= Solution
Put $D=1+i\hbar t/m$ and write the <Gaussian wave packet> as $\psi=\pi^{-1/4}D^{-1/2}e^{-x^2/(2D)}$. Differentiation gives
$$
\frac{\psi_t}{\psi}=\frac{i\hbar}{m}\left(-\frac1{2D}+\frac{x^2}{2D^2}\right),\qquad
\frac{\psi_{xx}}{\psi}=\frac{x^2}{D^2}-\frac1D.
$$
Hence $\boxed{i\hbar\psi_t=-(\hbar^2/2m)\psi_{xx}}$, the free <Time-dependent Schrödinger equation>.
With $\tau=\hbar t/m$, its <probability density> is
$$
|\psi|^2=\frac{1}{\sqrt{\pi(1+\tau^2)}}\exp\left(-\frac{x^2}{1+\tau^2}\right).
$$
The stated <Gaussian integral> shows its integral is one. Symmetry gives $\langle x\rangle=0$; differentiating that Gaussian integral with respect to its coefficient gives $\langle x^2\rangle=(1+\tau^2)/2$. Thus the position <standard deviation> is
$$
\boxed{\Delta x=\sqrt{\frac{1+(\hbar t/m)^2}{2}}.}
$$
\b[It is not a stationary state:] its probability density changes with time as the packet spreads. A <stationary state> would have only an overall time-dependent phase and a fixed position density.
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