I've seen solutions saying that an equation in the format:$$ \ln(x) – \frac{bx}{a} = – \frac{bc}{a} $$
can be solved using the Lambert W function and I am comfortable doing so. My equation however is in the form:
$$\ln(x) – {dx^2} – {bx} – \frac{bc}{a} = 0$$
I am aware of a specific solution where $b = 0$ but unfortunately I don't have that luxury. If a completely different function is needed instead of the Lambert W for a general solution to this, I don't know where to start looking
Best Answer
$\def\H{\operatorname H}$
Series solution 1
Let $x=e^w$ and apply Lagrange reversion to get a branch of the inverse function:
$$\ln(x)-ax^2-bx=c\iff w=c+ae^{2w}+be^w\implies x=e^c+\sum_{n=1}^\infty\frac1{n!}\frac{d^{n-1}}{dc^{n-1}}e^c(ae^{2c}+be^c)^n$$
Then, use Binomial theorem:
$$\frac{d^{n-1}}{dc^{n-1}}e^c(ae^{2c}+be^c)=\sum_{m=0}^n\binom mn\frac{d^{n-1}}{dc^{n-1}}e^{c(m+n+1)}$$
Therefore:
$$\bbox[3px,border:3px solid green]{\ln(x)-ax^2-bx=c\implies x=e^c+\sum_{n=1}^\infty\sum_{m=1}^n\frac{a^m b^{n-m}e^{(m+n+1)c}}{(n-m)!m!}(m+n+1)^{n-1}}$$
shown here
Series solution 2
Apply a Tschirnhaus transformation $2ax=w-b$ and Lagrange reversion:
$$\ln(x)-ax^2-bx=c\iff w=b+2a e^{c-\frac{b^2}{4a}}e^\frac{w^2}{4a}=b+ue^{-vw^2}=b+\sum_{n=1}^\infty\frac{u^n}{n!}\frac{d^{n-1}}{db^{n-1}}e^{-vnb^2}$$
Somehow, expanding $e^x$ as a Maclaurin series, evaluating derivatives, and evaluating this sum makes it hard to find the Hermite polynomials, but recalling their Rodrigues formula:
$$\H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$
and $\frac{d^n f(ax)}{da^n}=a^nf^{(n)}(ax)$ gives:
$$\bbox[3px,border:3px solid blue]{w=b+ue^{-vw^2}\implies w=b-\sum_{n=1}^\infty\frac{\left(-ue^{-vb^2}\right)^n}{n!} (vn)^\frac{n-1}2\H_{n-1}(\sqrt{vn}b)}$$
shown here