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?