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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.

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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?

No, there's no such rule because that would be a false statement: "odd" times "neither" isn't going to be odd (e.g., consider $x\cdot(x+1)$). But note that you do NOT have "a function times another function" here; instead, the second factor here is just a number, either $f'(a)$ or $f'(-a)$ in the two equations, obtained by plugging in a fixed value $a$ or $-a$ into $f'(x)$. So this statement is simply an observation that if you multiply or divide an odd function by a nonzero number, you get an odd function: for a constant $c\ne0$, $g(y)$ is odd iff $cg(y)$ is odd.

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?

To be honest, I only partially understand this part. Here's what I do understand. First of all, this equation comes from subtracting the two equations above from each other: $$g(y)f′(a)−g(y)f′(−a)=g(y)(f′(a)−f′(−a))=0$$ and subtracting the right-hand sides gives $0$. I presume the context suggests that $g(y)$ is not identically zero (it depends on $y$, and even if its value is zero for some $y$, it's not zero for all $y$), therefore $f′(a)−f′(−a)=0$ has to be true, which is equivalent to $f′(a)=f′(−a)$. This looks precisely like the definition of being even …

… but to me it looks that this equality $f′(a)=f′(−a)$ has been established for one input value $x=a$ only, while an even function must satisfy it for all $x$ in the domain. Unless $a$ can vary, I don't see how this argument is valid.

Or maybe it's just a poor exposition for an actually true conclusion? In fact, we don't need $f'$ to be an even or odd function. The condition $f′(a)=f′(−a)$ does not imply that, but it does imply what we want: that $C_3=0$. Here's how: $$\begin{gather} f′(a)=f′(−a) \\ C_3\sinh(ka)+C_4\cosh(ka)=−C_3\sinh(ka)+C_4\cosh(ka) \\ C_3\sinh(ka)=−C_3\sinh(ka) \end{gather}$$ from which $C_3\sinh(ka)=0$. Note that $\sinh(x)=0$ only at $x=0$. I presume from the context we know that $a\ne0$, therefore $\sinh(a)\ne0$, therefore $C_3=0$.

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    $\begingroup$ Thanks, it clarified things. I forgot to specify that the values for $x$ and $y$ are $-a<x<a$ and $-a<y<a$. One question though, I'm not clear on how you went from $C_3 \sinh(ka)=−C_3 \sinh(ka)$ to $C_3 \sinh(ka)=0$. To get to $C_3=0$ I set $C_3=-C_3$ therefore $C_3=0$ since this is the only answer that makes sense. Is that correct? $\endgroup$ – enea19 Feb 11 '19 at 8:13
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    $\begingroup$ Yes, it's pretty much correct. In more detail, add $C_3\sinh(ka)$ to both sides and divide by $2$. $\endgroup$ – zipirovich Feb 11 '19 at 15:17

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