Find solution of recurrence relation

Question

What is the solution of the recurrence relation of
$$x_n^2=x_{n-1}^2+6x_{n-2}^2+7^n$$
with $x_0=x_1=1$

Answer 1

As it was suggested in the comments and with quite a few shortcuts (i.e. wolfram links), substituting $a_n=x_n^2$ $$a_n=a_{n-1}+6a_{n-2}+7^n \tag{1}$$ this also means $$a_{n+1}=a_{n}+6a_{n-1}+7^{n+1} \tag{2}$$ we multiply $(1)$ by 7 $$7a_n=7a_{n-1}+42a_{n-2}+7^{n+1} \tag{3}$$ now subtract $(3)$ from $(2)$ $$a_{n+1}-7a_n=a_{n}+6a_{n-1}-7a_{n-1}-42a_{n-2} \iff \\ a_{n+1}-8a_n+a_{n-1}+42a_{n-2}=0$$ which is a linear homogeneous recurrence with characteristic polynomial $$x^3-8x^2+x+42=0$$ with roots $-2, 3, 7$ and general solution $$a_n=A(-2)^n+B3^n+C7^n \tag{4}$$ Now, from $x_0=x_1=1 \Rightarrow a_0=a_1=1$ and from $(1) \Rightarrow a_2=56$ leading to $$a_0=1=A+B+C$$ $$a_1=1=-2A+3B+7C$$ $$a_2=56=4A+9B+49C$$ solving this system of linear equations leads to $$a_n=\frac{67}{45}(-2)^n-\frac{37}{20}3^n+\frac{49}{36}7^n \tag{5}$$ (can be validated here and here) and $$x_n=\sqrt{\frac{67}{45}(-2)^n-\frac{37}{20}3^n+\frac{49}{36}7^n}$$

Answer 2

Hint:

Let $a_n=x_n^2$ then:

$$a_{n}=a_{n-1}+6a_{n-2}+7^n$$

Which we will call equation $1$. Suppose $a_{n}=c7^n$ is one solution to the recurrence above (a particular solution). Then,

$$c7^{n}=c7^{n-1}+6c7^{n-2}+7^{n}$$

Subtracting this equation from equation $1$ gives,

$$\left(a_{n}-c7^n \right)=\left(a_{n-1}-c7^{n-1} \right)+6 \left(a_{n-2}-c7^{n-2} \right)$$

Letting $b_{n}=a_{n}-c7^{n}$ gives,

$$b_{n}=b_{n-1}+6b_{n-2}$$