$$T_n = 6T_{n-1} - 13T_{n-2} + 12T_{n-3} - 4T_{n-4} + 5n^2 + 3n + 2 + 2^n + n2^n$$
The characteristic polynomial is $x^4 - 6x^3 + 13x^2 - 12x + 4 = 0$, or $(x-2)^2 (x-1)^2 = 0$. Therefore the homogeneous solution looks like:
$$T_n = \alpha_1 2^n + \alpha_2 n 2^n + \alpha_3 (1)^n + \alpha_4 n (1)^n$$
The non-homogeneous components $5n^2 + 3n + 2 + 2^n + n2^n$ have some interference with this when we create our trial solution. We cannot assume trial solution $an^2 + bn + c + d2^n + en2^n$.
The $an^2$ term clashes with nothing.
$bn$ clashes with $\alpha_4 n (1)^n$.
The constant $c$ clashes with $\alpha_3 (1)^n$.
$d2^n$ clashes with $\alpha_1 2^n$.
$en2^n$ also clashes with $\alpha_2 n 2^n$.
The trial terms need to each be multiplied by factors of $n$ until nothing is colliding with anything in the homogeneous piece (nor colliding with other non-homogeneous terms).
I changed the trial solution to $an^2 + bn^3 + cn^4 + dn^2 2^n + en^3 2^n$.
But when I plug it into the non-homogeneous recurrence, I get an unsolvable system for some reason, according to Mathematica. Where did I go wrong?
If $T_n = n$ for $1 \le n \le 4$, then use the recursive definition to get
$$T[1 \dots 9] = [1, ~2, ~3, ~4, 339, ~ 2658, ~12869, ~49362, ~164969]$$
Plug that into $$T_n = \sum_{k = 0}^3 \lambda_k~n^k~2^n + \sum_{k=0}^4 \lambda_{k + 4}~n^k$$
to get 9 equations and 9 variables. Solve for $\lambda$ to get:
$$\lambda[0 \dots 8] = \left[ -\frac{389}{2}, ~ -\frac{1}{6}, ~0, ~\frac{2}{3}, ~238, ~\frac{367}{6}, ~\frac{967}{12}, ~\frac{53}{6}, ~\frac{5}{12} \right]$$
Which is:
$$T_{n} = 2^n ~ \left(-\frac{389}{2} -\frac{1}{6}n + \frac{2}{3}n^3 \right) + 238 + \frac{367}{6}n + \frac{967}{12}n^2 + \frac{53}{6} n^3 + \frac{5}{12} n^4 \tag{S}$$
Evaluate $$S \vert_{n \to n} - 6~S\vert_{n \to n - 1} + 13~S\vert_{n \to n - 2} - 12~S\vert_{n \to n - 3} + 4~S\vert_{n \to n - 4}$$
and compare with the original recursion to check your answer.