The particular solution of the recurrence relation I cannot find out why the particular solution of $a_n=2a_{n-1} +3n$ is $a_{n}=-3n-6$ 
here is the how I solve the relation
$a_n-2a_{n-1}=3n$
as $\beta (n)= 3n$ 
using direct guessing 
$a_n=B_1 n+ B_2$
$B_1 n+ B_2 - 2 (B_1 n+ B_2) = 3n$
So $B_1 = -3$, $B_2 = 0$
the particular solution is $a_n = -3 n$
and the homo. solution is $a_n = A_1 (-2)^n$
Why it is wrong??

Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence without subtractions in indices:
$$
a_{n + 1} = 2 a_n + 3 n + 3
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums, particularly:
\begin{align}
\sum_{n \ge 0} z^n
  &= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
  &= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
  &= \frac{z}{(1 - z)^2}
\end{align}
to get:
$$
\frac{A(z) - a_0}{z} = 2 A(z) + \frac{3}{(1 - z)^2} + \frac{3}{1 - z}
$$
As partial fractions this gives:
$$
A(z)
  = \frac{a_0 + 6}{1 - 2 z}
      - \frac{3}{1 - z}
      - \frac{3}{(1 - z)^2}
$$
Using the generalized binomial theorem, in particular:
$$
\binom{-m}{n} = \binom{m + n - 1}{m - 1} (-1)^n
$$
you read off the coefficients:
\begin{align}
a_n &= (a_0 + 6) 2^n - 3 - 3 \binom{n + 2 - 1}{2 - 1} \\
    &= (a_0 + 6) 2^n - 3 - 3 (n + 1) \\
    &= (a_0 + 6) 2^n - 3 n - 6
\end{align}
