Consider the recurrence xk+2 = axk+1 + bxk + c where c may not be zero.
(a) If a + b ≠ 1 show that p can be found such that, if we set yk = xk + p, then yk+2 = ayk+1 + byk. [Hence, the sequence xk can be found provided yk can be found by the methods of this section (or otherwise).]
(b) Use (a) to solve the recurrence
xk+2 - xk+1 + 6xk + 5 where x0 = 1 and x1 = 1.