TITLE: General solution of $yy'' - (y')^2 + y' = 0$
QUESTION [6 upvotes]: How do you derive the general solution of
$$ y y'' - (y')^2 + y' = 0 $$
Thanks.
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Never mind.  Solved.
Let $u = y'$.  Then $y'' = u' = \frac{du}{dy}$ . $y' = \frac{du}{dy} . u$
Hence the equation is
$$ yu \cdot\frac{du}{dy} = u^2 - u $$
which is separable.

REPLY [10 votes]: Hint: Divide everything by $y^2$. Then use the identities 
$$
\left(\dfrac{y'}y\right)'=\dfrac{y''y-(y')^2}{y^2}\qquad\text{and}\qquad\left(\dfrac1y\right)'=\dfrac{-y'}{y^2}.
$$