How do i compute this differential equation?

Question

$$ y''(x) - \frac{B^2}{\cosh^2(Bx)}y(x)= Ay(x) $$

I know that can be solved analytically, but what method can I use?

Answer 1

Assume $B\neq0$ for the key case:

Hint:

$y''(x)-\dfrac{B^2}{\cosh^2(Bx)}y(x)=Ay(x)$

Let $r=\cosh(Bx)$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=B\sinh(Bx)\dfrac{dy}{dr}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(B\sinh(Bx)\dfrac{dy}{dr}\right)=B\sinh(Bx)\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)+B^2\cosh(Bx)\dfrac{dy}{dr}=B\sinh(Bx)\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}+B^2\cosh(Bx)\dfrac{dy}{dr}=B\sinh(Bx)\dfrac{d^2y}{dr^2}B\sinh(Bx)+B^2\cosh(Bx)\dfrac{dy}{dr}=B^2\sinh^2(Bx)\dfrac{d^2y}{dr^2}+B^2\cosh(Bx)\dfrac{dy}{dr}$

$\therefore B^2\sinh^2(Bx)\dfrac{d^2y}{dr^2}+B^2\cosh(Bx)\dfrac{dy}{dr}-\dfrac{B^2}{\cosh^2(Bx)}y=Ay$

$B^2(r^2-1)\dfrac{d^2y}{dr^2}+B^2r\dfrac{dy}{dr}-\left(A+\dfrac{B^2}{r^2}\right)y=0$

Let $s=r^2$ ,

Then $\dfrac{dy}{dr}=\dfrac{dy}{ds}\dfrac{ds}{dr}=2r\dfrac{dy}{ds}$

$\dfrac{d^2y}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{dy}{ds}\right)=2r\dfrac{d}{dr}\left(\dfrac{dy}{ds}\right)+2\dfrac{dy}{ds}=2r\dfrac{d}{ds}\left(\dfrac{dy}{ds}\right)\dfrac{ds}{dr}+2\dfrac{dy}{ds}=2r\dfrac{d^2y}{ds^2}2r+2\dfrac{dy}{ds}=4r^2\dfrac{d^2y}{ds^2}+2\dfrac{dy}{ds}$

$\therefore B^2(r^2-1)\left(4r^2\dfrac{d^2y}{ds^2}+2\dfrac{dy}{ds}\right)+2B^2r^2\dfrac{dy}{ds}-\left(A+\dfrac{B^2}{r^2}\right)y=0$

$4B^2r^2(r^2-1)\dfrac{d^2y}{ds^2}+2B^2(2r^2-1)\dfrac{dy}{ds}-\left(A+\dfrac{B^2}{r^2}\right)y=0$

$4B^2s(s-1)\dfrac{d^2y}{ds^2}+2B^2(2s-1)\dfrac{dy}{ds}-\left(A+\dfrac{B^2}{s}\right)y=0$

$\dfrac{d^2y}{ds^2}+\dfrac{2s-1}{2s(s-1)}\dfrac{dy}{ds}-\left(\dfrac{A}{4B^2s(s-1)}+\dfrac{1}{4s^2(s-1)}\right)y=0$

$\dfrac{d^2y}{ds^2}+\left(\dfrac{1}{2s}+\dfrac{1}{2(s-1)}\right)\dfrac{dy}{ds}-\left(\dfrac{A+B^2}{4B^2s(s-1)}-\dfrac{1}{4s^2}\right)y=0$

Let $y=s^ku$ ,

Then $\dfrac{dy}{ds}=s^k\dfrac{du}{ds}+ks^{k-1}u$

$\dfrac{d^2y}{ds^2}=s^k\dfrac{d^2u}{ds^2}+ks^{k-1}\dfrac{du}{ds}+ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u=s^k\dfrac{d^2u}{ds^2}+2ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u$

$\therefore s^k\dfrac{d^2u}{ds^2}+2ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u+\left(\dfrac{1}{2s}+\dfrac{1}{2(s-1)}\right)\left(s^k\dfrac{du}{ds}+ks^{k-1}u\right)-\left(\dfrac{A+B^2}{4B^2s(s-1)}-\dfrac{1}{4s^2}\right)s^ku=0$

$\dfrac{d^2u}{ds^2}+\dfrac{2k}{s}\dfrac{du}{ds}+\dfrac{k(k-1)}{s^2}u+\left(\dfrac{1}{2s}+\dfrac{1}{2(s-1)}\right)\dfrac{du}{ds}+\left(\dfrac{k}{2s^2}+\dfrac{k}{2s(s-1)}\right)u-\left(\dfrac{A+B^2}{4B^2s(s-1)}-\dfrac{1}{4s^2}\right)u=0$

$\dfrac{d^2u}{ds^2}+\left(\dfrac{4k+1}{2s}+\dfrac{1}{2(s-1)}\right)\dfrac{du}{ds}+\left(\dfrac{4k^2-2k+1}{4s^2}+\dfrac{B^2(2k-1)-A}{4B^2s(s-1)}\right)u=0$

Choose $4k^2-2k+1=0$ , i.e. $k=\dfrac{1\pm i\sqrt3}{4}$ , the ODE becomes

$\dfrac{d^2u}{ds^2}+\left(\dfrac{2\pm i\sqrt3}{2s}+\dfrac{1}{2(s-1)}\right)\dfrac{du}{ds}+\dfrac{B^2(-1\pm i\sqrt3)-2A}{8B^2s(s-1)}u=0$

Which reduces to Gaussian hypergeometric equation.