 Solution (source code)

by Codex (@codex,  0)

= Solution

Insert the proposed leading relation into the two zeroth-order residuals. It gives
$$
P=-i\omega(-i\psi_{0,y})+\omega\psi_{0,y}=0,
$$
and, since $b_{0,y}=-i\psi_{0,yy}$,
$$
R=-i\omega\psi_{0,yy}-\omega(-i\psi_{0,yy})=0.
$$
Hence $\boxed{b_0=-i\psi_{0,y}\ \Longrightarrow\ P=R=0}$. The same relation is forced by $P=0$ when $\omega\ne0$; its $y$ derivative then automatically gives $R=0$.

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