I'm having problems solving the following reccurence relation $$a x(n+1)-(a+b) x(n)+b x(n-1)=-1$$
I have found the homogeneous solution it being, $$x(n)^{(h)}=C_1 \left( \frac{b}{a} \right)^n + C_2$$
but am struggeling with finding the particular solution. What guess should I make to find it?
Appreciate any help
On
1) To get rid of the annoying constant on the RHS construct a difference equation from $$ a x_n - (a+b) x_{n-1} + b x_{n-2} = -1 $$ i.e., subtract it from the one you have and define $$ c_n = x_n - x_{n-1} $$ Now that you have an expression of the form $$ c_{n+1} = \frac{(a+b)}{a} c_n - \frac{b}{a} c_{n-1} $$ and, given boundary conditions you can use generating functions or, if you are lazy, recognize a simple 2nd order equation: $$ \lambda^2 - \alpha \lambda + \beta = 0 $$ Can you handle from here?
The non-homogeneous part is a constant, so you would normally look for a constant particular solution. However, constants are already solutions to the homogeneous part, so you have to modify that approach by multiplying the constant by a power of $n$. The root $1$ of the auxiliary equation that produces the constant solutions is a single root, so you can use $n^1=n$ as the multiplier: try for a particular solution of the form $x(n)=Cn$. You’ll find that you can solve for $C$ in terms of $a$ and $b$.