# Determining if a function is even or odd using a system of equations and solving for unknown constants

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.

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 $$C_3=0$$.

• 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? – enea19 Feb 11 '19 at 8:13
• Yes, it's pretty much correct. In more detail, add $C_3\sinh(ka)$ to both sides and divide by $2$. – zipirovich Feb 11 '19 at 15:17