Question

Which one of the options given is the inverse Laplace transform of \(\frac{1}{s^3 - s}\)?
\(u(t)\) denotes the unit-step function.

• A. \(\left(-1 + \frac{1}{2}e^{-t} + \frac{1}{2}e^t\right)u(t)\)
• B. \(\left(\frac{1}{3}e^{-t} - e^t\right)u(t)\)
• C. \(\left(-1 + \frac{1}{2}e^{-(t-1)} + \frac{1}{2}e^{(t-1)}\right)u(t-1)\)
• D. \(\left(-1 - \frac{1}{2}e^{-(t-1)} - \frac{1}{2}e^{(t-1)}\right)u(t-1)\)

Hint:

The cover-up method is a very fast way to find the coefficients for partial fractions when the denominator has distinct linear roots. To find the coefficient for the term \(\frac{A}{s-a}\), cover up the \((s-a)\) factor in the original fraction and substitute \(s=a\) into what's left.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the inverse Laplace transform of a rational function of \(s\), the standard method is to use partial fraction expansion. This breaks down the complex fraction into simpler terms whose inverse transforms are known.
Step 2: Key Formula or Approach:
1. Factor the denominator of the function \(F(s) = \frac{1}{s^3 - s}\).
2. Express \(F(s)\) as a sum of partial fractions.
3. Find the inverse Laplace transform of each term using standard transform pairs like \(\mathcal{L}^{-1}\{\frac{1}{s}\} = u(t)\) and \(\mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{at}u(t)\).
4. Combine the results.
Step 3: Detailed Calculation:
1. Factor the denominator: \[ s^3 - s = s(s^2 - 1) = s(s-1)(s+1) \] So, \(F(s) = \frac{1}{s(s-1)(s+1)}\). 2. Partial Fraction Expansion: \[ \frac{1}{s(s-1)(s+1)} = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s+1} \] We can find the coefficients A, B, and C using the cover-up method:
- For A (cover up \(s\), set \(s=0\)): \[ A = \frac{1}{(0-1)(0+1)} = \frac{1}{(-1)(1)} = -1 \] - For B (cover up \(s-1\), set \(s=1\)): \[ B = \frac{1}{1(1+1)} = \frac{1}{2} \] - For C (cover up \(s+1\), set \(s=-1\)): \[ C = \frac{1}{-1(-1-1)} = \frac{1}{(-1)(-2)} = \frac{1}{2} \] So the expansion is: \[ F(s) = -\frac{1}{s} + \frac{1/2}{s-1} + \frac{1/2}{s+1} \] 3. Inverse Laplace Transform:
Now we find the inverse transform of each term: \[ f(t) = \mathcal{L}^{-1}\left\{-\frac{1}{s} + \frac{1}{2}\frac{1}{s-1} + \frac{1}{2}\frac{1}{s-(-1)}\right\} \] \[ f(t) = -\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} + \frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} + \frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} \] Using the standard pairs: \[ f(t) = -1 \cdot u(t) + \frac{1}{2} e^{1t} u(t) + \frac{1}{2} e^{-1t} u(t) \] Factoring out \(u(t)\): \[ f(t) = \left(-1 + \frac{1}{2}e^t + \frac{1}{2}e^{-t}\right) u(t) \] Rearranging to match option (A): \[ f(t) = \left(-1 + \frac{1}{2}e^{-t} + \frac{1}{2}e^t\right) u(t) \] This can also be written using the definition of \(\cosh(t) = \frac{e^t + e^{-t}}{2}\) as \(f(t) = (-1 + \cosh(t))u(t)\). Step 4: Final Answer:
The inverse Laplace transform is \(\left(-1 + \frac{1}{2}e^{-t} + \frac{1}{2}e^t\right)u(t)\), which matches option (A).
Step 5: Why This is Correct:
The procedure of partial fraction decomposition is correctly applied to the given function, and the inverse Laplace transform of each resulting simple term is correctly identified from standard tables.