Letting $u=rR(r)$, the radial part of the SE becomes:
I am interested in obtaining the energy of the ground state (which I know is $3\hbar \omega/2$). As such, I set $l=0$ to get
which is identical the 1D harmonic oscillator problem. The lowest energy of the 1D oscillator is $\hbar \omega/2$, which is not the right energy for the 3D case. Why does this method not give me the proper energy for the 3D case? How can I find the ground state energy using the spherical equations?