Where is the mistake in solving this PDE? Given the initial condition $u(x,1)=x+1$ I'm asked to find the solution of the following PDE : 
$$(y+u)u_x+yu_y=x-y$$
Using the method of characteristics we know that the characteristic equations are : 
$$x'(t)=y+u\\y'(t)=y\\u'(t)=x-y$$
We can easily see that $y(t)=c_2e^t$. For $x$ and $u$ I tried the following: 
$$x'(t)-y'(t)=u=>(x(t)-y(t))'=u\\=>u''(t)=u\\=>u=c_3e^t$$
Similarly $y'(t)+u'(t)$ will give $x''(t)=x(t)=>x(t)=c_1e^t$
Applying the initial conditions to the $x(t)$,$y(t)$,$u(t)$ I get a wrong solution $u=x+y$ which doesn't satisfy the PDE. So, I assume I've made a mistake in the calculations above. Where is it?
 A: Hint for the second equation
Sum both equations
$$\frac {d(x+u)}{dt}={x+u}$$
$$\frac {d(x+u)}{x+u}=dt$$
Integrate 
$$\implies \ln (x+u)=t+K$$
$$x+u=Ke^t$$
$$ \implies K=\frac {x+u}y$$
A: $$(y+u)u_x+yu_y=x-y$$ 
The characteristic differential equations are :
$$\frac{dx}{dt}=y+u\\\frac{dy}{dt}=y\\\frac{du}{dt}=x-y$$
Or written on a summarized form :
$$\frac{dx}{y+u}=\frac{dy}{y}=\frac{du}{x-y}=dt$$
A first characteristic equation comes from $\frac{dx}{y+u}=\frac{du}{x-y}=\frac{dx+du}{y+u+x-y}=\frac{d(x+u)}{x+u}=\frac{dy}{y}$ which leads to :
$$\frac{u+x}{y}=c_1$$
A second characteristic equation comes from $\frac{dx}{y+u}=\frac{dy}{y}=\frac{dx}{y+(c_1y-x)}$. Solving it leads to : $(c_1+1)y^2-2xy=c_2$ and simplifying with $c_1=\frac{u+x}{y}$ gives the second characteristic equation :
$$(u+y-x)y=c_2$$
The general solution of the PDE expressed on the form of implicit equation is :
$$\Phi\left(\frac{u+x}{y}\:,\:(u+y-x)y\right)=0$$
where $\Phi$ is an arbitrary function of two variables.
Boundary condition :
$u(x,1)=x+1$ thus $\Phi\left(\frac{(1+x)+x}{1}\:,\:((1+x)+1-x)y\right)=\Phi\left(2x+1\:,\:2\right)=0.$
So, the function $\Phi$ is determined. In fact, this is a degenerate form where the function doesn't depends of the first variable and where the second variable $=2$ necessarily.
Thus, the particular solution fitting to the boundary condition is so that $(u+y-x)y=2$
$$u(x,y)=x-y+\frac{2}{y}$$
NOTE :
An alternative way consists in writing the general solution on the form
$$(u+y-x)y=F\left(\frac{u+x}{y}\right)$$
where $F$ is an arbitrary function.
The boundary condition $u(x,1)=x+1$ implies 
$((x+1)+1-x)1=F\left(\frac{(x+1)+x}{1}\right)=2=F(2x+1)$.
So, the function $F$ is the constant function$=2$. Putting it into the general solution gives the solution which fits the boundary condition :
$$(u+y-x)y=2$$
$$u=x-y+\frac{2}{y}$$
