Solving $yy''=(y')^2-y'$ I want to solve the ODE $yy''=(y')^2-y'$ with the initial conditions $y(0)=1, y'(0)=2$.
My attempt:
$$yy''=(y')^2-y'$$
$$(\frac {y'}y)'=(\frac 1y)'$$
$$\frac {y'}y=\frac 1y+c$$
This holds for all $x$. Plugging the initial conditions for $x=0$, we get $c=1$.
$$y'=1+y$$
Solving this got me to $y=-1$. It seems like a singular solution, but I used all methods I know and didn't get anything else.
Your ideas, please?

edit: I'm sorry about the silly question. I put my effort on getting through the hard part and lost my ability to perform simple calculations. Thank you for your answers.

Your solution of $y'=1+y$ is incorrect. Hint:
$$
{y'\over 1+y} = {d\over dx}\log(1+y)
$$
