How can I solve this equation, $$x = -c_1 e ^ x + c_2e ^{-x}, \;\;\; 0 < c_1, c_2 < 1$$ We can use $t = e^x$ which will result in, $$t \ln(t) + c_1 t ^ 2 - c_2 = 0, \;\;\; 0 < c_1, c_2 < 1$$ but how can I solve this one then?
Solution for $x = -c_1 e ^ x + c_2 e ^{-x}$
1
$\begingroup$
calculus
-
0That doesn't have the look of something with an analytical solution. Use Newton-Raphson or something like it. – 2011-04-18
-
0Ummm, I already did it. I am just wondered if it can be solved analytically or not. – 2011-04-18
-
5@Mohsen: Highly unlikely. If $c_1=0$ or $c_2=0$, then you need Lambert's W function, having both nonzero is only going to make it more difficult. It is equivalent to $xe^x = -c_1(e^2)^x + c_2$. – 2011-04-18
-
0@Arturo: You are right. Actually, I was solving a system of equations containing two Lambert's W functions. I simplified the equations and got the one I posted here. – 2011-04-18
-
0@Mohsen: Yeah, if you have a system of transcendental equations, the likelihood of obtaining an analytical solution is rather tiny. On the other hand, I'd rather do Newton-Raphson on a system with exponentials than a system with Lambert functions. – 2011-04-18
-
0@J. M.: I was thinking about the same issue :) Is there any reason not to apply Newton-Raphson to the Lambert function? – 2011-04-18
-
0Well, algorithms to compute the Lambert function use Newton-Raphson (or variants thereof) to begin with... so I'd say the Lambert function would be more expensive to compute than an exponential. – 2011-04-18
-
0You wrote $-c1$. Do you have any assumptions on these parameters (namely, positivity)? – 2011-04-18
-
0@Gortaur: I did edit the question and added a constraint. Both $c_1$ and $c_2$ must be in the range (0, 1). Thanks. – 2011-04-18