Non-linear coupled PDEs $(\alpha_y/\beta)_y = K\alpha\beta = -(\beta_x/\alpha)_x$ Studying an engineering problem I came up with the following system
$$
\begin{align}
\frac{\partial}{\partial y}\left(\frac{1}{\beta}\frac{\partial\alpha}{\partial y}\right)&=K\:\alpha\beta\\
\frac{\partial}{\partial x}\left(\frac{1}{\alpha}\frac{\partial\beta}{\partial x}\right)&=-K\:\alpha\beta,
\end{align}
$$
for the functions $\alpha=\alpha(x,y)$ and $\beta=\beta(x,y)$, and $K$ a fixed, real constant. At first it doesn't look like a general solution in terms of initial values of $\alpha$ and $\beta$ can be written, but maybe there's some transformation and/or change of variables that may help.
 A: Expanding the differentials gives us the coupled differential equations
$$
\begin{cases}
\alpha_{yy} - \frac{\beta_y}{\beta} \alpha_y - K\beta^2 \alpha = 0 \\
\beta_{xx} - \frac{\alpha_x}{\alpha} \beta_x + K \alpha^2 \beta = 0
\end{cases}
$$
which are both of second order with non-constant coefficients. The first equation might be viewed as an $x$-dependent ODE of the variable $y$ with variable coefficients, whereas the second equation might be viewed as a $y$-dependent ODE of the variable $x$ with variable coefficients. Formally, we need to solve two coupled differential equations, where the coupling takes place in the expression of the coefficients.
Case $K=0$. I don't know about a general theory for this kind of problem either. Nevertheless, in the special case $K=0$, we end up with the PDE system $\alpha_{yy}/\alpha_y = {\beta_y}/{\beta}$ and $\beta_{xx}/\beta_x = \alpha_x/\alpha$. Thus, we get the first-order differential system
$$
\begin{cases}
\alpha_y(x,y) = f(x)\, \beta(x,y) \\
\beta_x(x,y) = g(y)\, \alpha(x,y)
\end{cases}
$$
with arbitrary variable coefficients $f$, $g$. For instance, if $g$ is identically zero, then solutions are trivial: we find
$$
\alpha(x,y) = f(x) G(y) ,
\qquad
\beta(x,y) = G'(y)
$$ for arbitrary $G$, which is a separable solution (a similar solution is found if $f\equiv 0$). Otherwise, both functions $f$, $g$ are not identically zero, and we have $\beta = \alpha_y/f$ as well as $\alpha = \beta_x/g$. Substitution in the above system's l.h.s leads to
$$ \textstyle
\beta_{xy} - \frac{g'}{g} \beta_x = fg\, \beta , \qquad
\alpha_{xy} - \frac{f'}{f} \alpha_y = fg\, \alpha .
$$
Finally, we have managed to decouple the initial coupled PDEs into a set of two linear second-order PDEs of hyperbolic type. Hope you can take it from here. If $K$ is a small parameter, then one might expand an approximate solution based on the solution obtained for $K=0$ using asymptotic analysis.
