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 $x = -c_1 e ^ x + c_2 e ^{-x}$
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calculus
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0That doesn't have the look of something with an analytical solution. Use Newton-Raphson or something like it. – 2011-04-18
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0Ummm, I already did it. I am just wondered if it can be solved analytically or not. – 2011-04-18
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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
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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
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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
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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
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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
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0You wrote $-c1$. Do you have any assumptions on these parameters (namely, positivity)? – 2011-04-18
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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
1 Answers
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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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1I don't know what you are doing, but surely for $c_1\not= c_2$ $x=0$ is not a solution? – 2011-09-18
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0$x=0$ 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... – 2011-09-18
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0@J.M,you are right...obvious logical mistake – 2011-09-18