## Classifying PDE's

Hello,

I want to find the regions of the complex plan where the pde given by the equation $y u_{xx} -2u_{xy}+c^xu{yy}+x^2u_x-u=0$ is hyperbolic, parabolic, and elliptic.

I should just have to evaluate the discriminant, $b^2-ac$ (we wrote out our initial generic PDE as $a u_{xx} +2bu_{xy}+c^xu{yy}+du_x +eu_y+fu+g=0$ so there's no 4 in the discriminant) which is $$(-2)^2-yc^x=4-yc^x.$$ I'm starting to think I'm doing this wrong because the calculations are becoming horribly complicated. The discriminate would be $$4-yc^x.$$ Assume that $c \neq 0$.

First, $4-yc^x=0 \iff y=4 c^{-x}$ describes when the pde is parabolic.

Second, $4-yc^x<0$ decribes when the pde is elliptic. To get a better determination you have to argue on the parity of $c$. If $c>0$ then again we have, since $c^x>0 \forall x \in \mathbb{R}$, $y>4/c^x$. If instead $c<0$ then $$c^x=\begin{cases}c^x<0 & \text{when } x \text{ is odd} \\ c^x>0 & \text{when } x \text{ is even} \end{cases}.$$

On $x$ even we get $y>4/c^x$ as before. On $x$ odd we'd have $c^x<0$ so that $4-yc^x<0$ gives $4+|yc^x|<0$ when $y>0$ so $4<-|yc^x|$. If instead $y<0$ then we would have $4-|yc^x|<0$ or in other words $4<|yc^x|$.

I haven't even done the case where $4-yc^x>0$ yet...

Does this look vaguely correct?