TITLE: characteristic curves for second-order equations
QUESTION [12 upvotes]: Reading about characteristic curves for second-order equations, in particular semi-linear equations of second order with two independent variables: 
$a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{y,y}=f(x,y,u,u_{x},u_{y})$     $   $  $  $    $(1)$  
My book reference, define characteristic curve to $(1)$, as plane curves along which the PDE can be written in a form containing only total derivatives of $u_{x}$ and $u_{y}$. 
I do not understand this definition (only total derivatives of $u_{x}$ and $u_{y}$??? ), I do not know how to see it this, for example, in equation
$xu_{xx}+2xu_{xy}+xu_{yy}=u_{x}+u_{y}$.  
I read some questions about characteristic curves here, but did not help.
Can anyone help me?
Thank you.

REPLY [5 votes]: They just mean that the PDE can factored into something of the form $\prod_i\left(a_i\partial_t^{\alpha_i} +b_i\partial_x^{\beta_i}+c_i \right)$, where $\beta_i,\alpha_i\in\mathbb{N}$. So for example, the heat equation is not of this form, but the wave equation is. And once it's in this form, you just solve this nested sequence of characteristic problems.
If you want an example, see how Evans derives the solution for the easiest form of wave equation.
Since you haven't accepted, here are all the details. The example PDE you posted can be factored as $$(\partial_x+\partial_y)(x\partial_x+x\partial_y-1)u=0$$
Solve it by letting $v(x,y):=(x\partial_x+x\partial_y-1)u$. Then we have $v_x+v_y=0$ Solve this for $v$, and then solve $v=(x\partial_x+x\partial_y-1)u$ for u.