Solve $$(y-u)u_x + (u-x)u_y = x-y, \qquad u=0 \text{ when } xy=1.$$
I tried to solve the equation above using characteristic method $$\begin{cases} x'=y-u \\ y' = u - x \\ u' = x-y \end{cases}. \tag{1}$$
Adding together the first and the second equation in $(1)$ we get that $x'+y'=y-x$. Thus (using third equation in $(1)$) $$u = x+ y + C \implies C = u - x- y.$$
Solving the first and the second equation yields to $$\frac{1}{2} y^2 - uy = ux - \frac{1}{2}x^2 + \tilde{C} \implies \tilde{C} = \frac{1}{2}(y^2 +x^2)-u(x+y).$$
That gives us $$F\big(u-x-y, \frac{1}{2}(y^2 +x^2)-u(x+y) \big) = 0.$$
How can I finish this example?
$$(y-u)u_x+(u-x)u_y=x-y$$
System of characteristic ODEs (Charpit-Lagrange) : $$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}$$ A first characteristic equation comes from :
$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}=\frac{dx+dy+du}{(y-u)+(u-x)+(x-y)}=\frac{dx+dy+du}{0} \quad\implies\quad dx+dy+du=0$
$$u+x+y=c_1$$
A second characteristic equation comes from :
$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}=\frac{xdx+ydy+udu}{x(y-u)+y(u-x)+u(x-y)}=\frac{xdx+ydy+udu}{0} \quad\implies\quad xdx+ydy+udu=0$
$$x^2+y^2+u^2=c_2$$
General solution of the PDE expressed on the form of implicit equation $\Phi(c_1,c_2)=0$ or equivalently $c_2=F(c_1)$ :
$$x^2+y^2+u^2=F(x+y+u)$$ $\Phi$ or $F$ are undetermined functions until no condition is specified.
With condition $u=0$ when $xy=1$ :
Then $x^2+y^2+0=F(x+y+0)=x^2+\frac{1}{x^2}=F(x+\frac{1}{x})$
Let $X=x+\frac{1}{x} \quad\implies\quad x=\frac12\left(X\pm\sqrt{X^2-4}\right)$
$F(X)=\frac14\left(X\pm\sqrt{X^2-4}\right)^2+\frac{4}{\left(X\pm\sqrt{X^2-4}\right)^2}$ and after simplification : $$F(X)=X^2-2$$ Now $F(X)$ is determined. We put it into the above general solution where $X=x+y+u$ Then $F(x+y+u)=(x+y+u)^2-2$ .
$$x^2+y^2+u^2=(x+y+u)^2-2$$ Solving for $u$ leads to : $$\boxed{u(x,y)=\frac{1-xy}{x+y}}$$
NOTE:
Another method consists in the change of variables $\zeta=x+y$ ; $\eta=xy$ which leads to a separable PDE.