I am trying to solve the rational difference equation $\displaystyle x_{n+1}=\frac{a^2}{2a-x_{n}}$ using the roots of the characteristic equation.
The characteristic equation is $\displaystyle \lambda=\frac{a^2}{2a-\lambda}$, which can be rewritten as $\lambda^2-2a\lambda+a^2=0$ and $a$ is the repeated root.
Since the roots of the characteristic equation is a repeated root, I do not know how to proceed.
If the rational difference equation is $\displaystyle x_{n+1}=\frac{ab}{(a+b)-x_{n}}$, then the roots of the characteristic equation would be $a$ and $b$ and the substitution $\displaystyle y_n=\frac{x_n-a}{x_n-b}$ should work.
Is there any trick in the substitution? Or this approach does not work at all?
For reference, the solution of $\displaystyle x_{n+1}=\frac{a^2}{2a-x_{n}}$ is $\displaystyle x_n=a-\frac{a(a-x_1)}{na-(n-1)x_1}$.
Set $y_n=\dfrac{1}{x_n-a}$. After few computations, you find $y_{n+1}=y_n -a^{-1}$, which is easily solved.