How to find the solution of $u_x-u_y=u^2$ I want to find the solution $u(x,y)$ of the PDE
$$
  u_x-u_y=u^2 \quad \text{where} \quad u_x = \frac{\partial u}{\partial x}\,,~u_y = \frac{\partial u}{\partial u} \,
$$
The boundary condition at points on the boundary $\Gamma_u$ given by $y=2x-1$ is $u(x,y)=4$. 

My approach:
$\frac{dx}{dt}=1,\frac{dy}{dt}=-1, \frac{du}{dt}=u^2$
$x=t+a,y=-t+b,-u^{-3}=3(t+c)$
We want to construct the characteristic through each point of $\Gamma_u$. 
Suppose that a characteristic passes through $\Gamma_u$ at $P : (s,2s-1,4)$ at $t = 0$. 
Then
$x=a=s,y=b=2s-1,-4^{-3}=3c,c=-\frac{1}{192}$
$x=s,y=2s-1,u=4$
$x=t+s,y=-t+2s-1,-u^{-3}=3(t+-\frac{1}{192})$
We want to eliminate s and t from these equations.
$ y=-3t-1        $
$y=-3(\frac{-u^{-3}}{3}+\frac{1}{192})-1$ 
but answer at the end of my book says 
$u(x,y)=\frac{3}{y-2x+\frac{7}{4}}$
Could you tell me where the mistake is?

edit 1: $\frac{dx}{dt}=1,\frac{dy}{dt}=-1, \frac{du}{dt}=u^2$
$x=t+a,y=-t+b,-u^{-1}=t+c$
$x=a=s,y=b=2s-1,-4^{-1}=c$
$x=s,y=2s-1,u=4$
$x=t+s,y=-t+2s-1,-u^{-1}=t-\frac{1}{4})$
We want to eliminate s and t from these equations.
$s=x-t$,
$ y=2(x-t)-t-1        $
$y=-3(-u^{-1}+\frac{1}{4})+2x-1$ 
$u=\frac{3}{y+\frac{7}{4}-2x}$ 
thanks

Here's an outline of the solution. Use the method of characteristics and get by solving ODEs $x(t) = t + x_{0}$, $y(t) = -t + y_{0}$ and $u(t) = \frac{-1}{t+c}$. Now let $-t + y_{0} = y(t) = 2x(t) - 1 = 2t + 2x_{0} - 1$. Then $t = \frac{y_{0}-2x_{0}+1}{3}$. We then have $c = -1/4 - \frac{y_{0}-2x_{0}+1}{3} = -7/12 - (y+t)/3 + 2(x-t)/3$. So, $t+c = -7/12  - y/3 + 2x/3$. Finally, we have $u(x, y) = -(-7/12 - y/3 + 2x/3)^{-1} = \frac{3}{7/4 + y - 2x}$.
