How do I explain my steps in solving this PDE? Characteristics I'm taking an exams in which I'll be asked to solve a first order PDE. In solving it I will use the method of characteristics, which I know how to use but I can't explain my steps properly. For example let's take the following PDE,given some initial data as well: 
$$x^2u_x+uu_y=1\\u(x,1-x)=0$$ 
Now, I don't usually apply theorems and methods without understanding them and that is why I'm asking this question. But doing the steps mechanically I begin with this :
Initial Step
$$\frac{dx}{x^2}=\frac{dy}{u}=\frac{du}{1}     $$OR
$$x'(t)=x^2\\y'(t)=u\\u'(t)=1$$
If I recall correctly this are the characteristics equations. From here on, I have 2 ways of solving the problem. I will write both of them : 
Method 1
$$\frac{dx}{x^2}=du=>-1/x=u-c_1=>\\u+\frac{1}{x}=c_1$$
$$\frac{dy}{u}=\frac{du}{1}=>y=\frac{u^2}{2}-c_2\\=>\frac{u^2}{2}-y=c_2$$
Now I proceed to say that the following stands ( but why? ) : 
$$c_1=f(c_2)=>u+\frac{1}{x}=f(\frac{u^2}{2}-y)$$
This is the general solution where f is an arbitrary function. Now I can make use of the initial data and find f but that is not a problem . 
I'm now gonna show the other method which only stands when we have initial conditions.
Method 2
The differential equations including $x'(t)$, $y'(t)$,$u'(t)$ give the following : 
$$x(t)=\frac{-1}{t+c_1}\\u(t)=u+c_3\\y(t)=\frac{y^2}{2}+c_2t+c_3$$
Here is another key moment I don't understand. It seems like we parametrize the initial data and use these 3 conditions : $$x(s,t)=s\\y(s,t)=1-s\\u(s,t)=0$$
Applying these to the functions of $t$ we can get rid of $t$,$s$ and the constants and find a solution for $u$. But how did $x(t)$ just became $x(s,t)$. Also, when applying these conditions, I can use any value of $t$ I want. Mostly,0 is used. How can you do this so freely? 
 A: For the first method. You get two (uniparametric) families of surfaces ($\\u+\dfrac{1}{x}=c_1$ and $\dfrac{u^2}{2}-y=c_2$), the intersection of wich is a (two parameter) family of curves. Consider some, fixed by now, values of $c_1$ and $c_2$. They determine a curve into the surface of a solution brung about by some boundary conditions. So is, $c_1$ is not free, it has to correspond to some $c_2$ to make the curve pass for some point the boundary conditions impose. But it has to be so for every value of $c_1$ or, in other words, there must exist some functional relation between them: $c_1=f(c_2)$.
The second method works almost the same, but in this case we work directly with parametric equations for the surface (two parameters and three coordinates). Your equations are ok, but not the solution. It must be:
$$x(t)=\frac{-1}{t+x_0}\\u(t)=t+u_0\\y(t)=\frac{(t+u_0)^2}{2}+y_0$$
We can set $x_0=0$ and operate, yelding:
$$x(t)=\frac{-1}{t};t=\frac{-1}{x}\\u=t+u_0=-\dfrac{1}{x}+u_0;u_0=u+\dfrac{1}{x}\\y=\frac{(t+u_0)^2}{2}+y_0=\dfrac{u^2}{2}+y_0;y_0=-\left(\dfrac{u^2}{2}-y\right)$$
We have now exactly a family of curves depending on two parameters. The same argument used in the first method works here: we have some functional relation (too brung about by the boundary conditions) between these parameters: $u_0=g(y_0)$. And we recover the same general solution as with the first method: $u+\dfrac{1}{x}=g\left(\dfrac{u^2}{2}-y\right)$
In this case, the boundary conditions are $u(x,1-x)=0$, enough to determine $f$ in $u+\dfrac{1}{x}=f\left(\dfrac{u^2}{2}-y\right)$
$$\dfrac{1}{x}=f(-(1-x))$$
Making $w=-(1-x)$, we get $x=w+1$ and $f(w)=\dfrac{1}{w+1}$
Finally $u+\dfrac{1}{x}=\dfrac{1}{\dfrac{u^2}{2}-y+1}$
