How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved How are can the above recurrence relation be solved? 
I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$
And then here:
$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$
And we are given that these: $T(0) = 2, T(1) = 3$
So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants. 

Use "generatingfunctionology" techniques. Define $G(z) = \sum_{n \ge 0} T(n) z^n$, write your recurrence as:
$$
T(n + 2) = T(n + 1) + T(n) + 3 n + 3
$$
Multiply by $z^n$, add for $n \ge 0$. Remember that:
$$
\sum_{n \ge 0} (n + 1) z^n = \frac{1}{(1 - z)^2}
$$
and you get:
$$
\frac{(G(z) - T(0) - T(1) z}{z^2} 
  = \frac{G(z) - T(0)}{z} + G(z) + 3 \cdot \frac{1}{(1 - z)^2}
$$
Solve for $G(z)$, expand in partial fractions:
$$
G(z) = \frac{2 - 3 z + z^2 + z^3}{1 - 3 z + 2 z^2 + z^3 - z^4}
     = \frac{4 + 2 z}{1 - z - z^2} - \frac{1}{1 - z} - \frac{1}{(1 - z)^2}
$$
We have the last two terms, the first one is for Fibonacci numbers:
$$
T(n) = 4 F_{n + 1} + 2 F_n - 1 - (n + 1) = 4 F_{n + 1} + 2 F_n - n - 2
$$
