Non-linear first order PDE I am trying to solve a PDE
$x (u_x)^2 + u_y = y $
$u(x,0)=x$
using the method of characteristics (i.e. $p=u_x$, $q=u_y$ and the above PDE is $xp^2+q-y=0$. I have tried so many times but just cant get it! Help would be appreciated!
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
x\pars{\mrm{u}_{x}}^{2} + \,\mrm{u}_{y} = y \implies
\bbx{\ds{x\pars{\varphi_{x}}^{2} + \,\varphi_{y} = 0}}\,,\quad
\varphi\pars{x,y} \equiv \,\mrm{u}\pars{x,y} - {1 \over 2}\,y^{2}\,,\quad
\varphi\pars{x,0} = x 
\end{align}

$$
0 = x\,\mrm{f}'^{2}\pars{x}\,\mrm{g}^{2}\pars{y} + \,\mrm{f}\pars{x}\,\mrm{g}'\pars{y} \implies
x\,{\mrm{f}'^{2}\pars{x} \over \mrm{f}\pars{x}} = -\,{\mrm{g}'\pars{y} \over \,\mrm{g}^{2}\pars{y}} = \mu
$$

$$
{\mrm{f}'\pars{x} \over \mrm{f}^{1/2}\pars{x}} = {\root{\mu} \over x^{1/2}}\,,
\quad
{\mrm{g}'\pars{y} \over \mrm{g}^{2}\pars{y}} = -\mu
$$

$$
2\,\mrm{f}^{1/2}\pars{x} = 2\root{\mu}x^{1/2} + 2a
\quad
-\,{1 \over \mrm{g}\pars{y}} = -\mu y - b
$$

\begin{align}
\varphi\pars{x,y} & = {\pars{\root{\mu}x^{1/2} + a}^{2} \over \mu y + b} =
{\mu x + 2a\root{\mu}x^{1/2} + a^{2} \over \mu y + b}\implies
x = \varphi\pars{x,0} = 
{\mu x + 2a\root{\mu}x^{1/2} + a^{2} \over b}
\\[5mm] \implies & b = \mu\,,\quad a = 0
\implies \varphi\pars{x} = {x \over y + 1}\implies
\bbx{\ds{\mrm{u}\pars{x,y} = {x \over y + 1} + {1 \over 2}\,y^{2}}}
\end{align}
