Consider the definition of the Lambert-W function \begin{align} We^W=z. \end{align} Defining the parameter $\tau$ as follows we find a constant-coefficient second order linear ODE \begin{align} (z+e^W)W'_z&=1,\\ z''_{\tau\tau}-2z'_{\tau}+z&=0.\quad[z+e^W=z'_\tau] \end{align} This ODE is easily solved, giving our parametric representation \begin{align} z=(A+B\tau)e^\tau,\quad W=\tau+\log(B\tau), \end{align} for constants $A$ and $B$. This solution does not work with the original definition, though. Substitution yields \begin{align} A+B\tau=B\tau^2+B\tau\log(B\tau), \end{align} which is true only for constant $\tau$ or the trivial case of $A=B=0$. The only question with a parametric representation that I've found is this one. I'm not sure what is going awry here, could someone point the way?
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How did you derive $z_{\tau\tau}-2z_{\tau}+z=0$ ? – user619894 Nov 14 '22 at 08:31
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@user619894 $W=\log(z'-z)$, $\mathrm d_z=1/z'\tau \mathrm d\tau$, and $W'z=(z''{\tau\tau}-z'\tau)/(z'\tau(z'_\tau-z))$ – Eli Bartlett Nov 14 '22 at 08:52