Quick Method of Characteristis Question help please

[lackrange]

So I already did most of the problem, there is just one thing I am unsure of, the equation is:
$$xyu_x+(2y^2-x^6)u_y=0$$
I found the characteristics, they are
$$\frac{y^2+x^6}{x^4}=A$$ for constant A...so A=y(1)^2+1. Now I am given an initial condition $$u(x,\alpha x^n)=x^2$$ and am asked to find the explicit solution. I think I may just be very tired or something because this is usually easy, but I am having trouble. On my characteristic curve, I have that du/dx=0, so u is constant on the curve. I thought I would just find when the two curves are equal ie. $$\alpha x^n=\sqrt{Ax^4-x^6}$$
(on the right side I just solved my characteristic curve for y) but in any case, the problem is giving me trouble, and it's due in the morning. I know the general solution is
$$u(x,y)=g(\frac{x^6+y^2}{x^4})$$ Can someone help me please?

 

[lippyka]

The equation can be solved using the method of characteristics. In this case, the characteristic equations are \begin{cases} \frac{dx}{dt}=1 \\ \frac{dy}{dt}=2y^2-x^6\end{cases}The solution of the equation is then given by \begin{align*}u(x,y)&=g\left(\frac{x^6+y^2}{x^4}\right) \\g(A)&=x^2 \quad \text{when } \frac{x^6+y^2}{x^4}=A\end{align*}where $A$ is an arbitrary constant.For the given initial condition $u(x,\alpha x^n)=x^2$, we have \begin{align*}\frac{x^6+(\alpha x^n)^2}{x^4}&=A \\A&=\alpha^2x^{2n+2}+1\end{align*}Therefore, the explicit solution is \begin{align*}u(x,y)&=g\left(\frac{x^6+y^2}{x^4}\right) \\&=x^2 \quad \text{when } \frac{x^6+y^2}{x^4}=\alpha^2x^{2n+2}+1\end{align*}