If you have a second order PDE for a function [itex]\psi(x,t)[/itex], it is of the general form
[itex]F(\partial_{tt}\psi,\partial_{xx}\psi,\partial_{xt}\psi,\partial_{t} \psi,\partial_{x}\psi,\psi)=g(x,t)[/itex]
If the equation is linear, then [itex]F[/itex] is linear with respect to all its arguments, i.e.
[itex]F(\partial_{tt}\psi,\partial_{xx}\psi,\partial_{xt}\psi,\partial_{t} \psi,\partial_{x}\psi,\psi,x,t)\\=a(x,t)\partial_{tt}\psi+b(x,t) \partial_{xx} \psi+c(x,t)\partial_{xt}\psi+d(x,t)\partial_{t} \psi+e(x,t)\partial_{x}\psi+f(x,t)\psi[/itex]
There's no terms proportional to [itex](\partial_{x}\psi)^{2}[/itex] or something like that.
In the case of time-dependent SE for one 1D particle, we have [itex]d(x,t)=i\hbar[/itex], [itex]b(x,t)=\frac{\hbar^{2}}{2m}[/itex], [itex]f(x,t)=-V(x,t)[/itex] and other coefficients are zero.