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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!

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  • $\begingroup$ The next thing to do is to write out the characteristic equations. $\endgroup$ – Chee Han Apr 10 '17 at 21:46
  • $\begingroup$ Yes, I have done that, but I know I am not doing it correctly, I get $x'(s)=2xp$, $y'(s)=1$, $u'(s)=2xp^2+q$, $p'(s)=-2p^2$, $q'(s)=1$ I think I am doing something wrong with my intiial condition for p and q $\endgroup$ – Reya Apr 10 '17 at 21:54
  • $\begingroup$ Am i correct in saying $p(0)=1$ and $q(0)=-x(0)p(0)^2$ as $y(0)=0$ $\endgroup$ – Reya Apr 10 '17 at 21:59
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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