The Laplace transform of the function $f(t) = e^{-3t} \sin(4t)$ for $s > -3$ is given by:

(A) \frac{4}{(s+3)^2 + 16}

(B) \frac{s+3}{(s+3)^2 + 16}

(C) \frac{4}{(s-3)^2 + 16}

(D) \frac{s-3}{(s-3)^2 + 16}

Answer Key:

A

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Base Transform:** $\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}$. 2. **Stage 2: Frequency Shifting Property:** $\mathcal{L}\{e^{at} g(t)\} = G(s-a)$. 3. **Stage 3: Substitution:** With $a = -3, \omega = 4$, we obtain $\frac{4}{(s+3)^2 + 16}$. 4. **Stage 4: Selection:** Option A is correct.

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