Method of characteristics - solution doesn't seem consistent with original PDE As an example, a problem from the book Numerical solutions of Partial Differential Equations: Finite difference methods by G.D. Smith.
The PDE:
$$ \frac{\partial{U}}{\partial{x}} + \frac{x}{\sqrt{U}}\frac{\partial{U}}{\partial{y}} = 2x $$
With boundary condition $U(x,0) = 0$
I have some basic understanding of how to solve this:
$$ dx = \frac{\sqrt{U}}{x}dy = \frac{dU}{2x}$$
This yields $U = x^2+ Const_X$. For the characteristic going through $(x_R,0)$, we get $Const_X = -x_r^2$ since at $y=0$, $U=0$
We can also get:
$$ \frac{dy}{dU} = \frac{1}{2\sqrt{U}} \to y= \sqrt{U}$$
(The additional constant can be shown to be 0, since for $y=0$, $U=0$)
Giving us the characteristic curve $x^2-y^2 = x_R^2$
But when you plug $U = x^2 -x_r^2$ back to the PDE, you get that along this characteristic, $\frac{x}{\sqrt{U}}\frac{\partial{U}}{\partial{y}} = 0$
... Which contradicts $\sqrt{U} = y$ along the very same curve.
Given how little explanation was given in class for this method, I would've assumed I made a wrong assumption somewhere, except this is the exact solution given in the book (3rd edition, so if it was wrong, I'd assume someone would've figured it out by then)
 A: You obtained correctly a first family of characteristic curves : $$x^2-U=c_1$$ A second family of characteristic curves is : $$y-\sqrt{U}=c_2$$
The general solution is obtained on the form of implicit equation :
$$y-\sqrt{U}=F(x^2-U) \tag 1$$
where $F$ is an arbitrary function.
Condition : 
$U(x,0)=0=0-\sqrt{0}=F(x^2-0)\quad\implies\quad F=$constant function $=0$.
Putting this function into equation $(1)$ leads to the solution :
$\quad y-\sqrt{U}=0\quad\implies\quad U(x,y)=y^2$
So, the solution satisfying the PDE and the boundary condition is a function of $y$ only :
$$U(x,y)=y^2$$
Final checking :
$U_x=0$ and $U_y=2y\quad;\quad U_x+\frac{x}{\sqrt{U}}U_y=0+\frac{x}{y}2y=2x\quad$ is OK.
and the condition $U(x,0)=0^2=0\quad$ is satisfied.
NOTE IN ADDITION :
Another approach to answer to the question :
$$ \frac{\partial{U}}{\partial{x}} + \frac{x}{\sqrt{U}}\frac{\partial{U}}{\partial{y}} = 2x $$
The change of function and variable : 
$\quad\begin{cases}X=x^2\\ \sqrt{U(x,y)}=V(X,y) \end{cases}\quad$ leads to :
$$ V\frac{\partial{V}}{\partial{X}} + \frac{\partial{V}}{\partial{y}} = 1 $$
This a well-known Burger's equation with inhomogeneous RHS. The literature is extensive about these kind of equations, as well as a lot of questions in StackExchange. For example : Confused about Burger's Equation with an inhomogeneous RHS 
I hope that studying this topic of Burger's equations with examples will clarify and make you aware that there is no contradiction.
