Given the general partial differential equation
$$a \frac{\partial^2u}{\partial x^2} + 2b \frac{\partial^2 u}{\partial x \partial y} + c\frac{\partial^2u}{\partial y^2} = 0,$$
if $ac - b^2 > 0$, I want to show that the substitution $s = \frac{bx - ay}{\sqrt{ac - b^2}}$ and $t = y$ changes the above equation to
$$\frac{\partial ^2 u}{\partial s} + \frac{\partial^2u}{\partial t}.$$
My try:
So, just from inspection, it seems like somehow $c$ becomes $1$ because the last term in the original PDE just becomes the $\partial^2 u/\partial t$ term. Also, it looks like $b = 0 $ and $a = 1$ by similar reasoning. I have no clue how to prove this though. Also, I'm guessing that the entire $s$ term becomes $x$. I've been struggling with trying to prove this fact for a while now.
If you compute the second partials using the chain rule, starting with $$u_{x} = u_ss_x+u_tt_x$$ and likewise for $y$, you can substitute these into the PDE and the middle term will drop out