateral $\mathcal{Z}$-transform which is defined as $$X(z) = \sum_{n=0}^{+\infty}x[n]z^{-n}$$The equation is $$x[n+2] -2x[n+1] +x[n] = 8u[n]$$We can easily show that the following properties holds:$$\mathcal{Z}\{\alpha^nu[n]\} = \frac{1}{1-\alpha z^{-1}} \\ \mathcal{Z}\{x[n+1]\} = zX(z) - zx[0]\implies\mathcal{Z}\{x[n+2]\} = z^2X(z) - z^2x[0] - zx[1] \\ \mathcal{Z}\{nx[n]\} = -z\frac{dX(z)}{dz}\\ \mathcal{Z}\{x[n-1]\} = z^{-1}X(z) + x[-1] \\ \mathcal{Z}\{\delta[n]\}  = 1$$Applying the transform to the equation yields $$z^2X(z) - z^2x[0] - zx[1] - 2(zX(z) - zx[0]) +X(z) = \frac{8}{1-z^{-1}}$$For convenience, let $x[0] = a$ and $x[1] = b$. Partial fraction decomposition gives $$X(z) = a + \frac{8}{(z-1)^3} + \frac{8 - a + b}{(z-1)^2} + \frac{b}{z-1}$$Taking the inverse transform gives $$x[n] = a\delta[n] + bu[n-1] + (8 - a + b)(n-1)u[n-1] + 8(\frac{1}{2}(n-2)(n-1)u[n-1])$$Which can be written as $$x[n] = a - 4n - an + bn + 4n^2 = 4n^2 + n(b-a-4) + a$$
It can be verified by WA. Note that the counterpart of the Laplace transform for discrete-time signals is the $\mathcal{Z}$-transform.
