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The ODE $$−y'' (x)−\frac{2a^2}{\cosh^2 (ax)} y(x)=k^2 y(x)$$ can be made into the form $$\frac{\cosh^2(ax)}{k^2\cosh(ax) - a^2} = \frac{y}{y''}.$$

Observing that $y'' = k^2\cosh(ax) - a^2$, we get the solution.

The solution is $$y=k^2 \cosh(ax)−\frac{(ax)^2}{2}+c_1 x+c_2.$$

However, the $y$ on the numerator implies the solution is only $\cosh^2(ax)$.

Question Why does it include a $k^2$, three extra terms, and can $k$ be determined in terms of $a$?

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  • $\begingroup$ $a/b=c/d$ does not in any way, shape, or form suggest that $a=c$ and $b=d$. Instead, it suggests that $c=ra,d=rb$ for some $r$. $\endgroup$ – oldrinb Dec 21 '14 at 1:53
  • $\begingroup$ Correct @oldrinb, in which the number $r$ was carefully considered and is part of the coefficients $k, a$. So $k = rm$ and $a = rn$ for some $m, n$ $\endgroup$ – Don Larynx Dec 21 '14 at 1:57
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The general solution of the ODE : $$−y'' (x)−\frac{2a^2}{\cosh^2 (ax)} y(x)=k^2 y(x)$$ without the approximation for large $x$, involves the Gauss hypergeometric function. It is a complicated formula (joint page) :

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This differential equation arises in quantum mechanics: Let $E = k^2$ and

$$H = -\frac{d^2}{dx^2} - \frac{2a^2}{\cosh^2(ax)}$$

be the Hamiltonian ($\hbar = 2m = 1$) on $L^2(\mathbb{R})$ where $a \in \mathbb{R}$. The second term represents an attracting potential which drops down quickly for $x \to \pm \infty$. Then, the given differential equation is Schrödinger's equation.

Using the ansatz

$$\psi_k(x) = \left( 1 + i \frac{f(x)}{k} \right) e^{ikx},$$

we find

$$-\frac{i}{k} f''(x) + 2f'(x) - \frac{i}{k} \frac{2a^2}{\cosh^2(ax)} f(x) - \frac{2a^2}{\cosh^2(ax)} = 0.$$

Therefore, clearly

$$f(x) = a \tanh(ax).$$

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