Let $f:\mathbb{R} \to \mathbb{R}$ be a function which satisfies $e^xf(y)+e^yf(x)=2e^{x+y}-e^{x-y}$ for all real $x$ and $y$.
If I place $x=y$, I get $f(x)=e^x-\frac{1}{2}e^{-x}$ which does not satisfy the original equation.
Now instead, if I set$x=y=0$, I get $f(0)=\frac{1}{2}$. Now setting $y=0$ in the original equation, I get $f(x)=\frac{e^x}{2}$.
My question is that why do we get incorrect and different solutions if we proceed in the ways mentioned above? Also, what is the correct solution of this functional equation and how can I find it?
Suppose $g(x)f(y)+g(y)f(x)=2g(x+y)-g(x-y) $ where $g(0) \ne 0$.
Setting $x=0$, $g(0)f(y)+g(y)f(0)=2g(y)-g(-y) $.
Setting $y=0$, $g(x)f(0)+g(0)f(x)=2g(x)-g(x) =g(x) $ or $g(0)f(x) =g(x)(1-f(0)) $.
Setting $x=y=0$, $g(0)f(0) =g(0)(1-f(0)) $. Since $g(0) \ne 0$, $f(0) = \frac12$ and $f(x) =\frac{g(x)}{2g(0)} $.
Putting this back,
$\begin{array}\\ 2g(x+y)-g(x-y) &=g(x)f(y)+g(y)f(x)\\ &=g(x)\frac{g(y)}{2g(0)}+g(y)\frac{g(x)}{2g(0)}\\ &=\frac{g(x)g(y)}{g(0)}\\ or\\ g(0)(2g(x+y)-g(x-y)) &=g(x)g(y)\\ \end{array} $.
Setting $x=y$, $g(0)(2g(2x)-g(0)) = 2g^2(x) $. Setting $x=0$, $g(0) = 2g^2(0) $. Since $g(0) \ne 0$, $g(0) = \frac12 $.
Therefore $2g(x+y)-g(x-y) =2g(x)g(y) $.
This does not hold for $g(x) = ae^x+b$, so the original problem has no solution.
I don't know how to solve this, so I'll propose it as a question.
And here it is: Are there any solutions to $2g(x+y)-g(x-y) =2g(x)g(y)$ with $g(0) \ne 0$?