I came across this question. I understand the given answer except for one particular point in the solution.
The answerer uses a particular substitution: $x=-W(u)$
They then proceed to manipulate in a chain of steps, upon which one equality is: $[e^{W(u)}]^{-1}=[\frac{u}{W(u)}]^{-1}$.
I have been shown that this stems from $W(u)e^{W(u)}=u \Rightarrow e^{W(u)}=\frac{u}{W(u)}$, which now makes sense to me.
Were we to make such a substitution into an equation, how would we go about knowing what value of $u$ to use? Are there a number of $u$ that could work, and if so, how do we choose the best value?
Let's see if I can make my question a little more concrete with an example.
Say I want to solve: $x+ae^{(\frac{b}{x})}=c$
were $a$, $b$, and $c$ are constants. I can multiply both sides by $e$ to get
$e^x\cdot e^{ae^{(\frac{b}{x})}}=e^c$
then I can substitute the property from above to get
$\frac{W(u)}{u}e^{ae^{(\frac{b}{x})}}=e^c$
moving $\frac{W(u)}{u}$ to the other side and taking the natural log of both sides I get
$ae^{(\frac{b}{x})}=c\ln \left(\frac{u}{W(u)}\right)$
dividing by $a$ and taking the natural log again I get
$\frac{b}{x}=\ln \left(\frac{c}{a}\ln \left(\frac{u}{W(u)}\right)\right)$
rearranging for x, I get
$x=\frac{b}{\ln \left(\frac{c}{a}\ln \left(\frac{u}{W(u)}\right)\right)}$
Now that I have used the Lambert W function to help isolate $x$, how could I proceed to find $u$ and $W(u)$. I have yet to uncover a resource explaining it, so I request help from someone willing to
- provide a reference link or suggestion that will teach me how to solve this and other explain other intricacies of the Lambert W function
- point out my misgivings and suggest an alternative method to solve the equation (should one exist), linking a similar problem should they know of one
- explain in an answer how one would go about finding $u$