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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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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0@J.M,you are right...obvious logical mistake – 2011-09-18