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.
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Analytically solving an equation to validate the solution should not be done. You should simply plug the value back in and see if the equation is approximately true. – Simply Beautiful Art Oct 18 '20 at 02:25
4 Answers
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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.
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I fixed an error in the computation of $b$. Now the result is fully complaint with Wolfram. – Feb 02 '16 at 09:23
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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.
adjan
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This type of equations does not have analytic solution. You can use Taylor formula to estimate it.
corindo
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@Yves Daoust : Taylor's formula gives you polynomial approximation of the function and you can estimate the error by the error term. – corindo Feb 02 '16 at 08:34
