Is the PDE $$ (1+x^2)^2\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}+2x(1+x^2)\frac{\partial u}{\partial x}=0\text{ in }\Omega:=\mathbb{R}^2 $$ hyperbolic?
To answer this I set
$a(x,y):=(1+x^2)^2, b(x,y):=0, c(x,y)=-1$, then $b^2-ac=(1+x^2)^2 >0$
and this we had as a criterion for hyperbolicism.
Am I right?
You are right but $\Delta(x, y)=b^2-4ac$. If we have a second-order PDE $$a(x, y)u_{xx}+b(x ,y)u_{xy}+c(x, y)u_{yy}+ F(x, y, u, u_x, u_y)=0$$ (called semilinear equation), then we can classify the eqn. according to the sign of $\Delta(x, y).$ It is hperbolic on $\Omega$ if $\Delta(x, y)>0$ , parabolic on $\Omega$ if $\Delta(x, y)=0$ and elliptic on $\Omega$ if $\Delta(x, y)<0$ for all $(x, y)\in\Omega$ .