how to solve $x(e^{-{1\over x}}-1)=$ constant

Question

As mentionned in the title, how to solve analytically the equation

$x \cdot \left(e^{-\frac{c_1}{x}}-1\right)=c_2$ where $c_1$ and $c_2$ are known constants.

I can easily find a solution numerically, but i would like to validate it analytically.

Answer 1

Let $x=\dfrac1{at+b}$. Then

$$e^{-{c_1\over x}}=\frac{c_2}x+1$$

becomes

$$e^{-c_1(at+b)}=c_2(at+b)+1,$$

$$e^{-c_1at}e^{-c_1b}=c_2at+c_2b+1.$$

We choose $a=\dfrac1{c_1}$ and $b=-\dfrac1{c_2}$ and the equation simplifies to

$$e^{-t}e^{-c_1b}=c_2at,$$

$$\frac1{c_2a}e^{-c_1b}=te^t,$$ i.e.

$$\frac{c_1}{c_2}e^{c_1/c_2}=te^t.$$

Finally,

$$x=\frac{c_1c_2}{c_2W\left(\frac{c_1}{c_2}e^{c_1/c_2}\right)-c_1},$$ where $W$ denotes the Lambert function. You can't find a simpler form.

Answer 2

This type of equations does not have analytic solution. You can use Taylor formula to estimate it.

Answer 3

According to Wolfram Alpha the result is

$$x = \frac{c_1c_2}{ {-c_1 + c_2 \cdot W\left({c_1 e^{{c_1 \over c_2}}\over c_2}\right)}}$$

with $W$ being the product logarithm function, which cannot be expressed in terms of elementary functions.

Answer 4

The analytic solution of the equation is shown below :