heterogeneous recurrence with f(n) as constant How to solve this $s_{n+1}=4s_{n-1}-3s_n+5$ where f(n)=5 conditions $s_0=-3$ $s_1=3$
I calculated the general solution $s_n=c_1*(-4)^n+c_2*1^n$ of this recurrence. The roots are $q_1=-4$ and $q_2=1$ but I have problem with particular solution with method of prediction .
I have problem with this 5 as constant
solution(homo+heterogeneous) is $s(n) = n - (-4)^n - 2$
I learn in Polish so I can't find too much words to describe this
 A: $s_{n+1}=4s_{n-1}-3s_n+5$ 
$q^2-4+3q=0$
$\Delta=3^2-4*1*(-4)=25$ 
$\sqrt\Delta=5$
$q_1=\frac{-3-5}2=-4$
$q_2=\frac{-3+5}2=1$
homo general $s_n=c_1*1^n+c_2*(-4)^n$
$k=1$ where $k$ is multipiclity of root
hetero particular $s_n=Q(n)*q^n*n^k$
$s_n=A*1^n*n^1=An$; $1^n=1$ because $1^0=1$, $1^1=1$ and so on
$A(n+1)=4(A(n-1))-3An+5$
$An+A=4An-4A-3An+5$
$An+A=An-4A+5$
$A=1$
hetero particular $s_n=n$
hetero general $s_n=c_1*1^n+c_2*(-4)^n+n$
$\begin{cases} -3=s_0=c_1*1^0+c_2*(-4)^0=c_1+c_2 => c_1=-3-c_2=-3+\frac{6}{5}=\frac{-9}{5} 
3=s_1=c_1*1^1+c_2*(-4)^1=-3-c_2-4*c_2 => -5c_2=6 => c_2=\frac{-6}{5}
\end{cases}$
hetero general $s_n=\frac{-9}{5}*1^n-\frac{6}{5}*(-4)^n+n$
A: $$s_{n + 2} = -3~s_{n + 1} + 4~s_{n} + 5$$
$$s_{n + 3} = -3~s_{n + 2} + 4~s_{n + 1} + 5$$
So
$$s_{n + 3} - s_{n + 2} = (-3~s_{n + 2} + 4~s_{n + 1} + 5) - (-3~s_{n + 1} + 4~s_{n} + 5)$$
$$s_{n + 3} = -2~s_{n + 2} + 7~s_{n + 1}$$
The roots are $-4$, $1$ and $1$, so the final form is:
$$s_n = A~(-4)^n + B~1^n + C~n~1^n$$
$$s_n = A~(-4)^n + B + C~n$$
Use initial conditions to find $s_0$, $s_1$, and $s_2$ to find $A$, $B$, and $C$.
