I'm struggling to find solution of the below equation:
$$u_x^3 - u_y = 0$$ $$u(x,0) = 2x^{3/2}$$
I was taught about method of characteristics and how to parametrize the equations with $s$ and $t$ to get ODE's but in this case I'm struggling with what will be the coefficients and overall how to proceed to get a solution.
Note: This problem is from Fritz John's and answer given is $u = 2x^{3/2}(1-27y)^{-1/2}$
With $0=F(\overbrace{x,y}^{\mathbf{x}},u,\overbrace{p,q}^{\nabla u}):=p^3-q$, the method of characteristic gives $$ \left\lbrace \begin{aligned} \dot{x}&=\frac{\partial F}{\partial p}=3 p^2\\ \dot{y}&=\frac{\partial F}{\partial q}=-1\\ \dot{u}&=p\frac{\partial F}{\partial p}+q\frac{\partial F}{\partial q}=3p^3-q\\ \dot{p}&=-\frac{\partial F}{\partial x}-p\frac{\partial F}{\partial u}=0\\ \dot{q}&=-\frac{\partial F}{\partial y}-q\frac{\partial F}{\partial u}=0. \end{aligned} \right. $$
So the characteristics are lines $$ \left\lbrace \begin{aligned} x&=x_0+3p_0^2 t\\ y&=y_0-t\\ u&=u_0+(3p_0^3-q_0)t\\ p&=p_0\\ q&=q_0 \end{aligned} \right. $$
From $F=0$ we have $q=p^3$ (and let's drop the suffix 0 from $p,q$ which are constant in $t$). Let $y_0=0$, so $u_0=u(x_0,0)=2x_0^{3/2}$, $p=3x_0^{1/2}$, and $t=-y$. Hence $x=x_0+3p^2t=x_0-27x_0y=(1-27y)x_0$, $x_0=x/(1-27y)$ and \begin{align*} u&=2x_0^{3/2}+2p^3t\\ &=2x_0^{3/2}+2(3x_0^{1/2})^3(-y)\\ &=2x_0^{3/2}(1-27y)\\ &=2\left(\frac{x}{1-27y}\right)^{3/2}(1-27y)\\ &=2x^{3/2}(1-27y)^{-1/2}. \end{align*}