Simple linear recursion $x_n=\frac{x_{n-1}}{a}+\frac{b}{a}$ with $a>1, b>0$ and $x_0>0$
I tried to solve it using the generating function but it does not work because of $\frac{b}{a}$, so may you have an idea.

There are all sorts of ways to solve this. For instance, let $y_n=x_n-d$; then
$$y_n+d=\frac1a(y_{n-1}+d)+\frac{b}a=\frac{y_{n-1}}a+\frac{b+d}a\;,\tag{1}$$
so $$y_n=\frac{y_{n-1}}a+\frac{b+(1-a)d}a\;.$$ If we now set $d=\dfrac{b}{a-1}$, $(1)$ becomes $$y_n=\frac{y_{n-1}}a$$ with initial condition $y_0=x_0-d$. This is a simple exponential recurrence, so the solution is
$$y_n=\frac{y_0}{a^n}\;$$
and $$x_n=y_n+d=\frac{y_0}{a^n}+d=\frac{x_0-\frac{b}{a-1}}{a^n}+\frac{b}{a-1}=\frac{(a-1)x_0-b}{a^n(a-1)}+\frac{b}{a-1}\;.$$
