Suppose for $\xi=ct-x$ and $\psi=\psi(\xi)$ we have the ODE $$ \partial_{\xi\xi}\psi-c\partial_\xi\psi+f'(V(\xi))\psi=0.\quad (1) $$
Now, let
\begin{align*} V_i=V_i(x,t):=&V((-1)^i(x-d_i(t))-ct),\\ \psi_i=\psi_i(x,t):=&(-1)^i\psi((-1)^i(x-d_i(t))-ct). \end{align*}
It is claimed that $$ -\partial_t\psi_i=\partial_{xx}\psi_i+f'(V_i)\psi_i-\partial_x\psi_i d_i'\qquad (2) $$ since $\psi$ satisfies $(1)$.
I do not see this! Thus my question is: How to get (2)?
I think it might be useful to distinguish the cases that $i$ is even and odd.
For $i$ even, we have $$ \psi_i=\psi(x-d_i(t)-ct)\quad\text{and}\quad V_i=V(x-d_i(t)-ct). $$
For $i$ odd, we have $$ \psi_i=-\psi(x+d_i(t)-ct)\quad\text{and}\quad V_i=V(-x+d_i(t)-ct). $$
Attempt: I tried to use that, for even $i$, the argument of $\psi_i$ and $V_i$ is $$ y:=x-d_i(t)-ct=-\xi-d_i(t). $$ So maybe one can use scaling and shifting, i.e. $$ -y-d_i(t)=\xi, $$ to bring everything back to variable $\xi$ in order to be able to use equation $(1)$? But I have no idea if this helps or makes sense