Particular solution of non-homogeneous recurrence relation can somebody help me with my homework, please? I have to solve this: 
$a_{n}$ = $-a_{n-1}$ + $12a_{n-2}$ - 10n + 13 + $7.3^{n}$
$a_{0} = 3$, $a_{1}=24$.
I know to solve this (homogeneous equation):
$a_{n}$ = $-a_{n-1}$ + $12a_{n-2}$ 
$a_{n}$ = $x^{n}$
$x^{n}$ = -$x^{n-1}$ + $12.x^{n-2}$
$x^{n}$ + $x^{n-1}$ - $12.x^{n-2}$ = 0
(x + 4)(x - 3) = 0
$t_n$ = $(-4)^{n}.\alpha$ + $3^{n} \beta$
How can I find particular solution? ....using A,B,C... because i don't know solve this with sum (I found here something with sum, but we did't use sum on course).
Thanks everyone for your help :) 

You have:
$$a_n = -a_{n-1} + 12a_{n-2} - 10n + 13 + 7\cdot 3^n$$
$$a_{n+1} = -a_{n} + 12a_{n-1} - 10(n+1) + 13 + 7\cdot 3^{n+1}$$
Subtracting the first from the second equation you get:
$$a_{n+1} = 13a_{n-1} - 12a_{n-2} - 10 + 14 \cdot 3^n$$
Repeat the same trick to get rid of the constant and you have:
$$a_{n+2} = a_{n+1} + 13a_n - 25a_{n-1} + 12a_{n-2} + 28\cdot 3^n$$
Now subtract three times the above equation from:
$$a_{n+3} = a_{n+2} + 13a_{n+1} - 25a_{n} + 12a_{n-1} + 28\cdot 3^{n+1}$$
You will get a homogeneous linear recurrence relation
