Solving $ A\frac{\partial z}{\partial x} + B \frac{\partial^2 z}{ \partial x \partial y} + C \frac{\partial^3 z}{\partial x \partial^2 y} = 0 $? Non-mathematician here trying to find a hopefully analytic solution or any constructive directions  for solving differential equations of this particular form: Take a function $z(x,y)$, is there any solution structure for
$ A\frac{\partial z}{\partial x} + B \frac{\partial^2 z}{ \partial x \partial y} + C \frac{\partial^3 z}{\partial x \partial^2 y} = 0 $?
$A,B,C \in \Bbb{R} $
Thanks
 A: Here's an insight that could help you.
We could fix $x$ and define $u_x(y) := \partial_xz(x,y)$. According to Schwartz theorem, if $z$ is at least two times continuously differentiable, $u_x$ verifies:
$$
Au_x + Bu_x' + Cu_x'' = 0
$$
The general solutions of this equation are known and depend on the quantity $B^2 - 4AC$, look for "second order linear differential equation" on the web if you don't know the analytical solutions.
With that, you will find a general form of $u_x$, but it will not be sufficient for the next step if you don't have any initial or boundary conditions on $\partial_xz$ as a function of $y$.
Provided that you found your exact analytical solution of $u_x$, you can integrate with respect to $x$ and find your function $z$, again with boundary or initial conditions on $z$ as a function of $x$.
A: If you are satisfied with any solution, then you can consider
$$
Z(x,y)=X(x)Y(y).
$$
Substituting this into the equation you will get that for Y that satisfies
$$
 AY+BY’+CY’’=0
$$
Any X will work fine (except the constant, for a constant $X$ any $Y$ is ok).
In fact, at least for $z$ defined on a convex connected set that is pretty much all. To see that rewrite the equation as
$$
\partial_x(Az+B\partial_y z+C \partial_y^2 z)=0.
$$
It follows from this that
$$
Az+B\partial_y z+C \partial_y^2 z = f(y),
$$
for an arbitrary function $f$. If $z$ is independent on $x$ then we can get any dependence on $y$ tuning $f(y)$. If $z$ is $x$ dependent, then you can start solving this equation thinking about $x$ as about a parameter, it will enter the solution via the coefficients of homogeneous solutions, i.e. $z$ will be a sum of $X(x)Y(y)$ terms. Thus, solutions above are not just any solution but all you need to build up a general solution.
