Whispering wall function After visiting the Whispering wall where a whisper can be clearly transmitted between two locations on the surface of a dam wall over more than 100 metres, I was puzzled as to how this occurs. I decided to try to solve the following problem: What must be the shape of a curved wall such that any sound emitted towards the wall from one corner will be reflected directly to the opposite corner as in the following diagram (co-ordinates are $x$ and $y$)?

From this I derived the following differential equation (derivation below) but myriad substitutions and changes of variable have still left me puzzled as how to put it into a form which can be solved:
$$(y’)^2++y’\left(\frac{a^2+y^2-x^2}{ay}\right)+\frac{x}{a}=0$$
for $-a<x<0$ with $y(\pm a)=0$ and $y’(0)=0$, but I’m not sure that this is right. Can solve this equation or else find a mistake in it and find the correct equation to solve my geometrical problem?
Derivation (note that from the answers it would appear that this is slightly incorrect): Taking the following horrible paint diagram where $D$ is the point at which the sound wave is reflected and $AB$ is the tangent to the curve at that point, where we must have that the angles $BDC$ and $ADF$ marked $\theta$ are equal since a specular reflection is assumed to occur:

(Note that this is for negative $x$ only so $-x>0$; the vertical line is through the maximum of the curve). I then take the triangle $DBC$ and divide the lengths of the sides by $a-x$ to find that (using the relation between the tangent and the derivative):
$$\theta=\arctan{y’}+\arctan{\frac{y}{a-x}}$$
I then take the triangle $ADE$ and divide the lengths by $y$ to obtain:
$$\theta=\arctan{\frac{1}{y'}}-\arctan{\frac{a+x}{y}}$$
Setting these equal and taking $\tan$ of both sides and using the law for the tangent of a sum of angles to obtain two fractions which I cross-multiply and then collecting terms I get the differential equation shown above.
(Note that interesting facts about similar walls which might better explain the Whispering wall I originally mentioned can be found here, here and here, but I am now specifically looking for a solution to my problem. Also not that the Wikipedia page in my very first link mentioned a ‘parabolic effect’, but I could not see how that would be relevant to my problem)
 A: $$\arctan y'-\arctan \frac{y}{x-a}=\arctan \frac{y}{x+a}- \arctan y'$$
$$2\arctan y'=\arctan \frac{y}{x-a}+\arctan \frac{y}{x+a}$$
$$\frac{2y'}{1-y'^2}=\frac{\frac{y}{x-a}+\frac{y}{x+a}}{1-\frac{y^2}{x^2-a^2}}$$
$$\frac{2y'}{1-y'^2}=\frac{\frac{2xy}{x^2-a^2}}{\frac{x^2-a^2-y^2}{x^2-a^2}}$$
$$y'(x^2-y^2-a^2)=xy(1-y'^2)$$
$$y'^2+\frac{x^2-y^2-a^2}{xy}y'-1=0$$
A: Following @Djura Marinkov's remark, the correct equation to study is
\begin{equation}
y'^2 + \frac{x^2 - y^2 - a^2}{x y} y' -1 = 0. \tag{1}
\end{equation}
Using the substitution $y(x)^2 = z(\eta(x))$ where $\eta(x) = x^2$, we obtain
\begin{equation}
z'^2 + \left(1-\frac{z+a^2}{\eta}\right) z' - \frac{z}{\eta} = 0, \tag{2}
\end{equation}
which looks marginally easier than $(1)$, but not much. However, this changes when we take the derivative of $(2)$ to $\eta$: then we obtain
\begin{align}
 z'' =& \frac{-z(1+z') + z'(-a^2 + \eta(1+z'))}{\eta(-a^2+\eta-z+2 \eta z')}\\
 =& \frac{z'^2 + \left(1-\frac{z+a^2}{\eta}\right)z'-\frac{z}{\eta}}{-a^2+\eta-z+2 \eta z'}\\
=& 0,
\end{align}
where we used $(2)$ to obtain the last line. Therefore, $z$ must be a linear function, i.e.
\begin{equation}
 y^2 = \alpha x^2 + \beta \tag{3}.
\end{equation}
You can now substitute $(3)$ in $(1)$ to find (a relation between) $\alpha$ and $\beta$, in terms of $a$. Note that this answer reflects (yes) some useful properties of ellipses, see here, here or here.
