I'm following a solution to a problem but I wanted to ask about a particular step. I have the following equation
$$ g(y)f'(x)=(C_1\cos(\sqrt{G}ky)+C_2\sin(\sqrt{G}ky))(C_3\sinh(kx)+C_4\cosh(kx)) $$
where $\sqrt{G},k>0$ are known constants, $C_1$, $C_2$, $C_3$, and $C_4$ are unknown constants that need to be found, and $g(y)f'(-a)=g(y)f'(a)=2y$ are the boundary conditions. By plugging in the BSs we get this system of equations,
$$ g(y)f'(-a)=(C_1\cos(\sqrt{G}ky)+C_2\sin(\sqrt{G}ky))(-C_3\sinh(ka)+C_4\cosh(ka))=2y $$$$ g(y)f'(a)=(C_1\cos(\sqrt{G}ky)+C_2\sin(\sqrt{G}ky))(C_3\sinh(ka)+C_4\cosh(ka))=2y $$
The author of the solution states that the RHS of both equations are odd (i.e. $2y$ is odd, which makes sense) therefore $g(y)$ is odd. This confuses me a bit because from what I can see $f'(x)$ is neither even or odd so is it a rule that an odd function times a neither equals an odd?
Then the author says that since $\cos(k)$ is an even function, $C_1=0$ then $g(y)=C_2\sin(\sqrt{G}ky)$. This makes sense if the above assumption is true. Then the author states that
$$ g(y)(f'(a)-f'(-a))=0 $$
hence $f'(a)$ is even. This also confuses me because how was the above equation made and how can it be stated that $f'(a)$ is even? Isn't it a contradiction because $f'(-a)$ should instead be odd? I thought that $f'(x)$ is neither even or odd.
The author then says that since $f'(a)$ is even and $\sinh(ka)$ is odd, $C_3=0$. This also makes sense if the statement on $g(y)(f'(a)-f'(-a))=0$ is true.