# 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.

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: $${y'\over 1+y} = {d\over dx}\log(1+y)$$
The full solution with the homogeneous part is $$y=-1+C·e^x$$ and from the initial conditions $C=2$.
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 $$y = 2 e^x -1$$