Does $\frac{d^2x}{dt^2}=-Axe^{-x^2/2}$ have an elementary solution?

Question

I'm trying to find a solution to the following differential equation: $$ \frac{d^2x}{dt^2}=-Axe^{-x^2/2} $$ Or better yet if I can find some function $\xi(x)$ such that $$ \frac{d^2\xi}{dt^2}=-B\xi. $$ Somehow, $x(t)$ is supposed to be periodic (how would I know that in general, or maybe it's just from my inital conditions) and I'd like to map it onto a function $\xi(x)$ that is sinusoidal in time. I'm obviously not a math person, so what are some resources I can look into for doing something like this? (if it is even possible, analytically I mean...)

Is there away to get $\xi(x)$ directly, or would I just have to solve the first differential equation, hope there is some sinusoidal term in there and try to factor that out? Everything is real.

Edit: Notice that $\xi(t)=D\sin(\omega t+\phi)$, where $\omega=\sqrt{B}$ and $D,\phi$ are some horrendous mixture of constants. According to @Aryadeva's answer below, $$ \frac{dx}{dt}=\sqrt{C+2Ae^{-x^2/2}}. $$ Then, the $x$ derivative of the transformation $\xi(x)$ is \begin{align*} \frac{d\xi}{dx}&=\frac{d\xi}{dt}\frac{dt}{dx}=\frac{d\xi}{dt}\left(\frac{dx}{dt}\right)^{-1}\\ &=\frac{\omega D\cos{(\omega t+\phi)}}{\sqrt{C+2Ae^{-x^2/2}}}. \end{align*} It follows that \begin{align*} \xi(x)=\omega D\cos{(\omega t+\phi)}\int_{x_{0}}^{x}\frac{dq}{\sqrt{C+2Ae^{-q^2/2}}}+E. \end{align*} Or maybe if we wanted to get rid of that pesky looking time dependence, everyone's favorite trig identity implies \begin{align*} D^2\cos^2(\omega t+\phi)&=D^2(1-\sin^2(\omega t+\phi))\\&=D^2-\xi(x)^2 \end{align*} so returning a few steps above, \begin{align*} \frac{d\xi}{dx} &=\frac{\omega D\cos{(\omega t+\phi)}}{\sqrt{C+2Ae^{-x^2/2}}}\\&=\omega\frac{\sqrt{D^2-\xi(x)^2}}{\sqrt{C+2Ae^{-x^2/2}}} \end{align*} which returns the separable \begin{align*} \int\frac{d\xi}{\sqrt{D^2-\xi^2}} =\int\frac{\omega dx}{\sqrt{C+2Ae^{-x^2/2}}}. \end{align*} I don't know if I want to get into those inverse tangents to find out for sure (at least not now), but maybe there's a way to isolate $\xi(x)$ through this whole mess and solve this numerically.

Answer 1

$$\frac{d^2x}{dt^2}=-Ax e^{-x^2/2}~~~~(1)$$ Multiply by $2\frac{dx}{dt}$ on both sides $$\implies 2\frac{dx}{dt}\frac{d^2x}{dt^2}=-2A xe^{-x^2/2} \frac{dx}{dt}$$ $$\implies \frac{d}{dt}\left(\frac{dx}{dt}\right)^2=-2 A x e^{-x^2/2} \frac{dx}{dt}$$ Integrate w.r.t $t$ both sides $$\int \frac{d}{dt}\left(\frac{dx}{dt}\right)^2 dt=\left(\frac{dx}{dt}\right)^2=-\int 2 A x e^{-x^2/2} dx$$ $$\implies \frac{dx}{dt}=\pm \sqrt{B+ 2 A e^{-x^2/2}}$$ $$\implies \int \frac{dx}{\sqrt{B+ 2 A e^{-x^2/2}}}=\pm t +C$$ Here, $B$ and $C$ are two constants to be determined by the initial conditions.

Answer 2

$$\frac{d^2x}{dt^2}=-Axe^{-x^2/2}\implies \frac{\frac{d^2t}{dx^2}}{\Big[\frac{dt}{dx}\Big]^3}=Axe^{-x^2/2}$$

Reduction of order $p=\frac{dt}{dx}$ $$p^2=\frac{1}{2 A e^{-\frac{x^2}{2}}-c_1}$$

Answer 3

$$\frac{d^2x}{dt^2}=-Ax e^{-x^2/2}~~~~(1)$$ That you can rewrite as: $$\frac{d^2x}{dt^2}=A\dfrac {d e^{-x^2/2}}{dx}$$ $$\frac{dx}{dt}\frac{d^2x}{dt^2}=A\dfrac {d e^{-x^2/2}}{dx}\frac{dx}{dt}$$ $$ x'dx'=A{d e^{-x^2/2}}$$ $$\int x'dx'=A\int {d e^{-x^2/2}}$$ $$x'^2=C+2A { e^{-x^2/2}}$$ It's separable but not easy to integrate.