Recovering the definition of \(W(x)\)
Just to check we are on the right planet set \(a=1\), \(c=1\), \(b=\ln(x)\) and \(d=0\) to get \begin{equation} x e^{-f(x)} = f(x) \end{equation} or equivalently \( f(x)e^{f(x)} = x\) which is the definition of the Lambert W function. Sure enough those substitutions recover $$ f(x) = W(x) $$ as expected. A minor modification uses \(a(x)=1\), \(c(x)=1\), \(b(x)=0\) but leaves \(d(x)\) general.
The solution to \( e^{-f(x)} - f(x) = d(x) \)
is \begin{equation} f(x) = W \left( e^{d(x)} \right) \end{equation} Only slightly more generally, set \(a(x)=a\), \(b=\ln(g(x)), c=1, d=1\) for those cases where \(g(x)\) is on the wrong side, as it were.
The solution to \( e^{-af(x)} = f(x) g(x) \)
or equivalently \( f(x) e^{af(x)} g(x) = 1 \) is \begin{equation} f(x) = \frac{1}{a} W\left( a g(x) \right) \end{equation} In particular,
The solution to \( x^k e^{-af(x)} f(x) = 1 \)
which seems to crop up a fair bit for your author is \begin{equation} f(x) = \frac{1}{a} W\left( a x^{-k} \right) \end{equation}
Similarly setting \(a=-1\) ...
The solution to \( x e^{f(x)} f(x) = 1 \)
(i.e. where \(x\) is on the wrong side but we otherwise have the Lambert W definition) must be \begin{equation} f(x) = - W\left( -\frac{1}{x} \right) \end{equation} We might also take \(b = \ln(g(x))\) and thus
The solution to \( g(x) e^{-af(x)} = f(x) \)
or equivalently \( f(x)e^{af(x)} = g(x) \) is \begin{equation} f(x) = \frac{1}{a} W\left( a g(x) \right) \end{equation} which reduces to the Lambert W function for \(a=1\), as we expect. It is also pretty obvious from first principles, because if we multiply both sides by \(a\) we have \begin{equation} (af) e^{(af)} = ag \end{equation} and thus \(af = W(ag)\). Next suppose we want a power of \(f\) to appear. Let \(b=\beta(x)/k\), \(c = \gamma^{1/k}\), \(a = k/\alpha\) and \(d=0\). And then raise both sides to the power \(k\). It follows that...
The solution to \( e^{\alpha f(x) + \beta(x)} = \gamma f(x)^k \)
is \begin{equation} f(x) = \frac{\alpha}{k}W\left( \frac{k e^{\frac{1}{k}\beta(x)} }{\alpha \gamma^{1/k}} \right) \end{equation} and if, in particular, \(\beta(x) = -\ln(g(x))\) then
The solution to \( e^{\alpha f(x)} = g(x) \gamma f(x)^k \)
is \begin{equation} f(x) = \frac{\alpha}{k}W\left( \frac{k }{\alpha g(x)^{1/k} \gamma^{1/k}} \right) \end{equation} and if we take \(c = \frac{1}{\gamma}\) and \(g(x)=x\) and \(\alpha = \frac{s}{2}\) and \(k=2\) then p>
The solution to \( e^{-\frac{s}{2}f(x)} x f(x)^2 = c \)
is \begin{equation} f(x) = \frac{s}{2k}W\left( \frac{2k}{s} \sqrt{ \frac{c}{x}} \right) \end{equation}
