Finding a characteristic equation of second order PDE? How to find the characteristic equation of the following PDE  
$$PDE: (\sin^2 {x} ) u_{xx}+ (\sin {2x})u_{xy}+(\cos^2x)u_{yy}=x$$
 A: Define $A = \sin^2 x$, $B = \sin{2x}$ and $C = \cos^2{x}$ so the solution to the characteristic equation reads:
$$\frac{\xi_x}{\xi_y} = \frac{-B  \pm \sqrt{B^2 -4AC}}{2A} = \ldots = -\cot{x}  $$
Note that your equation is parabolic since $B^2 - 4AC = 0$ (use the double angle formula), so the other characteristic may be choosen arbitrarily. Also note that for $x=0$, $\cot{x}$ blows up, but this is no problem since for $x=0$, the original PDE reduces to $u_{yy} = 0$.
A: Since it would be too long (and a waste of time) to repeat the theory which can be found in the books, the calculus below refers to a paper (pages 32-41) http://www.math.psu.edu/wysocki/M412/Notes412_5.pdf 
The symbols used here are exactly those used of the paper.
The second order linear PDE considered is :
$$a\;u_{xx}+ 2b\;u_{xy}+ c\;u_{yy}+ d\;u_{x}+ e\;u_{y}+fu=g$$
In the present case : 
$$\sin^2(x)u_{xx}+2\sin(x)\cos(x)u_{xy}+\cos^2(x)u_{yy}=x \quad \quad \text{(1)}$$
$a=\sin^2(x)$ ; $b=\cos(x)\sin(x)$ ; $c= cos^2(x)$ ; $d=e=f=0$ ; $g=x$
We observe that $b^2-ac=0$ and so, the PDE is of parabolic kind.
A change of variables $(x,y)\to (\xi,\eta)$ 
$$u(x,y)=u\left(x(\xi,\eta),y(\xi,\eta)\right)=w(\xi,\eta)$$ 
transforms $(1)$ into a PDE on the form :
$$A\; w_{\xi\xi}+2B\; w_{\xi\eta}+C\; w_{\eta\eta}+ D\; w_{\xi}+ E\; w_{\eta} = G$$
In the present case, there is no more term because $d=e=f=0$.
The method consists in choosing $\xi(x,y)$ and $\eta(x,y)$ so that  $A(\xi,\eta)= B(\xi,\eta)=0$ which will simplify a lot the PDE (i.e.: the canonical form). Following the principle explained, 
$$A=a\xi_x^2+2b\xi_x\xi_y+c\xi_y^2=a(\xi_x -\mu\xi_y)^2=0$$
where $\mu=-\frac{b}{a}=-\frac{\cos(x)}{\sin(x)}$
We look for the solution of the first order linear PDE 
$$\xi_x-\mu\xi_y=0$$
The solution $\xi$ is constant along each characteristic determined by :
$$\frac{dy}{dx}=-\mu= \frac{\cos(x)}{\sin(x)}$$
$$y-\ln\left(\sin(x)\right)=constant$$
This solution of the characteristic equation defines the variable $\xi$
$$\xi=y-\ln\left(\sin(x)\right)$$
Of course, we could chose any function $F$ and  $\xi=F\left(y-\ln\left(\sin(x)\right)\right)$, but why not choosing the simplest one ?
The second new variable $\eta(x,y)$ is chosen so that 
$$B=a\xi_x\eta_x+b\left(\xi_x\eta_y+\xi_y\eta_x \right)+c\xi_y\eta_y=0$$
In case of parabolic EDP, this condition is fulfilled insofar $A=0$ since with $B^2-AC=0$ which implies $B=0$. As a consequence, $\eta(x,y)$ can be any function except a function of $y-\ln\left(\sin(x)\right)$. Generally, in the examples given in the books, one simply choose $\eta=x$, or $\eta=y$. 
However, it is possible to achieve an ultimate simplification in taking advantage of the freedom in the choice of $\eta(x,y)$. For example, it is possible to chose $\eta$ so that $C=1$
$$C=a\eta_x^2+2b\eta_x\eta_y+c\eta_y^2=1$$ 
This is obtained with $\eta=\eta(x)$ and  $\sin^2(x)\left(\frac{d\eta}{dx}\right)^2=1$
$$\frac{d\eta}{dx}=\frac{1}{\sin(x)}$$
$$\eta=\int \frac{dx}{\sin(x)}$$
In summary, with the new variables
  \begin{cases}
   \xi=y-\ln\left(\sin(x)\right)  \\
    \eta=\int \frac{dx}{\sin(x)}=-\ln\left(\cot(x)+\csc(x) \right)  \\
  \end{cases}
and considering the homogeneous EDP $$\sin^2(x)u_{xx}+2\sin(x)\cos(x)u_{xy}+\cos^2(x)u_{yy}=0 $$
it is transformed into a very simple form of EDP:
$$\frac{\partial u}{\partial \xi}+\frac{\partial^2  u}{\partial \eta^2}=0$$
Of course, in order to find the solution of the non-homogeneous EDP $(1)$, one have finally to add a particular solution, for example $U(x)$ 
$$\sin^2(x)\frac{d^2U}{dx^2}=x$$
$$U(x)=\int\int \frac{x\: dx}{\sin^2(x)}dx $$
