Solve PDE using method of characteristics The following problem I find challeging. Can you help me find a solution? The question is as follows:

Determine the solution (in explicit form), using the method of characteristics, of the following initial-value problem for $u=u(x,y)$:
  $$\left\{
  \begin{array}{l l}
    u_xu_y=xy \\
    u(x,y)=x \quad for \quad y=0 
\end{array} \right. $$

Cheers
 A: As doraemonpaul remarked, the method of characteristics is given on this Wikipedia page for fully nonlinear equations. We let $x_1 = x, x_2 = y, p_1 = u_x, p_2 = u_y$, then 
$$ F(x_1,x_2, u, p_1, p_2) = p_1p_2 - x_1 x_2 $$
the Lagrange-Charpit equations simplify to 
$$ \frac{\dot{x_1}}{p_2} = \frac{\dot{x_2}}{p_1} = \frac{\dot{p_1}}{x_2} = \frac{\dot{p_2}}{x_1} = \frac{\dot{u}}{2p_1 p_2} $$
Cross-multiplying we get
$$ \frac{d}{ds}(x^2) = \frac{d}{ds}(u_y^2) \qquad \frac{d}{ds}(y^2) = \frac{d}{ds}(u_x^2) $$
Now let $(x(s),y(s))$ be the integral curve starting at $(x_0,0)$. Note that $u_x(x_0,0) = 1$ and $u_y(x_0,0) = 0$ by the initial condition and the equation. We integrate along the curve to get
$$ [u_y(x(s),y(s))]^2 = x(s)^2 - x_0^2 \qquad [u_x(x(s),y(s))]^2 = 1 + y(s)^2 $$
So using the equation again we have that
$$ x^2 y^2 = (x^2 - x_0^2)(1 + y^2) \implies 0 = x^2 - x_0^2(1+ y^2) $$
Solving for $x_0$ and replacing we have that
$$ u_y^2 = \frac{x^2 y^2}{1+y^2} \qquad u_x^2 = 1 + y^2 $$
and by the equation we have finally that either
$$ u_x = \sqrt{1+y^2} \qquad u_y = \frac{xy}{\sqrt{1+y^2}} $$
or 
$$ u_x = - \sqrt{1+y^2} \qquad u_y = - \frac{xy}{\sqrt{1+y^2}}$$
The latter is inadmissible by the boundary condition. Integrating in $y$ and using the initial value we have that 
$$ u =  x\sqrt{1+y^2} $$
is the final solution. 
A: I have no idea about how to solve this nonlinear PDE directly by using the method of characteristics. I only know about the following facts about this nonlinear PDE:
Even http://eqworld.ipmnet.ru/en/solutions/fpde/fpdetoc3.htm have not mentioned about this type of PDE.
Let $\begin{cases}p=\dfrac{x^2}{2}\\q=\dfrac{y^2}{2}\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=x\dfrac{\partial u}{\partial p}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=y\dfrac{\partial u}{\partial q}$
$\therefore xu_pyu_q=xy$
$u_pu_q=1$
This belongs to a PDE of the type http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3308.pdf. But even this website is not quite useful as it cannot know its general solution.
