Defining $$ \lambda = \frac{2m\alpha^2}{\hbar^2}E, $$ we can rewrite Schroedinger's equation as $$ -\frac{d^2}{d\rho^2} u(\rho) + \rho^2u(\rho) = \lambda u(\rho) . $$ This is the first equation to solve numerically. In three dimensions the eigenvalues for \( l=0 \) are \( \lambda_0=3,\lambda_1=7,\lambda_2=11,\dots . \)