First order PDE problem $yu_{x} - xu_{y} = x^2$ I am a beginner to PDE, trying to solve $$yu_{x} - xu_{y} = x^2$$ using characteristic line, first I have \begin{align*} 
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2} 
\end{align*}
Then I get the characteristic line is given by $$C = \frac{1}{2}y^2 - \frac{1}{2}x^2$$ 
Next, I solve the first term and third term, I have \begin{align*}
\frac{du}{x^2} &= \frac{dx}{y} \\
du &= \frac{x^2}{y}dx
 \end{align*}
Here is my problem, that is we can not integrate right-hand side without eliminating the variable $y$, but if we try to replace $y$ in terms of $x$ and $C$, the result does not look integrable and really messy.
Can someone help me, please? Thanks!

$$yu_x-xu_y=x^2$$
Charpit-Lagrange system of characteristic ODEs:
$$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{x^2}$$
A first characteristic equation comes from $\frac{dx}{y}=\frac{dy}{-x}$
$$x^2+y^2=c_1$$
A second characteristic equation comes from $\frac{dx}{\sqrt{c_1-x^2}}=\frac{du}{x^2}$
$du=\frac{x^2dx}{\sqrt{c_1-x^2}}\quad\implies\quad u=\int\frac{x^2dx}{\sqrt{c_1-x^2}}=-\frac{xy}{2}+\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right)+c_2$
$$u+\frac{xy}{2}-\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) =c_2$$
The solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is :
$$u+\frac{xy}{2}-\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) =F(x^2+y^2)$$
$$\boxed{u(x,y)=-\frac{xy}{2}+\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) +F(x^2+y^2)}$$
$F$ is an arbitrary function.
The function $F$ has to be determined according to some boundary condition which is missing in the wording of the question.
