Are there any solutions to $2g(x+y)-g(x-y) =2g(x)g(y)$ with $g(0) \ne 0$?

Question

Are there any solutions to $2g(x+y)-g(x-y) =2g(x)g(y)$ with $g(0) \ne 0$?

This came up (by replacing $e^x$ with $g(x)$) in an attempt to generalize this: Solving functional equation gives incorrect function

I have easily shown that $g(x) = ae^x+b$ is not a solution.

I also have unsuccessfully tried to get a differential equation for $g(x)$.

Answer 1

Note that the RHS is symmetric in $x$ and $y$ but the LHS not. So we get $$2g(x+y)-g(x-y)=2g(x+y)-g(y-x)$$

Therefore $g(q)=g(-q)$ for all $q \in \mathbb R$.

Note that $g(0)=\frac12$.

Let $y=x$, then we get $2g(2x)-g(0)=2g(x)^2$

Let $y=-x$, then we get $2g(0)-g(2x)=2g(x)g(-x)=2g(x)^2$

Thus we get $2g(2x)-\frac12=1-g(2x)$, thus $g(2x)=\frac12$ for all $x$.

This function indeed statistifies the given equation.

Conclusion. $g(x)=1/2$ is the only function statistifying the functional equation. $g(x)=0$ is another solution when removing the requirement $g(0)\neq0$, and those are the only ones.

Answer 2

$g(x)=\frac 12$ is a solution, and I would assert is the only solution

This can be established by writing out the results of putting $x=0$, $ y=0$, $x=y$ and $x=-y$