Suppose we are given a difference equation, for instance:
$z(k+2) - 2z(k+1) +z(k) = 0$
and the two solutions to the difference equation:
$z_1(k) =1$
$z_2(k) = k$
How am I supposed to check that the given solutions are indeed solutions to the difference equation? For differential equations, we differentiate the solutions and substitute it back into the equation and check the LHS = RHS. I am looking for an analogous approach for difference equations. I was hoping the answer to this question could use the example given to demonstrate the answer.
You check by substituting $z_1$ into the original equation, and check that it satisfies the equation.
So, let $k$ be arbitrary. Then $$z_1(k+2)-2z_1(k+1)+z_1(k) = 1 - 2 \cdot 1 + 1 = 0$$ which means that indeed, $z_2$ satisfies the initial equation for all $k$.