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!
 A: \begin{align*} 
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2} 
\end{align*}
From the first two ratios, 
$$\frac{dx}{y} = -\frac{dy}{x} \implies x~dx~+~y~dy~=0 $$
Integrating, $~x^2+y^2=c^2~\tag1$where $~c~$ is integrating constant.
From the last two ratios, $$ -\frac{dy}{x} = \frac{du}{x^2} \implies  du=-x~dy\implies du=-\sqrt{c^2-y^2~}~dy \qquad\text{[using equation $(1)$]}$$
Integrating, $$u=-\dfrac 12~y\sqrt{c^2-y^2}~-~\dfrac {c^2}{2} \sin^{-1}\dfrac yc~+~d\qquad\text{(using direct formula)
}$$
$$u=-\dfrac 12~xy~-~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)~+~d\qquad\text{[using equation $(1)$]}$$where $d$ is integrating constant.
Hence the general solution is of the form$$u=-\dfrac 12~xy~-~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)~+~\phi\left(\sqrt{x^2+y^2~}\right)$$where $~\phi~$ is arbitrary function of $~x,~y~$.
or, in the form of$$f\left(\sqrt{x^2+y^2}~,~u~+~\dfrac 12~xy~+~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)\right)=0$$where $~f~$ is arbitrary function of $~x,~y~$.

Integration formula :
$$\int\sqrt{a^2-x^2~}~dx=\dfrac 12~x\sqrt{a^2-x^2~}+\dfrac{a^2}{2}~\sin^{-1}\left(\dfrac{x}{a}\right)+c$$ where $~c~$ is a constant of integration.
A: $$yu_x-xu_y=x^2$$
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.
