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.

You got the hard part right.  But the solution to 
$$y'=1+y$$ is
$$\frac{dy}{1+y}=dx \\x + \ln K 1+y\\ y = K e^x -1
$$
And the initial conditions force $K=2$ so the answer is 
$$y = 2 e^x -1$$
