Question Bank • QB_EM_ADV_010 NAT • 2 Marks

Section 1: Engineering Mathematics

Differential Equations

Subtopic: Laplace transforms and their inverse

Question Statement:

The Laplace transform of $f(t) = t e^{-2t} \sin(3t)$ is evaluated at $s = 2$. The value of $1000 \times F(2)$ is ________ (round off to two decimal places).

Answer Key:

38.40

Step-by-Step Solution & Explanation

1. Standard Laplace transform: $\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2 + 9}$ 2. Multiplication by $t$: $\mathcal{L}\{t \sin(3t)\} = -\frac{d}{ds}\left( \frac{3}{s^2 + 9} \right) = \frac{6s}{(s^2 + 9)^2}$ 3. First shifting theorem (multiplication by $e^{-2t}$): $F(s) = \mathcal{L}\{t e^{-2t} \sin(3t)\} = \frac{6(s+2)}{((s+2)^2 + 9)^2}$ 4. Evaluating at $s = 2$: $s + 2 = 4$ $(s+2)^2 + 9 = 4^2 + 9 = 25$ $F(2) = \frac{6(4)}{(25)^2} = \frac{24}{625} = 0.0384$ $1000 \times F(2) = 1000 \times 0.0384 = 38.40$
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