Two PDE Questions 1.
Solve the equation 
$$
u_x^3-u_y=0~,
$$ with 
$u(x,0)=2x^\frac{3}{2}$
2.
Solve the equation 
$$
u=xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2)~,
$$ with
$u(x,0)=\dfrac{1}{2}(1-x^2)$
How I solve these problems if I learnt PDE three weeks ago?
 A: $1.$
$u_x^3-u_y=0$
$u_{xy}-3u_x^2u_{xx}=0$
Let $v=u_x$ ,
Then $v_y-3v^2v_x=0$ with $v(x,0)=3\sqrt x$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dx}{dt}=-3v^2=-3v_0^2$ , letting $x(0)=f(v_0)$ , we have $x=-3v_0^2t+f(v_0)=-3v^2y+f(v)$ , i.e. $v=F(x+3v^2y)$
$v(x,0)=3\sqrt x$ :
$F(x)=3\sqrt x$
$\therefore v=3\sqrt{x+3v^2y}$
$v^2=9(x+3v^2y)$
$v^2=9x+27v^2y$
$v^2-27v^2y=9x$
$v^2(1-27y)=9x$
$v^2=\dfrac{9x}{1-27y}$
$v=\pm\dfrac{3\sqrt x}{\sqrt{1-27y}}$
$u_x=\pm\dfrac{3\sqrt x}{\sqrt{1-27y}}$
$u(x,y)=\pm\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}+g(y)$
$u_y=\pm\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}+g_y(y)$
$\therefore\pm\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}\mp\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}-g_y(y)=0$
$g_y(y)=0$
$g(y)=C$
$\therefore u(x,y)=\pm\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}+C$
$u(x,0)=2x^\frac{3}{2}$ :
$C=0$ and the negative part reject
$\therefore u(x,y)=\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}$
