Characteristic equation: r2 - 5r + 6 = 0,
Solving the equation gives r = 2, 3.
f(r) = 2r+ r,
Particular solution, a(p) = A2(r)(2r) + A1r + A0
Putting this trial solution into the recurrence equation gives
A2r2r + A1r + A0 - 5[A2(r-1)2r-1 + A1(r-1) + A0] + 6[A2(r-2)2r-2 + A1(r-2) + A0] = 2r + r
or, (2r-2)(A2r22 - 5A2r.2 + 5A2.2 + 6A2.r - 12A2) + r(A1 - 5A1 + 6A1) + (A0 +5A1 - 5A0 - 12A1 +6A0) = 2r-2(22) + r(1) + (0)
Matching the coefficients give,
A2r22 - 5A2r.2 + 5A2.2 + 6A2.r - 12A2 = 4 => A2 = -2
A1 - 5A1 + 6A1 = 1 => A1 = 1/2
A0 +5A1 - 5A0 - 12A1 +6A0 = 0 => A0 = 7/4
So the particular solution is
a(p) = -2(r)(2r) + (1/2)r + 7/4