<s>[INST] Solve the Following First Order PDE Solve
$$F\frac{\partial F}{\partial x} - \frac{\partial F}{\partial y} = y$$
subject to $F(s,0) = s^2$.
This is the first time I am using the method of characteristics, so I would like to know if I have made any errors in my working. I have
$$
\frac{dx}{dt} = z, \quad \frac{dy}{dt} = -1, \quad \frac{dF}{dt} = y
$$ 
Solving these gives me
$$
y = -t + y_0, \quad F  =-\frac{t^2}{2} + y_0t + F_0, \quad x = -\frac{t^3}{6}+\frac{y_0 t^2}{2} + F_0t + x_0
$$
Then assume without loss of generality that $y_0 = 0$, so
$$
y = -t, \quad F  =-\frac{t^2}{2}+ F_0, \quad x = -\frac{t^3}{6} + F_0t + x_0
$$ 
The characteristic curves are therefore curves of the form
$$
t \mapsto \left(-\frac{t^3}{6} + F_0t + x_0,\quad-t,\quad -\frac{t^2}{2}+ F_0 \right)
$$
I can now use the initial condition $F(s,0) = s^2$. This means I substitute $x = s, y = 0$ and $F = s^2$ and solve for $x_0, F_0$ and $t$ . So
$$
y = 0 \implies t = 0, \quad s = x_0, \quad s^2 = F_0
$$
So the parametrisation of the solution surface is
$$
(s,t) \mapsto \left(-\frac{t^3}{6} + s^2t + s,\quad-t,\quad -\frac{t^2}{2}+ s^2 \right)
$$
To get this in terms of $x$ and $y$ I know $-y = t$, hence
$$
x = \frac{y^3}{6} - s^2y + s \implies s = \frac{1\pm\sqrt{1-4y\left(x-\frac{y^3}{6} \right)}}{2y}
$$
Thus
$$
F(x,y) = -\frac{y^2}{2} + \left( \frac{1\pm\sqrt{1-4y\left(x-\frac{y^3}{6} \right)}}{2y}\right)^2
$$
This function is not well defined when $y = 0$ or when the term inside the square root is negative.
Have I made a mistake somewhere or is the working fine?
 [/INST] Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=-1$ , letting $y(0)=0$ , we have $y=-t$
$\dfrac{dF}{dt}=y=-t$ , letting $F(0)=F_0$ , we have $F=F_0-\dfrac{t^2}{2}=F_0-\dfrac{y^2}{2}$
$\dfrac{dx}{dt}=F=F_0-\dfrac{t^2}{2}$ , letting $x(0)=f_1(F_0)$ , we have $x=f_1(F_0)+F_0t-\dfrac{t^3}{6}=f_1\biggl(F+\dfrac{y^2}{2}\biggr)-\biggl(F+\dfrac{y^2}{2}\biggr)y+\dfrac{y^3}{6}=f_1\biggl(F+\dfrac{y^2}{2}\biggr)-\dfrac{y^3}{3}-Fy~,~\text{i.e.}~F=f\biggl(x+\dfrac{y^3}{3}+Fy\biggr)-\dfrac{y^2}{2}$
$F(s,0)=s^2$ :
$f(s)=s^2$
$\therefore F=\biggl(x+\dfrac{y^3}{3}+Fy\biggr)^2-\dfrac{y^2}{2}$
</s>