find the general solution and the particular solution of the first order partial differential equation 
Consider the first order partial differential equation
  $$ x(y-u)u_x + y (u-x)u_y = u(x-y) $$
  Find the general solution of  this equation.
  Also, given
  $$ u(x,x+1) = x $$
  find the particular solution where the answer should come to
  $$ u = \frac{1}{27xy} (x+y+u-1)^2(x+y+u+2) $$

I have tried to use the characteristic equation to work out the general solution because it is a quasi-linear equation.
I've found
$$
\frac{dx}{dt}= xy-ux,\qquad
\frac{dy}{dt}= uy-xy \qquad\text{and}\qquad
\frac{du}{dt}=ux-uy 
$$
To find the first integers I found that 
$$
-\frac{dx}{dt}-\frac{dy}{dt} = \frac{du}{dt}
$$
I then integrated that equation to get 
$-x-y-u= c1(s)$.
But not sure if that's correct and if so not sure what to do after that
 A: Although I have detailed a solution in the comments, I think it is probably good to provide the full solution here with an explanation.
Eliminating $dt$ in your system of ODEs yields the following form
$$\frac{dx}{x(y-u)} = \frac{dy}{y(u-x)} = \frac{du}{u(x-y)}$$
Noting that $dA/A = d \ln A$, this can be simplified to 
$$\frac{d \ln x}{y-u} = \frac{d \ln y}{u-x} = \frac{d \ln u}{x-y}$$
Now, what the expression above is saying is that the vectors $(d \ln x, d \ln y, d \ln u)$ and $(y - u, u - x, x - y)$ are proportional to eachother i.e 
$$(d \ln x, d \ln y, d \ln u) \propto (y - u, u - x, x - y)$$
You might notice that the RHS can be written as the curl of two vectors
$$(y - u, u - x, x - y) = (x, y, u) \times (1, 1, 1)$$
so what we really have is 
$$(d \ln x, d \ln y, d \ln u) \propto (x, y, u) \times (1, 1, 1)$$
Remembering that the cross product of two vectors $a$ and $b$ yields a third vector $c$ which is orthogonal to both $a$ and $b$, the expression above really says that the tangent of the log of the solution curve is orthogonal to the vectors $(x, y, u)$ and $(1, 1, 1)$. So, if we take the dot product of the LHS with both vectors $(x, y, u)$ and $(1, 1, 1)$, we can get our integral curves
\begin{align}
(d \ln x, d \ln y, d \ln u) \cdot (x, y, u) &= ((x, y, u) \times (1, 1, 1)) \cdot (x, y, u) \\ 
&= 0 \\
\implies x d \ln x + y d \ln y + u d \ln u &= dx + dy + du \\
&= 0 \\
\implies x + y + u &= c_{1} \\\\
(d \ln x, d \ln y, d \ln u) \cdot (1, 1, 1) &= ((x, y, u) \times (1, 1, 1)) \cdot (1, 1, 1) \\ 
&= 0 \\
\implies d \ln x + d \ln y + d \ln u &= 0 \\
\implies \ln x y u &= c_{2} \\\\
\therefore \ln x y u &= f(c_{1}) \\ 
&= f(x + y + u)
\end{align}
and from here the result follows by applying the initial conditions as stated in the comments.
Note that this approach works well here because of the symmetry in the coefficients of the original PDE. Often times it can be quite difficult to determine the form of the two vectors whose curl is proportional to the tangent vector. However, I think this method gives a nice geometric view of what is happening.
