I can not to solve the following equation $$(*) \qquad u''(r) +2n\coth(r)\,u'(r)+ (n^2+\lambda^2) \, u(r)= 0 \quad \mbox{with} \, r>0$$ where $n\in \mathbb N$ and $\lambda \in \mathbb C $.
That I have the transformed into the equation in $z$ follows $$(* *) \qquad z^{ ''}(r) + \left( \frac{n(1-n)}{\sinh^{2}(r)}+\lambda^{2} \right) z(r) = 0, \quad \mbox{with} \, r>0$$ with $u(r)=g(r) z(r)$, where $g(r) = \dfrac{1}{\sinh^{n}(r)}$.
Someone can help me!!!
Thanks in advance
Use the substitution $s(r) = \cosh r$ and write $z(r) = Z(s(r))$. Then you obtain for $Z$ the differential equation \begin{equation} (1-s^2)Z''(s) - s Z'(s) + \left(\frac{n(n-1)}{1-s^2}+\lambda^2\right) Z(s) = 0, \end{equation} which is (a form of the) the associated Legendre equation. In this case, you can obtain the `proper' associated Legendre equation by introducing $\zeta(s) = (s^2-1)^{-\frac{1}{4}} Z(s) = \frac{1}{\sqrt{\sinh r}} \,Z(\cosh r)$. Eventually, you will obtain \begin{equation} z(r) = \sqrt{\sinh r} \left[c_1 P_{i \lambda - \frac{1}{2}}^{n-\frac{1}{2}}(\cosh r) + c_2 Q_{i \lambda - \frac{1}{2}}^{n-\frac{1}{2}}(\cosh r) \right], \end{equation} where $P_\nu^\mu(z)$ and $Q_\nu^\mu(z)$ are the standard solutions to the associated Legendre equation; see the DLMF link for relevant transformation rules which might simplify your expressions.