Solve recurrence with characteristic polynomial $a_n=7\cdot a_{n-1} -7\cdot a_{n-2}+175\cdot a_{n-3}+450\cdot a_{n-4}+(5+13\cdot n)\cdot9^n $ The equation
$$a_n=7\cdot a_{n-1} -7\cdot a_{n-2}+175\cdot a_{n-3}+450\cdot a_{n-4}+(5+13\cdot n)\cdot9^n \enspace,$$
where
$a_0=148, a_1=144, a_2=-55, a_3=-61$.
I assume that a solution will look like $a^s_n+a^h_n$ , where $a^s_n$. I solve like 
$$x^4-7x^3+7x^2-175x-450=0 \enspace,$$
then get $x_1=-2,\ \ x_2=9,\ \ x_3=-5i,\ \ x_4=5i\ \ $
and don't know what next.
But the real problem is with the second part $(5+13\cdot n)\cdot9^n$. I completely don't know what do with that.

Considering the homogeneous
$$
a_n-7\cdot a_{n-1}+7\cdot a_{n-2}-175\cdot a_{n-3}-450\cdot a_{n-4}=0
$$
after substituting $a_n = C_0 \beta^n$ we obtain
$$
C_0(\beta -9) (\beta +2) \left(\beta ^2+25\right) \beta ^{n-4}=0
$$
so we have the set of exponent solutions (for $n > 4$ assuming $\beta\ne 0$)
$$
\{-2,9,\pm i 5\}
$$
then
$$
a_n^h = C_1(-2)^n+C_29^n+C_3 (i 5)^n+C_4 (-i 5)^n
$$
for the particular we adopt
$$
a_n^p = (c_1+c_2n+c_3n^2)9^n
$$
and after substitution we obtain
$$
a_n^h = 9^n \left(c_1+\frac{16996635 n}{1359556}+\frac{9477 n^2}{2332}\right)
$$
and finally
$$
a_n =  C_1(-2)^n+C_29^n+C_3 (i 5)^n+C_4 (-i 5)^n + 9^n \left(c_1+\frac{16996635 n}{1359556}+\frac{9477 n^2}{2332}\right)
$$
