This question isn't necessarily that much related to quantum theory.
Say we have the wave function $~\psi(x)~$ such that $$\frac{-h}{2m}~\frac{d^2}{dx^2}~\psi(x)+V(x)~\psi(x)=E~\psi(x)$$ where $~E~$ is the energy to the system (we treat it as a constant here) and $~V(x)~$ is a function of $~x~$ as the potential.
Would this imply that $$\frac{-h}{2m}~\frac{d^2}{dx^2}~\psi(-x)+V(-x)~\psi(-x)=E~\psi(-x)~?$$ I rather get the feeling that this will not be the case but is there a way for $~\psi(-x)~$ to satisfy the ODE in some way?
Just substitute $y := -x$. The operator $\frac{d^2}{dx^2}$ becomes $\frac{d^2}{dy^2}$.