Question

The inverse Laplace transform of $$ \frac{1}{2s^2 + 3s + 1} $$ 

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

Hint:

Always factor the denominator and apply partial fraction decomposition to simplify Laplace inverse calculations.

Answer 1

The Correct Option is A

To find the inverse Laplace transform of \[ \frac{1}{2s^2 + 3s + 1} \] Step 1: Factor the quadratic expression in the denominator: \[ 2s^2 + 3s + 1 = (2s + 1)(s + 1) \] Step 2: Apply partial fraction decomposition: \[ \frac{1}{(2s + 1)(s + 1)} = \frac{A}{2s + 1} + \frac{B}{s + 1} \] Step 3: Solve for \( A \) and \( B \): Multiplying both sides by \((2s + 1)(s + 1)\) and equating coefficients gives: \[ 1 = A(s + 1) + B(2s + 1) \] Solve for \( A = 1 \) and \( B = -1 \). Step 4: Write the expression: \[ \frac{1}{(2s + 1)(s + 1)} = \frac{1}{2s + 1} - \frac{1}{s + 1} \] Step 5: Take the inverse Laplace transform: \[ \mathcal{L}^{-1} \left( \frac{1}{2s + 1} \right) = e^{-t/2}, \quad \mathcal{L}^{-1} \left( \frac{1}{s + 1} \right) = e^{-t} \] Thus, the inverse Laplace transform is: \[ e^{-t/2} - e^{-t} \] Conclusion: The correct inverse Laplace transform is given by option (A).