# Quick Method of Characteristis Question help please!

1. Sep 28, 2011

### 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?

Last edited: Sep 28, 2011