Discussion
Which one of the options given is the inverse Laplace transform of \(\dfrac{1}{s^{3} - s}\)?
\(u(t)\) denotes the unit-step function.
Answer: A
The inverse transform is \(\left(-1 + \frac{1}{2}e^{-t} + \frac{1}{2}e^{t}\right)u(t)\).
Factor the denominator and split into partial fractions:
\[ \begin{aligned} \frac{1}{s^{3} - s} &= \frac{1}{s(s - 1)(s + 1)} \\ &= -\frac{1}{s} + \frac{1/2}{s - 1} + \frac{1/2}{s + 1} \end{aligned} \]
The coefficients come from the cover-up rule: at \(s = 0\) the rest is \(1/((-1)(1)) = -1\); at \(s = 1\) it is \(1/(1 \times 2) = 1/2\); at \(s = -1\) it is \(1/((-1)(-2)) = 1/2\). Inverting term by term gives \(-1 + \frac{1}{2}e^{t} + \frac{1}{2}e^{-t}\) for \(t \ge 0\), that is \((\cosh t - 1)u(t)\).
The two options with \(u(t - 1)\) are delayed by one unit, which would need a factor \(e^{-s}\) in the transform; there is none. The option \(\left(\frac{1}{3}e^{-t} - e^{t}\right)u(t)\) fails a quick check: since the transform falls like \(1/s^{3}\), the function must start at 0, but that option starts at \(1/3 - 1 = -2/3\).
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