Solving simple functional relation

Question

Satisfying the boundary conditions

$$ y(0)=1, y(1)=2 $$

What general /particular functions obey $$ 1) \quad y(x) \,y(x+1)= 2,$$

and

$$2)\quad \dfrac{y(x)}{y(x+1)}=\dfrac{1}{2}? \;$$

Answer 1

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.