"\\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"
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