# Solution for $x = -c_1 e ^ x + c_2 e ^{-x}$

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?

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That doesn't have the look of something with an analytical solution. Use Newton-Raphson or something like it. – J. M. Apr 18 '11 at 4:45
Ummm, I already did it. I am just wondered if it can be solved analytically or not. – Helium Apr 18 '11 at 4:52
@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 $xe^x = -c_1(e^2)^x + c_2$. – Arturo Magidin Apr 18 '11 at 4:54
@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. – Helium Apr 18 '11 at 5:02
@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. – J. M. Apr 18 '11 at 5:14

Let's find first derivative of the both sides of the equation:

$x'=(-C_1e^x)'+(C_2e^{-x})'$

$1=-C_1e^x - C_2e^{-x}$ and now let's find first derivative of the left and right side:

$(1)'=(-C_1e^x)' - (C_2e^{-x})'$

$0=-C_1e^x + C_2e^{-x} \Rightarrow C_1e^x=C_2e^{-x}$ , which means that:

$x=-C_1e^x + C_1e^x$

$x=0$

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I don't know what you are doing, but surely for $c_1\not= c_2$ $x=0$ is not a solution? – Thomas Rot Sep 18 '11 at 13:50
$x=0$ would be right if $c_1$ and $c_2$ were the same, but your method is not a proper method for a general solution. Differentiation "loses" constants... – J. M. Sep 18 '11 at 13:52
@J.M,you are right...obvious logical mistake – pedja Sep 18 '11 at 14:04