$u_{n+1} = a u_n +b u_{n-1} +c$ Is there a closed form formula for the general term of a recurrence sequence satisfying $$u_{n+1} = a u_n +b u_{n-1} +c$$               The sequence I am interested in satisfies $u_{n+1} = 8 u_n - 3 u_{n-1}  -4$.
 A: You can solve any non-homogeneous, linear recurrence of the above form by converting it to homogeneous form.
(I'll use $n$ instead of $n+1$, doesn't matter anyway!)
$$u_{n}=au_{n-1}+bu_{n-2}+c$$
$$u_{n-1}=au_{n-2}+bu_{n-3}+c$$
Subtract them to get:-
$$u_{n}=(a+1)u_{n-1}+(b-a)u_{n-2}-bu_{n-3}$$
You can then use characteristic polynomial to find the answer.

Speaking strictly for the equation you wrote, the characteristic polynomial is
$$z^3=9z^2-11z+3$$ which has the roots
$$z=1, 4+\sqrt{13}, 4-\sqrt{13}$$ 
giving $$u_n=a_1+a_2{(4+\sqrt{13})}^n+a_3{(4-\sqrt{13})}^n$$, where $a_i$s depend on the initial condition.
A: This is a linear difference equation so can be stated as
$$
u = u_h + u_p
$$
with
$$
u_h(n+1)-au_h(n)-bu_h(n-1) = 0\\
u_p(n+1)-au_p(n)-bu_p(n-1) = c\\
$$
now making $u_h(n) = \phi^n$ and substituting we have
$$
\phi^n\left(\phi-a-b\phi^{-1}\right) = 0
$$
so
$$
\phi = \frac{1}{2} \left(a\pm\sqrt{a^2+4 b}\right)
$$
and
$$
u_h(n) = \frac{C_1}{2^n} \left(a-\sqrt{a^2+4 b}\right)^n+\frac{C_2}{2^n} \left(a+\sqrt{a^2+4 b}\right)^n
$$
and also
$$
u_p(n) = \frac{c}{1-a-b}
$$
so finally
$$
u(n) =  \frac{C_1}{2^n} \left(a-\sqrt{a^2+4 b}\right)^n+\frac{C_2}{2^n} \left(a+\sqrt{a^2+4 b}\right)^n + \frac{c}{1-a-b}
$$
