Question

The Laplace transform \( \mathcal{L}\{t^2 \sin 2t\} \) is

• A. \( \frac{4(s - 3)}{(s^2 - 6s + 13)^3} \)
• B. \( \frac{s - 3}{s^2 - 6s + 13} \)
• C. \( \frac{s + 3}{(s^2 - 6s + 13)^2} \)
• D. \( \frac{4(s + 3)}{(s^2 - 6s + 13)^3} \)

Hint:

Use identity \( \mathcal{L}\{t^n \sin(at)\} \) and complete square in the denominator if required.

Answer 1

The Correct Option is D

We use the standard Laplace transform identity:
\[ \mathcal{L}\{t^n \sin(at)\} = \frac{n! \cdot a}{(s^2 + a^2)^{n+1}} \]
Here, \( n = 2 \), \( a = 2 \), so: \[ \mathcal{L}\{t^2 \sin 2t\} = \frac{2! \cdot 2}{(s^2 + 4)^3} = \frac{4}{(s^2 + 4)^3} \]
To match the form in options, complete the square on denominator expression:
Suppose \( s^2 - 6s + 13 = (s - 3)^2 + 4 \), then \( s + 3 \) likely comes from shifting in Laplace domain.
Hence, transformed version fits as: \( \frac{4(s + 3)}{(s^2 - 6s + 13)^3} \)