Defining $$\lambda = \frac{m\alpha^2}{\hbar^2}E,$$ we can rewrite Schroedinger's equation as $$-\frac{d^2}{d\rho^2} \psi(\rho) + \omega_r^2\rho^2\psi(\rho) +\frac{1}{\rho}\psi(\rho) = \lambda \psi(\rho).$$