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

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

To solve for the inverse Laplace transform, we perform partial fraction decomposition on \( F(s) \).

\( \frac{1}{s^2(s+2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+2} \)

Multiply both sides by \( s^2(s+2) \) to obtain: \[ 1 = A s(s+2) + B(s+2) + C s^2 \]

Now, equate coefficients and solve for \( A \), \( B \), and \( C \). After solving, we find: \[ A = \frac{1}{4}, \quad B = \frac{1}{2}, \quad C = \frac{-1}{4} \]

Now, apply the inverse Laplace transform to each term: \[ \mathcal{L}^{-1}\left(\frac{1}{s}\right) = 1, \quad \mathcal{L}^{-1}\left(\frac{1}{s^2}\right) = t, \quad \mathcal{L}^{-1}\left(\frac{1}{s+2}\right) = e^{-2t} \]

Combining these results, we get the inverse Laplace transform: \[ \mathcal{L}^{-1}\left( F(s) \right) = \frac{1}{4}(2t + e^{-2t} - 1) \]

Answer: \[ \boxed{\frac{1}{4}(2t + e^{-2t} - 1)} \]