Identifying the constant of integration A short extract from a book of mine states that:

If $$\color{red}{A(x,y)\frac{\partial p}{\partial
 x}+B(x,y)\frac{\partial p}{\partial y}=0\tag{A}}$$ where $p=p(x,y)$
   and $A$ and $B$ are also functions of $x$ and $y$.
Also $$\color{blue}{\rm d p=\frac{\partial p}{\partial x}\rm d
 x+\frac{\partial p}{\partial y}\rm d y=0\tag{B}}$$
  The forms of $\color{red}{(\rm A)}$ and $\color{blue}{(\rm B)}$ are
  very alike and become the same if we require that
  $$\color{#180}{\frac{\rm d x}{A(x,y)}=\frac{\rm d y}{B(x,y)}\tag{C}}$$ By integrating this expression the form of $p$ can be found.


I need to apply the above method to solve $$x\frac{\partial u}{\partial x}-2y\frac{\partial u}{\partial y}=0$$

The book solution tells me that:

If we seek a solution of the form $u(x,y)=f(p)$, we deduce from $\color{#180}{(\rm C)}$ that $u(x,y)$ will be constant along lines of $(x,y)$ that satisfy $$\frac{\rm d x}{x}=\frac{\rm d y}{-2y}$$ which on integrating gives $x=cy^{-1/2}$. $\color{purple}{\text{Identifying the constant of integration}}$ $\color{purple}{c}$ $\color{purple}{\text{with}}$ $\color{purple}{p^{1/2}}$ (to avoid fractional powers), $\color{purple}{\text{we conclude that}}$ $\color{purple}{p=x^2y}$. Thus the general solution of the PDE is $$u(x,y)=f(x^2y)$$ where $f$ is an arbitrary function.


I have understood every step of this proof with the exception of the part marked $\color{purple}{\rm purple}$. Specifically, I don't understand why the constant of integration is $p^{1/2}$. 
So basically; Could someone please explain to me why on earth $c=p^{1/2}$?
Or put in another way, Can I make $c=p$ such that $p=x\sqrt{y}$. So the general solution of the PDE is $$u(x,y)=f(x\sqrt{y})$$ where $f$ is an arbitrary function? 
Many thanks.
 A: Using $p^{1/2}$ is simply a label to make the result look cleaner. Specifically, the choice eliminates a fractional power of $y$ in the result.
Consider the curves $x=cy^{-1/2}$, which you have already understood are curves upon which the function $u(x,y)$ is constant. Thus, you can see that for some function $g$:
$$u(x,y) = g(c)$$
That is, $u$ is simply dependent upon the constant $c$, because the value of $u$ only depends upon which curve you are on. Now, let's just decide to define another constant $p$ such that $c=p^{1/2}$. Since $u$ is a function of $c$ then clearly $u$ is also a function of $p$ so we can say that for some function $h$:
$$u(x,y) = h(p)$$
But, we know that:
$$p = c^2 = (xy^{1/2})^2 = x^2y$$
Thus, relabeling the function $h$ by $f$, we have the result:
$$u(x,y) = f(x^2y)$$
Note that it would have also been correct to not worry about the business of redefining a new constant $p$ and to simply say that $u(x,y)=g(c)=g(xy^{1/2})$. It's just that whoever wrote this solution wanted to eliminate the fractional exponent.
