My question follows this question: Naive interpretation of Galilean invariance of the TDSE
Essentially, I'm not sure how to proceed mathematically.
We have the transformations:
$$\begin{cases}x'=x-vt\\t'=t\end{cases}.\tag{1}$$
Then, the TDSE for a free particle is
$$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}\tag{2}$$
which after transforming the coordinates becomes
$$i\hbar\left(\frac{\partial\psi}{\partial t'}-v\frac{\partial\psi}{\partial x'}\right)=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x'^2}\tag{3}$$
For simplification, we look at the plane wave
$$\psi\left(x,t\right)=e^{i(px-Et)/\hbar}$$
using
$$x'=x-vt$$
we get
$$p'=p-mv$$ so that $$p'x'-E't' = p'x' - \frac{p'^2}{2m}t' =(p-mv)(x-vt) - \frac{(p-mv)^2}{2m}t=px-Et-mvx+\frac{mv^2}{2}t$$
then the wavefunction in the new frame is
$$\psi'(x',t')=e^{i(p'x'-E't')/\hbar}=e^{i(px-Et)/\hbar}e^{-i(mvx-\frac{mv^2}{2}t)/\hbar}$$
My question is how do I show that the new wavefunction satisfies schroedinger's equation. I tried plugging it into both of (2) or (3), it doesn't satisfy either, so what am I doing wrong?
