Partial differential equation solution $x u_x + y u_y = \frac{1}{\cos u}$ I have a problem with a following task:
Let us consider an equation $x u_x + y u_y =  \frac{1}{\cos u}$. Find a solution which satisfies condition $u(s^2, \sin s) = 0$. You can
write down the solution in the implicit form $F(x,y,u)=0$. Find some domain of
$s$ values for which there exists a unique solution.
@edit:My progress so far: 
System of characteristics:
$\begin{cases} x'(s,\tau) = x \\ y'(s, \tau) = y \\ u'(s,\tau)=\frac{1}{\cos(u)}\end{cases}$
(derivatives are with respect to $s$)
Initial conditions:
$\begin{cases} x(0,\tau)=\tau^2 \\ y(0, \tau) =\sin(\tau) \\ u(0, \tau) = 0 \end{cases}$
We calculate general solutions for those 3 equations, apply initial conditions, and we obtain:
$\begin{cases} x(s,\tau) = \tau^2 e^s \\ y(s, \tau)=\sin(\tau) e^s \\ u(s, \tau) = \arcsin(s) \end{cases}$
How can I now write down the solution in the form of $F(x,y,u)$?
I would be very grateful for any help ;-)
Best regards,
David.
 A: What you have so far is correct. Your last system of equations yields $s = \sin u$, hence $x = \tau^2 e^{\sin u}$ and $y = \sin \tau e^{\sin u}$. So, 
$$|\tau| = \sqrt{xe^{-\sin u}}, \quad  \tau = \arcsin (y\,e^{-\sin u})$$
which leads to 
$$
\sqrt{xe^{-\sin u}} = |\arcsin (y\,e^{-\sin u})|
$$
There you have it, an implicit equation $F(x,y,u)=0$. 
Of course, this computation is a minefield of non-invertible functions. But it seems that everything works as long as $|s|\le 1$, and there are issues after that, mostly due to the division in $1/\cos u$.
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0x$
$\dfrac{du}{dt}=\dfrac{1}{\cos u}$ , we have $\sin u=t+f(y_0)=\ln x+f\left(\dfrac{y}{x}\right)$
Case $1$: $u(x=s^2,y=\sin s)=0$ :
$f\left(\dfrac{\sin s}{s^2}\right)=-\ln s^2$
$\therefore\sin u=\ln x+f\left(\dfrac{y}{x}\right)$ , where $f(s)$ is the solution of $f\left(\dfrac{\sin s}{s^2}\right)=-\ln s^2$
Case $2$: $u(y=s^2,x=\sin s)=0$ :
$f\left(\dfrac{s^2}{\sin s}\right)=-\ln\sin s$
$\therefore\sin u=\ln x+f\left(\dfrac{y}{x}\right)$ , where $f(s)$ is the solution of $f\left(\dfrac{s^2}{\sin s}\right)=-\ln\sin s$
