Finding the closed form for $x_n^2 = -x_{n-1}^2+6x_{n-2}^2+n$ 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.

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