Non-constant-coefficient second-order linear ODE

Question

I came across this equation in my research (related to reaction diffusion system): $$\frac{d^2y}{dr^2}+B\operatorname{sech}^2(r) \frac{dy}{dr} + Cy = 0$$ where $B$ and $C$ are constants. Can it be solved analytically?

Answer 1

Making the change of the independent variable $x=e^{it}$, we obtain $$x^2w''+xw'+\frac{4iBx^3}{(x^2+1)^2}w'-Cw=0.$$ This equation has $2$ regular singularities $0,\infty$, and irregular at $\pm i$. Therefore for generic $B$ and $C$ its solutions cannot be expressed in terms of functions of hypergeometric type, or any other usual special functions.

Answer 2

Mathematica cannot solve this ODE even when $B=C=1$:

So, it seems very unlikely that this ODE can be solved explicitly.

Answer 3

Maple gives the solution in terms of the Heun confluent function. $$ y \! \left(r \right) = c_1 \mathit{HeunC}\! \left(2 B , \mathrm{i} \sqrt{C}, \mathrm{i} \sqrt{C}, -2 B , -\frac{C}{2}+B , \frac{\tanh \! \left(r \right)}{2}+\frac{1}{2}\right) \left(\cosh^{\mathrm{-i} \sqrt{C}}\left(r \right)\right)+c_2 \mathit{HeunC}\! \left(2 B , \mathrm{-i} \sqrt{C}, \mathrm{i} \sqrt{C}, -2 B , -\frac{C}{2}+B , \frac{\tanh \! \left(r \right)}{2}+\frac{1}{2}\right) \left(\frac{\tanh \! \left(r \right)}{2}+\frac{1}{2}\right)^{-\frac{\mathrm{i}}{2} \sqrt{C}} \left(-\frac{1}{2}+\frac{\tanh \! \left(r \right)}{2}\right)^{\frac{\mathrm{i}}{2} \sqrt{C}}$$

Answer 4

Like Josif Pinelis, I use the substitution $e^{2r}=x$. This leads to the equation $$C(1 + x)^2 y + 4 x (1 + 2 (1 + B) x + x^2) y' + 4x^2 (1 + x)^2 y''=0.$$ This has regular singular points at 0 and $\infty$ and an irregular singular point at $-1$, so it is a confluent Heun equation.