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.
 A: Your differential equations are linear, with constant coefficients, for which there are well-established methods of solution. First write them in matrix form as:
$$
\pmatrix{\frac{dx}{ds}\\
    \frac{dy}{ds}\\
   \frac{du}{ds}}=\pmatrix{0&1&1\\
                           1&0&1\\
                           1&1&0}\pmatrix{x\\y\\u}\ .
$$
Next, find the eigenvalues of the matrix $\ \pmatrix{0&1&1\\1&0&1\\1&1&0}\ $, which are $\ -1,-1,2\ $, and a complete set of eigenvectors, which can be taken to be
$$
\pmatrix{1\\-1\\0}, \pmatrix{0\\-1\\1}\ \text{ and }\ \pmatrix{1\\1\\1}\ .
$$
These allow you to diagonalise the matrix thus
$$
\pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{0&1&1\\1&0&1\\1&1&0} \pmatrix{1&0&1\\-1&-1&1\\0&1&1}=\pmatrix{-1&0&0\\0&-1&0\\0&0&2}\ .
$$
This, if you put
$$
z= \pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{x\\y\\u}\ ,
$$
you can write your differential equations in the form
\begin{align}
    \frac{dz}{ds}=& \pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{0&1&1\\1&0&1\\1&1&0} \pmatrix{1&0&1\\-1&-1&1\\0&1&1}z\\
   =& \pmatrix{-1&0&0\\0&-1&0\\0&0&2}z\ ,
\end{align}
which have solutions
\begin{align}
z_1&=Ae^{-s}\\
z_2&=Be^{-s}\\
z_3&=Ce^{2s}\ ,
\end{align}
and so we get
\begin{align}
\pmatrix{x\\y\\u}&= \pmatrix{1&0&1\\-1&-1&1\\0&1&1}z\ \text{, or}\\
\\
x&=Ae^{-s}+Ce^{2s}\\
y&=Ce^{2s}-(A+B) e^{-s}\ ,\text{ and}\\
u&=B e^{-s}+Ce^{2s}\ .
\end{align}
A: $$ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$$
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 $u$.
