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.