Solving characteristic base curves initial value PDE I'm trying to solve the characteristic base curves of an initial value problem.
$$
\left\{ \begin{matrix}\ xy\frac{\partial u}{\partial x} + (2y^2 - x^6)\frac{\partial u}{\partial y} = 0 ; x>0,y\geq0 \\ u(x(t),y(t)) = t^2\end{matrix}\right.
$$
And after that I need to investigate whether or not three given initial curves if the problem can be solved. (When it is possible show the solution). And investigate where the solution is unique.
Given initial curves:
$$
l_1 : (x(t),y(t)) = (t,\alpha t) \\
l_2 : (x(t),y(t)) = (t,\alpha t^2) \\
l_3 : (x(t),y(t)) = (t,\alpha t^3)
$$
I have come up with:
$$
\frac{dx}{xy}=\frac{dy}{2y^2-x^6}=\frac{du}{0}
$$
From that equation I derived that 
$$
(2y^2-x^6)dx = xydy = \text{Constant}\\
2y^2x -\frac{x^7}{7} = \frac{xy^2}{2}= \text{Constant}
$$
Giving the curve:
$$
y=\sqrt{\frac{2}{21}}x^3
$$
This seems to be a nice curve, but I don't know if this is the right way to go into the problem. If it is true does that mean that initial curve $l_3$ can solve the initial value problem?

The computation of the characteristic curves is wrong. Its differential equation is
$$
\frac{dx}{x\,y}=\frac{dy}{2\,y^2-x^6}\implies\frac{dy}{dx}=\frac{2}{x}\,y-\frac{x^5}{y}.
$$
This is a Bernoulli equation, whose solution is
$$
y=\pm\, x^2\,\sqrt{C-x^2\,},\quad C>0.
$$
