Equations similar to Lambert-$W$ with quadratic exponents

Question

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

Answer 1

$\def\H{\operatorname H}$

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

Answer 2

We see, both your equations are polynomial equations of more than one algebraically independent monomials ($\ln(x),x$) and with no univariate factor. We therefore don't know how to rearrange the equations for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

$$\ln(x)-\frac{bx}{a}=-\frac{bc}{a}$$ $x\to e^t$: $$(bc+at)e^{-t}=b$$

We see, both the factor and the exponent of the exponential term are linear functions of the solution variable ($t$). Such equations can be solved in terms of Lambert W.

$$\ln(x)-dx^2-bx-\frac{bc}{a}=0$$ $x\to e^t$: $$d(e^t)^2+be^t-t+\frac{bc}{a}=0\tag{1}$$

We see, because this equation is an irreducible polynomial equation of $t$ and different powers of $e^t$ greater than 1, the equation seems to be not in a form for applying Generalized Lambert W or Hyper Lambert W.

We see, for algebraic $a,b,c,d$, equation (1) is an irreducible algebraic equation in dependence of both $t$ and $e^t$ simultaneously. Algebraic equations of this kind cannot have solutions except $0$ that are elementary numbers. See Ferng-Ching Lin 1983 and Timothy Chow 1999.