Question

Find the Laplace Transform $L[e^{-2t}t \sin(2t) \cos(3t)]$.

Answer 1

$(s+2) \left( \frac{5}{(s^2 + 4s + 29)^2} - \frac{1}{(s^2 + 4s + 5)^2} \right)$

Answer 2

To find the Laplace Transform of $f(t) = e^{-2t}t \sin(2t) \cos(3t)$, we follow these steps:

1. Trigonometric Simplification: Use the product-to-sum identity $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$. $\sin(2t)\cos(3t) = \frac{1}{2}[\sin(2t+3t) + \sin(2t-3t)] = \frac{1}{2}[\sin(5t) + \sin(-t)] = \frac{1}{2}[\sin(5t) - \sin(t)]$.

2. Rewrite the function: $f(t) = e^{-2t} t \left( \frac{1}{2}[\sin(5t) - \sin(t)] \right) = \frac{1}{2} e^{-2t} (t \sin(5t) - t \sin(t))$.

3. Apply Linearity: The Laplace Transform is linear. $L[f(t)] = \frac{1}{2} \left( L[e^{-2t} t \sin(5t)] - L[e^{-2t} t \sin(t)] \right)$.

4. Laplace Transform of $t \sin(\omega t)$: Use the property $L[t g(t)] = - \frac{d}{ds} G(s)$, where $G(s) = L[g(t)]$. We know $L[\sin(\omega t)] = \frac{\omega}{s^2 + \omega^2}$. $L[t \sin(\omega t)] = - \frac{d}{ds} \left( \frac{\omega}{s^2 + \omega^2} \right) = \frac{2s\omega}{(s^2 + \omega^2)^2}$.

5. Calculate individual transforms:

   • For $t \sin(5t)$ ($\omega=5$): $L[t \sin(5t)] = \frac{2s(5)}{(s^2 + 5^2)^2} = \frac{10s}{(s^2 + 25)^2}$.
   • For $t \sin(t)$ ($\omega=1$): $L[t \sin(t)] = \frac{2s(1)}{(s^2 + 1^2)^2} = \frac{2s}{(s^2 + 1)^2}$.

6. Apply First Shifting Theorem: $L[e^{at} g(t)] = G(s-a)$. Here $a=-2$.

   • $L[e^{-2t} t \sin(5t)] = \frac{10(s+2)}{((s+2)^2 + 25)^2}$.
   • $L[e^{-2t} t \sin(t)] = \frac{2(s+2)}{((s+2)^2 + 1)^2}$.

7. Combine results: $L[f(t)] = \frac{1}{2} \left( \frac{10(s+2)}{((s+2)^2 + 25)^2} - \frac{2(s+2)}{((s+2)^2 + 1)^2} \right)$.

8. Simplify: $L[f(t)] = (s+2) \left( \frac{5}{((s+2)^2 + 25)^2} - \frac{1}{((s+2)^2 + 1)^2} \right)$. Substitute $(s+2)^2 = s^2 + 4s + 4$: $L[f(t)] = (s+2) \left( \frac{5}{(s^2 + 4s + 4 + 25)^2} - \frac{1}{(s^2 + 4s + 4 + 1)^2} \right)$ $L[f(t)] = (s+2) \left( \frac{5}{(s^2 + 4s + 29)^2} - \frac{1}{(s^2 + 4s + 5)^2} \right)$.