I this trying to solve y'' = yy' with y(0) = 1, y'(0) = 1.
I know y' = 1/x', y'' = -x''/(x')^3 . Then I tried substitute u = x'. So du = x''. y'' = -du/u^3. Then \int y'' = \frac{1}{2u^2}. I am a bit confused at this point. Could someone point out how to proceed?