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Solving the differential equation: $y'x\log y =1$ we easly find : $$ y(\log y-1)=\log x +c $$

I search an explicit solution $y=f(x)$ and WolframAlpha gives: $$ y=\dfrac{\log x+c}{W\left( \dfrac{\log x +c}{e}\right)} $$

Where $W$ is the Lambert function. I know that this function is defined such that $W(ze^z)=z$, but I don't see how this can give the Wolfram result.

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Hint: Let $y=e^t$, then divide both sides with e.

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  • $\begingroup$ From your hint I find: $\log y -1=W\left( \dfrac{\log x +c}{e}\right)$ , but how I can find the result of WolfranAlpha? $\endgroup$
    – Emilio Novati
    Commented Apr 29, 2015 at 21:45
  • $\begingroup$ Sorry: I see !!! Thank you! $\endgroup$
    – Emilio Novati
    Commented Apr 29, 2015 at 21:48

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