Does $\frac{d^2x}{dt^2}=-Axe^{-x^2/2}$ have an elementary solution? 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.
 A: $$\frac{d^2x}{dt^2}=-Ax e^{-x^2/2}~~~~(1)$$
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.
A: $$\frac{d^2x}{dt^2}=-Axe^{-x^2/2}\implies \frac{\frac{d^2t}{dx^2}}{\Big[\frac{dt}{dx}\Big]^3}=Axe^{-x^2/2}$$
Reduction of order $p=\frac{dt}{dx}$
$$p^2=\frac{1}{2 A e^{-\frac{x^2}{2}}-c_1}$$
A: $$\frac{d^2x}{dt^2}=-Ax e^{-x^2/2}~~~~(1)$$
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.
