Solving simple functional relation Satisfying  the boundary conditions
$$ y(0)=1, y(1)=2 $$
What general /particular functions obey
$$ 1) \quad y(x) \,y(x+1)= 2,$$
and
$$2)\quad \dfrac{y(x)}{y(x+1)}=\dfrac{1}{2}? \;$$
 A: Although not explicitly mentioned in the OP, should we take 1) and 2) simultaneously or separately, taking this equations simultaneously don't make sense, as we have $\left(y(x)\right)^2=1$ after multiplying them, and this holds for any $x$, together with $y(1)=2$ makes a contradiction. So we consider 1) and 2) separately.
Following this answer after taking log of both parts we have
$$\hbox{1) }f(x)+f(x+1)=\ln 2\\
\hbox{2) }f(x)-f(x+1)=-\ln 2$$
First we homogenize it:
$$\hbox{1) }f(x)+f(x+1)=0\\
\hbox{2) }f(x)-f(x+1)=0$$
For the second it's clear that an arbitrary periodic function with the period $\frac{1}{n},\,n\in \mathbb{N}$ will do.
Then we set $f(x)=e^{i\pi x}g(x)$ for the first one:
$$e^{i\pi x}g(x)-e^{i\pi x}g(x+1)=0$$
So we have $f(x)=e^{i\pi x}g(x)$ for arbitrary periodic $g(x)$ with period of $\frac 1n$.
Now, as the particular solutions should be linear to have $f(0)=0,\,f(1)=\ln 2$ we're mostly done.
Answer:
$$\hbox{1) }y(x)=2^{x-1}\cdot e^{(e^{i\pi x})}\cdot g(x)\\
\hbox{2) }y(x)=2^{x-1}\cdot g(x)$$
for an arbitrary periodical function $g(x)$ with period $\frac 1n,\,n\in\mathbb{N}$.
And I'll add Linear Difference and Functional Equations with One Independent Variable for further reference if I have mistaken somewhere, that link could be of more help.
