Solution for $x = -c_1  e ^ x + c_2  e ^{-x}$ How can I solve this equation,
$$x = -c_1 e ^ x + c_2e ^{-x}, \;\;\; 0 &lt; c_1, c_2 &lt; 1$$
We can use $t = e^x$ which will result in,
$$t \ln(t) + c_1 t ^ 2 - c_2 = 0, \;\;\; 0 &lt; c_1, c_2 &lt; 1$$
but how can I solve this one then?
Let's find first derivative of the both sides of the equation:

$x&#39;=(-C_1e^x)&#39;+(C_2e^{-x})&#39;$

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

$(1)&#39;=(-C_1e^x)&#39; -  (C_2e^{-x})&#39;$

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

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

$x=0$
Let's find first derivative of the both sides of the equation:

$x&#39;=(-C_1e^x)&#39;+(C_2e^{-x})&#39;$

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

$(1)&#39;=(-C_1e^x)&#39; -  (C_2e^{-x})&#39;$

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

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

$x=0$