<s>[INST] Method of characteristics Given the equation
$y^2 u_x + x  u_y = \sin(u^2)$ with
initial condition $u(x,0) = x$,
determine the values of $u_x$, $ u_y$, $u_{xx}$, $u_{yy}$, $u_{xy}$, $u_{yx}$ on the $x$-axis.
I tried the following:
$$\begin{cases}
\frac{dx}{dt} = y^2 \\\\
\frac{dy}{dt} = x \\\\
\frac{dz}{dt} = \sin(z^2)
\end{cases}$$
However, the first two coupled ODEs are really difficult to solve. So I assume there is an easier way to do this.
 [/INST] Hint:
$$\begin{align}
\frac{dx}{dt} &= y^{2} \ \ \ (1)\\
\frac{dy}{dt} &= x \ \ \ \ \ (2)\\\\
(2) \implies \frac{d}{dt} \bigg( \frac{dy}{dt} \bigg) &= \frac{d^{2}y}{dt^{2}} \\
&= \frac{dx}{dt} \\
&= y^{2} \\
&= (1)
\end{align}$$
EDIT:
We have
$$y'' = y^{2}$$
Multiply through by $y'$ and integrating
$$\begin{align}
y' y'' &= y' y^{2} \\
\implies \frac{d}{dt} \bigg( \frac{1}{2} y'^{2} \bigg) &= y' y^{2} \\
\implies \frac{1}{2} y'^{2} &= \frac{y^{3}}{3} + C_1 \\
\implies y'^{2} &= \frac{2y^{3}}{3} + C_1 \\
\implies y' &= \pm \sqrt{\frac{2y^{3}}{3} + C_1} \\
\end{align}$$
Separating and integrating
$$\int \frac{dy}{\sqrt{\frac{2y^{3}}{3} + C_1}} = \pm \int dt$$
Can you take it from there?
EDIT 2:
From our PDE we have
$$y^{2}u_x + xu_y = \sin(u^{2})$$
Separating into parts, we have
$$\frac{dx}{y^{2}} = \frac{dy}{x} = \frac{du}{\sin(u^{2})}$$
From this we find two equations, take
$$\frac{dy}{dx} = \frac{x}{y^{2}} \ \ \ \ \ (*) \\
\frac{du}{dy} = \frac{\sin(u^{2})}{x} \ \ \ \ \ (**)$$
From $(*)$ we find
$$\begin{align}
y^{2}dy &= xdx \\
\implies x &= \sqrt{\frac{2y^{3}}{3} + C_1} \\
\end{align}$$
Hence, using $(**)$ we see
$$\begin{align}
\frac{du}{\sin(u^{2})} &= \frac{dy}{\sqrt{\frac{2y^{3}}{3} + C_1}} \\
\end{align}$$
However, the $u$ integration is ridiculous and can't be done in terms of elementary functions.
</s>