Solving a differential equation of order two

Question

I can not to solve the following equation $$ y^{ ''}(t) + \left( \lambda^{2} - \frac{2}{\sinh^{2}(t)} \right) y(t) = 0, \quad \mbox{with} \, t>0 $$ where $\lambda \in \mathbb C$.

Someone can help me!!!

Thanks in advance

Answer 1

Using the substitution $s(t) = \cosh(t)$ and write $y(t) = Y(s(t))$. We thus find for $Y$ the differential equation \begin{equation} (1-s^2)Y''(s) - s Y'(s) + \left(\frac{2}{1-s^2}+\lambda^2\right) Y(s) = 0, \end{equation} which is (a form of the) the associated Legendre equation. In this case, we can obtain the ‘proper’ associated Legendre equation by introducing $$\zeta(s) = (s^2-1)^{-\frac{1}{4}} Y(s) = \frac{1}{\sqrt{\sinh t}} \,Y(\cosh t)$$ In fact, plugging $Y(s) = (s^2-1)^{\frac{1}{4}} \zeta(s)$ into the equation for $Y$ yields the associated Legendre equation, multiplied by a prefactor $-(s^2-1)^\frac{1}{4}$, with $\nu(\nu+1) = -\lambda^2-\frac{1}{4}$ and $\mu^2 = \frac{9}{4}$.

Eventually, we obtain \begin{align} y(t) &= \sqrt{\sinh t} \left[c_1 P_{i \lambda - \frac{1}{2}}^{2-\frac{1}{2}}(\cosh t) + c_2 Q_{i \lambda - \frac{1}{2}}^{2-\frac{1}{2}}(\cosh t) \right]\\ &= \sqrt{\sinh t} \left[c_1 P_{i \lambda - \frac{1}{2}}^{\frac{3}{2}}(\cosh t) + c_2 Q_{i \lambda - \frac{1}{2}}^{\frac{3}{2}}(\cosh t) \right] \end{align} where $P_\nu^\mu(z)$ and $Q_\nu^\mu(z)$ are the standard solutions to the associated Legendre equation given by $$\left( \left(1-x^{2}\right) \frac{d^2}{dx^2}-2x\frac{d}{dx} +\left(\nu(\nu+1) -\frac{\mu^2}{1-x^2} \right) \right)w=0$$ Finally, My question, is what we can write this solution in terms of elementary functions ($\sinh, \cosh ..)$, without intervening the Legendre function.

Thanks in advance