Differential equation $x'(t) e^{-x'(t)^2} = c$ with Lambert W function. Let $x(t)$ be a smooth function, find a solution of $x'(t) e^{-x'(t)^2} =c$.
I first saw this DE in a question of
Frederic Chopin (Integral involving piecewise continuous function), when I started working on the problem I consulted Wolfram Alpha.
Wolfram Alpha suggests a solution for the differential equation $x'(t) e^{-x'(t)^2} = c$ that has to do with Lambert W functions (https://en.wikipedia.org/wiki/Lambert_W_function). 
The solution is of the form $x(t) = k \pm \frac{1}{\sqrt{2}} it\sqrt{W(-2c^2)}$.
They start by saying $x'(t) = \pm \frac{1}{\sqrt{2}} i \sqrt{W(-2c^2)}$. If this is true I believe what they say. However, I do not understand how they found this.
Can anybody clarify the decision of Wolfram Alpha? And if false, does anybody has a suggestion for solving this DE.

Square the equation and multiply with $-2$,
$$
(-2x'^2)e^{-2x'^2}=-2c^2.
$$
Then apply Lambert-W as the inverse of the function $ve^v=u$,
$$
-2x'^2=W(-2c^2).
$$
Now select one of the square roots so that the sign of $x'$ is the sign of $c$ and integrate.
