Solve Partial differential equation(geometric optics) Solve $x^2((u_x)^2+(u_y)^2)=1$ , $u(x,0)=0$
Use the characteristic equation
The solution is 
$u(x,y)=-\ln\dfrac{\sqrt{x^2+y^2}+y}{x}$
I drove 
$\dfrac{dx}{dt}=2x^2p$
$\dfrac{dy}{dt}=2x^2q$
$Z=2t$
$\dfrac{dp}{dt}=\dfrac{2}{x}$
$dq=0$
From above equations, I got
$q=\dfrac{1}{s}$
$p=\sqrt{-\dfrac{1}{s^2}+\dfrac{1}{x^2}}$
Then I lost. Please help
 A: $x^2((u_x)^2+(u_y)^2)=1$
$(u_x)^2+(u_y)^2=\dfrac{1}{x^2}$
$(u_y)^2=\dfrac{1}{x^2}-(u_x)^2$
$u_y=\pm\sqrt{\dfrac{1}{x^2}-(u_x)^2}$
$u_{xy}=\pm\dfrac{-\dfrac{2}{x^3}-2u_xu_{xx}}{2\sqrt{\dfrac{1}{x^2}-(u_x)^2}}$
$u_{xy}=\mp\dfrac{x^3u_xu_{xx}+1}{x^3\sqrt{\dfrac{1}{x^2}-(u_x)^2}}$
Let $v=u_x$ ,
Then $v_y=\mp\dfrac{x^3vv_x+1}{x^3\sqrt{\dfrac{1}{x^2}-v^2}}$ with $v(x,0)=0$
$x^3vv_x\pm x^3\sqrt{\dfrac{1}{x^2}-v^2}~v_y=-1$ with $v(x,0)=0$
$-x^3v_x\mp\dfrac{x^3}{v}\sqrt{\dfrac{1}{x^2}-v^2}~v_y=\dfrac{1}{v}$ with $v(x,0)=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dv}{dt}=\dfrac{1}{v}$ , letting $v(0)=0$ , we have $v^2=2t$
$\dfrac{dx}{dt}=-x^3$ , letting $x(0)=x_0$ , we have $x^2=\dfrac{x_0^2}{2x_0^2t+1}=\dfrac{x_0^2}{x_0^2v^2+1}$
$\dfrac{x^2}{1-x^2v^2}=x_0^2$
$\dfrac{dy}{dt}=\mp\dfrac{x^3}{v}\sqrt{\dfrac{1}{x^2}-v^2}=\mp\dfrac{\sqrt{x_0}}{\sqrt{2t}(2x_0^2t+1)^\frac{3}{2}}$ , letting $y(0)=f(x_0^2)$ , according to http://integrals.wolfram.com/index.jsp?expr=c%5E%281%2F2%29%2F%28%282x%29%5E%281%2F2%29%282c%5E2x%2B1%29%5E%283%2F2%29%29&random=false, we have $y=f(x_0^2)\mp\sqrt{\dfrac{2x_0t}{2x_0^2t+1}}=f\left(\dfrac{x^2}{1-x^2v^2}\right)\mp\sqrt xv\sqrt[4]{1-x^2v^2}$
$v(x,0)=0$ , i.e. $y(x,v=0)=0$ :
$f(x^2)=0$
$f(x)=0$
$\therefore y=\mp\sqrt xv\sqrt[4]{1-x^2v^2}$
$y^4=x^2v^4(1-x^2v^2)$
$x^4v^6-x^2v^4+y^4=0$
