Solving transcendental equation involving exponential function

Question

I want to find analytical solutions of the equation $$(αx+β)=δe^{γx}, \qquad (1)$$ which is exactly this one: Solving transcendental equation involving exponential functions

The substitution $y = -\gamma x - \beta \gamma / \alpha$ transforms it into $$ y = \mu \cdot e^{-y}, \qquad (2)$$ where $\mu = -\gamma/\alpha \cdot e^{-\gamma \beta/\alpha}$.

When $\mu \geq 0 $, this equation has only one root, because functions $z = y$ and $z = \mu e^{-y}$ intersect in one point. Then I can treat this root as a funtion of $\mu$, that is $y = f(\mu)$. I can expand this function into the series about $\mu = 0$ by differentiating (2) and treating $y$ as an implicit function of $\mu$. I have a feeling though, that this is a reinvenging of wheel, and $y(\mu)$ must be a well-known special function.

Can someone point me to right direction where I can find an analytical representation of $y(\mu)$ through known (special) functions?

Answer 1

The equation

$$ y e^y = \mu $$

Has the solution in the Lambert W function

$$ y = W(\mu) $$

which is just defined as the inverse of the function $$ f(y) = ye^y $$

For $ -1/e < \mu < 0 $ the inverse is not one-to-one and you get two real solutions.

There does exist a Taylor series of the main branch ($y \ge -1$) around $0$ that converges for $-1/e < \mu < 1/e $

$$ W_0(x) = \sum_{n=1}^{\infty}\frac{(-n)^{n-1}}{n!}x^n $$

For larger values you get a rather complicated series involving $\ln x$ and $\ln(\ln x)$. Details are all in the Wiki page.

Answer 2

Substituting back in Dylan's answer. The solution $x$ to $$ \alpha x+\beta = \delta e^{\gamma x} $$ is $$ x = -{\frac {1}{\gamma\alpha} \left( \alpha\,W \left(-{\frac { \delta\gamma}{\alpha}{e^{-{{\beta\gamma}/{\alpha} }}}}\right)+\beta\,\gamma \right) } $$ using the Lambert W function.