We have thus $$-\frac{d^2}{d\rho^2} u(\rho) + \frac{mk}{\hbar^2} \alpha^4\rho^2u(\rho) = \frac{2m\alpha^2}{\hbar^2}E u(\rho) .$$ The constant $\alpha$ can now be fixed so that $$\frac{mk}{\hbar^2} \alpha^4 = 1,$$ or $$\alpha = \left(\frac{\hbar^2}{mk}\right)^{1/4}.$$ 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 $\lambda_0=3,\lambda_1=7,\lambda_2=11,\dots .$