Solving a PDE and finding the jump condition (method of characteristics)

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Here I have my PDE:

http://desmond.imageshack.us/Himg718/scaled.php?server=718&filename=pde.png&res=medium

I have found the solution by using the method of characteristics two times, one for x<0 and the other for x>0.

I have: U(x,y) = o for x<0 and U(x,y) = Uo(x-1)/(1+Uo*y) for x>0

The trouble I am having is finding the jump curve, x = \xi(y),

Im not sure if I am doing this correctly, but I keep getting the natural log of something, while my book says the correct answer is 1 - sqrt(1+Uo*y) for the jump curve...

can anyone help me?

much appreciated
 
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nevermind, i have figured out the problem

thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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