\[\frac{{\mathrm{d}}^{2}u}{{\mathrm{d}\zeta}^{2}}=\frac{1}{u}\left(\frac{\mathrm%
{d}u}{\mathrm{d}\zeta}\right)^{2}-\frac{1}{\zeta}\frac{\mathrm{d}u}{\mathrm{d}%
\zeta}+\frac{u^{2}(\alpha+\gamma u)}{4\zeta^{2}}+\frac{\beta}{4\zeta}+\frac{%
\delta}{4u},\]