solving second-order linear difference equation

I am having trouble solving:

$$a\frac{x_{n-1}}{n-1}-\frac{x_{n}}{n}+(1-a)\frac{x_{n+1}}{n+1}=0$$

So far I have tried eliminating the n-1,n,n+1 terms by multiplying them out but that doesn't feel right. Intuitively I think the solution would be to set some new variable $y_n=\frac{x_n}{n}$ and do it as usual with the characteristic equation but I'm not sure if I can do that.

Thanks.

edit: Background:I've just started difference equations and need it for a particular probability problem

-

Of course you can set $y_n = \frac{x_n}{n}$. You then get
$$(1-a)y_{n+1} - y_n + ay_{n-1} = 0$$
Hence the polynomial $(1-a)X^2 - X + a$. Compute the roots $r_1,r_2$, and the solution will be
$$y_n = \lambda_1 r_1^n + \lambda_2 r_2^n$$
Use the initial conditions (ie. values of $y_0,y_1$) to compute $\lambda_1,\lambda_2$. And you then get the expression for $x_n$.