 Solution

Integrating $F=-kx$ gives $V=k{x}^2/2$ up to a constant. Thus $\boxed{-{\hslash }^2{\varphi }^{\prime \prime }/(2m)+m{\omega }^2{x}^2\varphi /2=E\varphi }$, with $\omega =\sqrt{k/m}$. The potential is a symmetric upward parabola and tends to infinity in both directions; normalizable states decay at large $|x|$.