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**Defunkt** Hi, I would appreciate any help in solving this:

$\displaystyle y''(x) \cdot y(x) = (y'(x))^2 - y'(x)$

With initial conditions $\displaystyle y(0) = 1, y'(0) = 2$

It may be helpful to note the fact that $\displaystyle (y(x) \cdot y'(x))' = (y'(x))^2 + y(x) \cdot y''(x)$

or that $\displaystyle \left( \frac{y'(x)}{y(x)} \right) ' = \frac{y''(x) \cdot y(x) - (y'(x))^2}{(y(x))^2}$ (we get this if we divide both sides of the equations by y^2)

But I was not able to solve this using either.

Thanks for any help.