$8y(t+2)-6y(t+1)+y(t)=2^t$ via Annihilator Method This question is taken directly from "Difference Equations: An Introduction with Applications",second edition,  by Walter G. Kelley and Allan C. Peterson:
Solve 
$$8y(t+2)-6y(t+1)+y(t)=2^t$$
via the annihilator method.
Could someone show this in great detail, especially finding the homogenous equation solution for $2^t$?
 A: With constant coefficients these are similar to differential equations. 
The roots of $ 8 \lambda^2 - 6 \lambda + 1 = 0$ are $\lambda = 1/2, 1/4.$ Instead of $e^{\lambda t},$ you just take $\lambda^t.$ So our general solution with a $0$ on the right hand side is
$$ A \left( \frac{1}{2} \right)^t +  B \left( \frac{1}{4} \right)^t.  $$
Furthermore, any two solutions to the problem with $2^t$ on the right hand side differ precisely by such a sum. 
Next, we get some good luck with $2^t.$ As we can check (and have already shown in general) the number 2 is not one of the $\lambda.$ As a result, if we add in $C \cdot 2^t$ to $y(t),$ we get something nonzero, in particular
$$ C \cdot 8 \cdot 4 \cdot 2^t - C \cdot  6 \cdot 2 \cdot 2^t + C \cdot  2^t = C \cdot  ( 32 - 12 + 1 ) \cdot 2^t = 21 \cdot C \cdot 2^t. $$ We then need $21 C = 1.$ So, with completely general constants $A,B,$ we get 
$$ y(t) =  A \cdot \left( \frac{1}{2} \right)^t +  B \cdot \left( \frac{1}{4} \right)^t + \left( \frac{1}{21} \right) \cdot 2^t  $$
A: Three standard operators. Identity
$$  If(t) = f(t).   $$
Translation
$$ Ef(t) = f(t+1) $$
Difference
$$ \Delta f(t) = f(t+1) - f(t).   $$
Relation
$$  \Delta = E - I.  $$
Your book is using $1$ instead of $I,$ no real problem.
$$  E 2^t = 2^{t+1} = 2 \cdot 2^t  $$
In general,
$$  E \lambda^t = \lambda^{t+1} = \lambda \cdot \lambda^t.  $$
$$  \Delta \lambda^t = ( \lambda - 1) \cdot \lambda^t.  $$
For a constant funtion of $t,$
$$   E 1 = 1,$$
$$ I 1 = 1.$$
$$  \Delta 1 = E1 - I1 = 0.$$
$$ Et = t+1,   $$
$$ It = t,$$
$$ \Delta t = t+1 - t = 1.$$
So
$$ \Delta^2 t = 0.$$
A: Just in case anyone's wondering, here's what I get via Kelley and Peterson's "Annihilator Method":
The equation gives us an equation in shift operators (i.e. Jagy's translation operator):
$$(8E^2-6E+I)y(t)=0$$.
$$8(E-1/2)(E-1/4)=0$$
which is similar to Dr. Jagy's answer.
Then $2^t$ satisfies the equation:
$$(E-2)(2^t)=0$$
So now we have, combining the two results:
$$8(E-1/2)(E-1/4)(E-2)y(t)=0$$
...which leads to...
$$y(t)=c_1(1/2)^t+c_2(1/4)^t+c_3(2)^t$$
We then see that $c_1(1/2)^t+c_2(1/4)^t$ satisfies the homogeneous portion of the equation.  So we substitute $y(t)=c_3(2)^t$ into the original equation/question:
$$8c_3(2)^{t+2}-6c_3(2)^{t+1}+c_3(2)^t=2^t$$
$$c_3=1/21$$
Then we plug this in:
$$y(t)=c_1(1/2)^t+c_2(1/4)^t+\frac{1}{21}(2)^t$$
