TITLE: Finding the closed form for $x_n^2 = -x_{n-1}^2+6x_{n-2}^2+n$
QUESTION [5 upvotes]: I'm trying to find a closed form for the recurrence relation $x_n^2 = -x_{n-1}^2+6x_{n-2}^2+n$, with $x_1 = \frac{1}{4}, x_2=\frac{\sqrt{13}}{4}$ and $x_i \in \mathbb R^+$. 
My attempt was to let $z_n=x_n^2$ and then transform the recurrence relation to a 2nd order non-homogeneous linear recurrence relation. Substitution gives:
$$z_n = -z_{n-1}+6z_{n-2}$$.
Now the above recurrence relation has complementary function $$z'_n = L2^n +K(-3)^n$$ for $L,K$ arbitrary constants. 
I also found that the new recurrence relation has a particular solution given by 
$$z''_n = \frac{11-n}{4}$$. 
Thus the general solution is $$z_n = L2^n +K(-3)^n + \frac{11-n}{4}$$
By substitution I then found $L=\frac{-7}{8}, K = \frac{11}{48}$. So the unique solution for our recurrence relation is: $$z_n = \frac{-7}{8}2^n +\frac{11}{48}(-3)^n + \frac{11-n}{4}$$ Now, for $n=3$ I obtain a negative value of $z_n$ indicating that my solution is invalid. I've checked my solution numerous times and I'm pretty sure my numbers are right so I'm wondering if my method here is wrong.

REPLY [3 votes]: Try with this particular solution instead $z''_n = -\dfrac{n}{4}-\dfrac{11}{16}$, the one you used does not work.