Parametric solution of the PDE $-yu_x+xu_y=0,\:u(x,x^2)=x^3$ Find the parametric form of the solution of the following PDE using method of characteristics: $$-yu_x+xu_y=0\qquad u(x,x^2)=x^3$$
I assume that $(x,y)$ are functions of a parameter, say, $t$, i.e. $(x(t), y(t))$. In that case, the chain rule gives
$$\frac{du}{dt}=\frac{\partial u}{\partial x}.\frac{d x}{dt}+\frac{\partial u}{\partial y}.\frac{dy}{dt}=0$$
Comparing the equations, we need to solve the ODEs to find the characteristic curves.
$$\frac{du}{dt}=0,\quad\frac{d x}{dt}=-y \qquad\&\quad\frac{d y}{dt}=x$$
I get the solution, $x(t)=C_{11}\cos(t)+C_{12}\sin(t)$, $y(t)=C_{21}\cos(t)+C_{22}\sin(t)$ and $u(t)=C$.
Using initial conditions $x(0,s)=s$, $y(0,s)=s^2$ and $u(0,s)=s^3$ we get, $x(t)=s\cos(t)+C_{12}\sin(t)$,  $y(t)=s^2\cos(t)+C_{22}\sin(t)$ and $u(t)=s^3$. But how to remove $C_{12},C_{22}$ constants without initial conditions?
I noticed they used one more extra parameter, $s$. Why it was needed? Like why, we re-interpret the $x(t)$ as $x(t,s)$?
I am new to this method. It will be a great help if anyone provide some insight along the solution.
Thanks in advance.
 A: $$-yu_x+xu_y=0$$
The charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{-y}=\frac{dy}{x}=\frac{du}{0}$$
A first characteristic equation comes from $du=0$
$$u=c_1$$
which is an obvious particular solution of the PDE.
A second characteristic equation comes from solving $\frac{dx}{-y}=\frac{dy}{x}$:
$$x^2+y^2=c_2$$
The general solution of the PDE$\quad c_1=F(c_2)\quad$is :
$$\boxed{u(x,y)=F(x^2+y^2)}$$
$F$ is an arbitrary function until no boundary condition is taken into account.
The most interesting part of the problem is to find the particular solution(s) which satisfy the condition $u(x,x^2)=x^3$.
$$u(x,x^2)=F(x^2+(x^2)^2)=x^3$$
Let $X=x^2+(x^2)^2$.
Solving the quadratic eq. $(x^2)^2+(x^2)-X=0$ for $(x^2)$ then for $x$ :
$x=\pm \left(\frac{-1\pm \sqrt{1+4X}}{2} \right)^{1/2}\quad$ with
$\quad X(0)=0\quad\implies\quad x=\pm \left(\frac{-1+ \sqrt{1+4X}}{2} \right)^{1/2}$
$$F(X)=x^3\quad\implies\quad\begin{cases}
F(X)=+\left(\frac{-1+\sqrt{1+4X}}{2} \right)^{3/2}\quad \text{if}\quad x>0 \\
F(X)=0\quad \text{if}\quad x=0 \\
F(X)=-\left(\frac{-1+\sqrt{1+4X}}{2} \right)^{3/2}\quad \text{if}\quad x<0
\end{cases}$$
Now the function $F(X)$ is determined. We put it into the above general solution where $X= x^2+y^2$ :
$$\begin{cases}
u(x,y)=\left(\frac{-1+\sqrt{1+4(x^2+y^2)}}{2} \right)^{3/2}\quad \text{if}\quad x>0 \\
u(x,y)=0\quad \text{if}\quad x=0 \\
u(x,y)=-\left(\frac{-1+\sqrt{1+4(x^2+y^2)}}{2} \right)^{3/2}\quad \text{if}\quad x<0
\end{cases}$$
A: You have
$$
y=-\frac{dx}{dt}=C_{11}\sin(t)-C_{12}\cos(t)
$$
which eliminates the other two constants, as $C_{21}=-C_{12}$ and $C_{22}=C_{11}$.
The aim is to find the value $u$ for a given point $(x,y)$ in the domain. The characteristic curve says that $u$ is constant on the circle
$$
\pmatrix{x\\y}
=
\pmatrix{\cos(t)&-\sin(t)\\\sin(t)&\cos(t)}
\pmatrix{x_0\\y_0}
$$
The initial condition demands that $x_0=s$, $y_0=s^2$. So one has to find $s$ and $t$ so that the parametrization matches the point. From being on a circle centered at the origin we know that the angle $t$ can be eliminated by considering the radius
$$
x^2+y^2=s^2+s^4\implies s^2=\pm\sqrt{\frac14+x^2+y^2}-\frac12
\\
\implies s = \pm\sqrt{\sqrt{\frac14+x^2+y^2}-\frac12}.
$$
As that still gives 2 real intersection points with different values for $u$, the domain has to be further restricted, probably to $x\ge 0$.
