Solution
= Solution
Let $L[y]_n=y_{n+3}-3y_{n+1}+2y_n$. For the exponential trial sequence $D2^n$,
$$
L[D2^n]=(8-6+2)D2^n=4D2^n.
$$
Thus $D=1/4$ gives a <particular solution>. The difference of any two solutions of the <inhomogeneous linear recurrence> satisfies the homogeneous <linear recurrence relation> in part (i). \b[Therefore]
$$
\boxed{y_n=A+Bn+C(-2)^n+\frac{2^n}{4}.}
$$