Solution of an initial value problem (MCQ) (CSIR DEC 2015) The solution of the initial value problem
$ (x-y) u_{x} + (y-x-u) u_{y} = u $
with the initial condition $u(x,0) = 1$ satisfies


*

*$ u^2(x-y+u) + (y-x-u) = 0$

*$ u^2(x+y+u) + (y-x-u) = 0$

*$ u^2(x-y+u) - (x+y+u) = 0$

*$ u^2(y-x+u) + (x+y-u) = 0$


This is what I am able to do
The characteristic equations are :
$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} $
From this we get
$dx + dy + du = 0$
Integrating we get a characteristic curve
$x+y+u =C_{1}$
I am unable to get a second characteristic curve. 
Please help.
The correct answer is 2.
Thanks in advance!

Second characteristic curve :
$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} =\frac{dx-dy+du}{(x-y)-(y-x-u)+u} = \frac{d(x-y+u)}{2(x-y+u)}  = \frac{du}{u} $
$\frac{1}{2}\ln|x-y+u|-\ln|u|=$constant $\quad\to\quad \frac{x-y+u}{u^2}=C_2$
The implicit solution is $\quad x+y+u=F\left(\frac{x-y+u}{u^2}\right) \quad $any differentiable function $F$. 
The condition $u(x,0)=1$ determines the function $F(x+1)=x+1 \quad\to\quad F(X)=X$.
$F\left(\frac{x-y+u}{u^2}\right)=\frac{x-y+u}{u^2}$
$$x+y+u=\frac{x-y+u}{u^2}$$
