Solution to first order partial Cauchy problem? Solve the Cauchy Problem :
$$(x+y)u_{x} + (x-y)u_y = 1$$ with $ \; \; u(1,y)=\frac{1}{\sqrt{2}}$
My attempt:
We have the following characteristic equations : 
$$\frac{dx}{x+y} = \frac{dy}{x-y} = \frac{du}{1}$$
Solving
$$\frac{dx}{x+y} = \frac{dy}{x-y} \Rightarrow xdx - ydy = xdy + ydy \Rightarrow x^2 - y^2 -2xy = c_1 $$ where $c_1$ is a constant
I am unable to find another pair of characteristic equations to solve.
In order to reduce it to canonical form, if I use transformation $\phi(x,y)= x^2 - y^2 -2xy $ and $\gamma(x,y)= y$ 
I am getting $(x+y)u_{\gamma} = 1$ which is again of no help.
How do I proceed?

Your first characteristic equation is correct :
$$x^2-y^2-2xy=c_1$$
From this : $\quad x-y=\pm\sqrt{c_1+2y^2}$
Second characteristic, from $\quad\frac{dy}{x-y}=\frac{du}{1}=\frac{dy}{\pm\sqrt{c_1+2y^2}}$
$u-\int\frac{dy}{\pm\sqrt{c_1+2y^2}}=c_2$
$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2(c_1+2y^2)}+2y\bigg|=c_2$
$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2(x^2-y^2-2xy+2y^2)}+2y\bigg|=c_2$
$$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2}(x-y)+2y\bigg|=c_2$$
