Solving the functional equation $f(x + y) + g(x-y) = lambda g(x) f(y)$

Let $\lambda$ be a nonzero real constant. Find all functions $f,g : \mathbb R \rightarrow \mathbb R$ that satisfy the functional equation
for all $x,y \in\Bbb R$:
$$f(x + y) + g(x-y) = \lambda g(x) f(y)$$

I tried this:

$y=0$ implies $f(x)+g(x)$=$\lambda g(x)f(0)$.
Here we have two cases depending on $f(0)$:

1. $f(0)=0$ or


2. $f(0) \neq 0$.

If this is true, what next?

Answer

Let $x=y=0$ then the equation becomes
$$
f(0)+g(0)=λg(0)f(0)
$$
Case I. If $f(0)=0$ then $g(0)=0$ and vice versa. Setting $y=0$ in the original equation gives
$$
f(x)+g(x)=\lambda g(x)f(0)=0 \\
\therefore\:f(x)=-g(x)
$$
Similarly setting $x=0$ we get $f(y)=-g(-y)$ and so $f(x)=-g(x)=-g(-x)$. Now setting $y=x$ in the equation we get

$$
f(2x)+g(0)=\lambda g(x)f(x) \\
\therefore\:f(2x)=-\lambda f^2(x) \\
\therefore\:f(x) \text{ must be of the form } ca^x
$$
Substituting in above equation we get that either $c=0 \mbox{ or } c=-1/\lambda$. $c=0 \mbox{ gives } f(x)=g(x)=0$ for all $x$. For the other value of $c$ we would need $a=0$ to satisfy $f(0)=0$. Even if we assume that $f(x)$ is discontinuous at $x=0$ plugging $f(x)=ca^x$ in the above equation results in $a^y=0$ for all $y$. Hence $a$ has to be zero.

Case II. If $\lambda f(0)=1$ this would result in $f(0)=0$. So $g(0)=0$ and the analysis becomes the same as above

Case III. $f(0)\ne0$, $g(0)\ne0$, $\lambda f(0)\ne1$ and $\lambda g(0)\ne1$.

From setting $x=0,y=0$ we get
$$
\lambda = \frac{1}{f(0)}+\frac{1}{g(0)}
$$
Set $y=0$ and the equation becomes
$$
f(x)+g(x)=\lambda g(x)f(0) \\
\therefore\:f(x)=g(x)(\lambda f(0)-1) \\
=g(x) \frac{f(0)}{g(0)}
$$

Similarly setting $x=0$ yields
$$
g(-y)=f(y)(\lambda g(0)-1) \\
\therefore\:g(x)=f(-x)(\lambda g(0)-1) \\
=f(-x)\frac{g(0)}{f(0)}
$$
Combining the above two equations yields
$$
f(x)=f(-x)
$$

The original equation then becomes
$$
f(x+y)+f(x-y)\frac{g(0)}{f(0)}=\lambda f(x)\frac{g(0)}{f(0)}f(y)
$$
Replacing $y$ by $-y$ we get
$$
f(x-y)+f(x+y)\frac{g(0)}{f(0)}=\lambda f(x)\frac{g(0)}{f(0)}f(y)
$$
This implies that either $f(0)=g(0)$ or $f(x+y)=f(x-y)$. The second scenario results in $f(x)=g(x)=2/\lambda$. If $f(0)=g(0)$ the equation reduces to
$$

f(x-y)+f(x+y)=\lambda f(x)f(y)
$$ The only way to satisfy this is if $f(x)=\text{a constant}$. Hence in this case too $f(x)=g(x)=2/\lambda$.