Question

Consider the Schrödinger equation for a particle in the presence of the potential
of the form

Answer 1

$\frac{h^2d^2\psi(x)}{2m{dx^2}}+V(x)\psi(x)= E\psi(x)$

and

$\psi(x) = Ax{e^\frac{-x^2}{c^2}}$ having $E = 0$

Therefore; $\frac{h^2d^2\psi(x)}{2m{dx^2}}+V(x)\psi(x)= 0$