But if the PDE is any hyperbolic second order linear PDE with constant coefficients (e.g. u_xx- 4 u_yy = 0), then my argument above works, right?
Then we can immeidately say that the general solution is u=f(y-2x)+g(y+2x), right?
b^2 - 4ac = x^2 >0 (since we are given x>0) as well
So it falls into your case, but it does not reduce to the canonical form u_rs =0. WHY?
Is there any reason why the canonical form is different for constant (e.g. u_xx- x^2 u_yy = 0) and non-constant coefficients (ue.g. _xx- 4 u_yy = 0) hyperbolic PDEs?
If the PDE is where A, B and C are functions of x and y, then with a change of variables is
Substituting and re-arranging gives
To transform to your hyperbolic form you want to choose and such that
noting that since (it's hyperbolic), you can factor these choosing one factor for and the other for . Both are linear PDEs!
With this choice, the lower order terms above (in blue) are typically not zero.