Question

Obtain the inverse Laplace transform of \( F(s) = \frac{1}{s^2(s+2)} \)

• A. \( \frac{1}{2}(2t+e^{-2t}) \)
• B. \( \frac{1}{4}(2t-1) \)
• C. \( \frac{1}{2}(e^{-2t}-1) \)
• D. \( \frac{1}{4}(2t+e^{-2t}-1) \)

Hint:

For inverse Laplace of rational functions, use partial fractions.
Remember standard pairs: \(\mathcal{L}^{-1}\{1/s\}=u(t)\), \(\mathcal{L}^{-1}\{1/s^2\}=t u(t)\), \(\mathcal{L}^{-1}\{1/(s-a)\}=e^{at}u(t)\).

Answer 1

The Correct Option is D

Use partial fraction decomposition for \( F(s) = \frac{1}{s^2(s+2)} \). Let \(\frac{1}{s^2(s+2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+2}\). \(1 = As(s+2) + B(s+2) + Cs^2\). Set \(s=0 \Rightarrow 1 = B(2) \Rightarrow B = \frac{1}{2}\). Set \(s=-2 \Rightarrow 1 = C(-2)^2 = 4C \Rightarrow C = \frac{1}{4}\). Equate coefficients of \(s^2\): \(A+C = 0 \Rightarrow A = -C = -\frac{1}{4}\). So, \(F(s) = -\frac{1}{4s} + \frac{1}{2s^2} + \frac{1}{4(s+2)}\). Inverse Laplace Transform: \(\mathcal{L}^{-1}\{F(s)\} = -\frac{1}{4}\mathcal{L}^{-1}\{\frac{1}{s}\} + \frac{1}{2}\mathcal{L}^{-1}\{\frac{1}{s^2}\} + \frac{1}{4}\mathcal{L}^{-1}\{\frac{1}{s+2}\}\) \(f(t) = -\frac{1}{4}(1) + \frac{1}{2}(t) + \frac{1}{4}(e^{-2t})\) for \(t \ge 0\). \(f(t) = \frac{1}{4}(-1 + 2t + e^{-2t}) = \frac{1}{4}(2t + e^{-2t} - 1)\). \[ \boxed{\frac{1}{4}(2t+e^{-2t}-1)} \]