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How do you derive the general solution of $$ y y'' - (y')^2 + y' = 0 $$ Thanks.


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.

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up vote 7 down vote accepted

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

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Note that $y=0$ is a solution too. You may lose it after division. – Norbert Jan 2 '12 at 14:10
That's elegant. Thanks. – Simon S Jan 2 '12 at 16:21

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