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.