Question

Find the Laplace transform of t^2 e^-3t sin 2t.​

Answer 1

Answer:

please mark me the brainiest

Step-by-step explanation:

Answer 2

Laplace transform of $t^2 e^{-3t} sin 2t$ is $\dfrac{4(3s^{2} + 18s + 23)}{(s^{2} + 6s + 13)^3}$

Step-by-step explanation:

Given: $f (x) = t^2 e^{-3t} sin 2t$

To Find: Laplace transformation of $f(x)$

Solution:

• Laplace transform of $f(x) = t^2 e^{-3t} sin 2t$

The Laplace transform of $f(x) = t^2 e^{-3t} sin 2t$ can be determined by,

$\Rightarrow L[sin2t] = \dfrac{2}{s^{2} + 4 } \ \ \ \ \because L[sinat] = \dfrac{a}{s^{2} + a^{2} }$

$\Rightarrow L[e^{-3t} sin2t] = \dfrac{2}{(s+3)^{2} + 4 } \ \ \ \ \because L[e^{at} f(t)] = F(s-a)$

$\Rightarrow L[t^{2} e^{-3t} sin2t] = \dfrac{d^2}{ds^2} \Big( \dfrac{2}{(s+3)^{2} + 4 } \Big) \ \ \ \ \because L[t^{n} f(t)] = (-1)^n \dfrac{d^n [f(n)]}{ds^n}$

$\Rightarrow L[t^{2} e^{-3t} sin2t] = \dfrac{d}{ds} \Big( \dfrac{d}{ds} \Big( \dfrac{2}{(s+3)^{2} + 4 } \Big) \Big)$

$\Rightarrow L[t^{2} e^{-3t} sin2t] = \dfrac{d}{ds} \Big( \dfrac{-4(s+3)}{((s+3)^{2} + 4)^2 } \Big)$

$\Rightarrow L[t^{2} e^{-3t} sin2t] = \Big( \dfrac{-4(3s^2+18s+23)}{(s^{2} + 6s +13)^3 } \Big)$

Hence, the Laplace transform of $t^2 e^{-3t} sin 2t$ is $\dfrac{4(3s^{2} + 18s + 23)}{(s^{2} + 6s + 13)^3}$