How to solve $ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$ via method of characteristics? How to solve $ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$ via method of characteristics?
My attempt.
These are equations with which I begin:
$\dfrac{dx}{ds} = y+u $;
$\dfrac{dy}{ds} = x+u $;
$\dfrac{du}{ds} = x+y $.
However, I am stuck, because I can not solve the equations for $\dfrac{dx}{ds}$ and $\dfrac{dy}{ds}$ because we have a dependence on $u$.
Thanks for any help.

$$ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$$
Charpit-Lagrange system of characteristic ODEs :
$$ds=\frac{dx}{y+u}=\frac{dy}{x+u}=\frac{du}{x+y}=\frac{dx-dy}{(y+u)-(x+u)}=\frac{dx+dy+du}{(y+u)+(x+u)+(x+y)}$$
$$\frac{dx-dy}{y-x}=\frac{dx+dy+du}{2(x+y+u)}$$
$$-\ln|x-y|=\frac12\ln|x+y+u|+\text{constant}$$
A first characteristic equation is :
$$(x+y+u)(x-y)^2=c_1$$
A second characteristic equation comes from
$$\frac{dx}{y+u}=\frac{dy}{x+u}=\frac{du}{x+y}=\frac{dx-du}{(y+u)-(x+y)}=\frac{dy-du}{(x+u)-(x+y)}$$
$$\frac{dx-du}{u-x}=\frac{dy-du}{u-y}$$
$$\ln|u-x|=\ln|u-y|+\text{constant}$$
A second characteristic equation is :
$$\frac{u-x}{u-y}=c_2$$
The general solution of the PDE expresed on the form of implicit equation $c_2=F(c_1)$ is :
$$\boxed{\frac{u-x}{u-y}=F\big((x+y+u)(x-y)^2\big)}$$
F is an arbitrary function until no boundary condition is specified.
Note that the same general solution could be expressed on a number of equivalent forms, for example $u=-x-y+\frac{1}{(x-y)^2}G\left(\frac{u-x}{u-y}\right)$ where G is  an arbitrary function.
Of course the PDE has an infinity many solutions. Among them the linear one : $u=\frac{1}{1-c}(x-cy)$ which coresponds to the above second characteristc equation. Or for example another solution $u=-x-y+\frac{c}{(x-y)^2}$ which corresponds to the above first characteristic equation.
Depending on the kind of boundary condition, the function F could be (or not) determined explicitely. Then putting it into the above general solution the equation could be (or not) solved explicitely for $u$.
