Suppose $f$ is a given smooth function on $\mathbb{R}^2$. I want to show that for $a,b,c \in \mathbb{R}$ such that $b^2 - ac > 0$ there exists a smooth function $u$ such that $$ a\frac{\partial^2 u}{\partial x^2} + 2b\frac{\partial ^2 u}{\partial x \partial y} + c\frac{\partial^2 u}{\partial y^2} = f $$ My attempt at solving this is as follows: since $b^2 - ac > 0$, there exist $p,q,r,s \in \mathbb{R}$ such that $$ a\frac{\partial^2 u}{\partial x^2} + 2b\frac{\partial ^2 u}{\partial x \partial y} + c\frac{\partial^2 u}{\partial y^2} = XYu $$ where $X := \left(p\frac{\partial}{\partial x} - q\frac{\partial }{\partial y}\right)$ and $Y := \left(r\frac{\partial }{\partial x} - s\frac{\partial }{\partial y}\right)$. We can then easily check that $[X,Y] = 0$ so $X$ and $Y$ are commuting vector fields on $\mathbb{R}^2$ and therefore any point in $\mathbb{R}^2$ is the center of a coordinate chart $(s,t)$ for which $X = \frac{\partial }{\partial s}$ and $Y = \frac{\partial }{\partial t}$. Then the equation above becomes $\frac{\partial^2 u}{\partial s\partial t} = f$. Integrating twice with respect to $t$ and then $s$ yields a local solution $u$.
I'm having a problem patching these local solutions into a global solution though. Can anyone give me a hint on how to do that please? Is this even a good approach?
$a, b, c$ are constants, so the coordinate transformations that make $X = \partial/\partial s$ and $Y = \partial/\partial t$ are linear and global. Thus integrating with respect to $t$ and $s$ gives you a global solution.