Given the first-order difference equation
$y(k) = u(k) + y(k-1)$ for $k = 1, 2, 3, \dots$
with the input signal u(k) = k, and the initial condition y(–1) = 0. I am trying to verify that its solution also satisfies the second-order difference equation
$y(k) = 2y(k-1) - y(k-2) + 1$
with the initial conditions $y(0) = 0$ and $y(–1) = 0$.
Using the method of undetermined coefficients I attained
$y(k) = -(-1)^k + k^2$ (solution for the 1st order difference equation)
I'm having trouble finding a solution to the 2nd order difference equation. How do I solve for its particular solution?