Discussion
\(\frac{d^{2} f}{dx^{2}} + 2 \operatorname{sech}^{2} x\, f - f = 0.\)
One of the solution is :
Answer: A
One solution is sech x.
Try \(f = \operatorname{sech} x\). Using \((\operatorname{sech} x)' = -\operatorname{sech} x \tanh x\) and \((\tanh x)' = \operatorname{sech}^{2} x\):
\[ \begin{aligned} f' &= -\operatorname{sech} x \tanh x \\ f'' &= \operatorname{sech} x \tanh^{2} x - \operatorname{sech}^{3} x \end{aligned} \]
Substituting:
\[ \begin{aligned} &f'' + 2\operatorname{sech}^{2}x\, f - f \\ &= \operatorname{sech} x\,(\tanh^{2} x + \operatorname{sech}^{2} x - 1) = 0 \end{aligned} \]
since \(\tanh^{2} x + \operatorname{sech}^{2} x = 1\). The same substitution with \(\operatorname{sech}^{2} x\), \(\tanh x\) or \(\operatorname{sech}^{3} x\) leaves terms that do not cancel; for example \(f = \tanh x\) has \(f'' = -2\operatorname{sech}^{2}x\tanh x\), which cancels the middle term and leaves \(-\tanh x \neq 0\).
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