Problem 04 | Laplace Transform of Derivatives | Advance Engineering Mathematics Review at MATHalino

Problem 04
Find the Laplace transform of $f(t) = t \, \sin t$ using the transform of derivatives.

Solution 04

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$f(t) = t \, \sin t$ ..........

$f(0) = 0$

$f'(t) = t \, \cos t + \sin t$ .......... $f'(0) = 0$

$f''(t) = (-t \, \sin t + \cos t) + \cos t = -t \, \sin t + 2\cos t$

$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$

$\mathcal{L} (-t \, \sin t + 2\cos t) = s^2 \mathcal{L} (t \, \sin t) - s \, (0) - (0)$

$-\mathcal{L} (t \, \sin t) + 2\mathcal{L} (\cos t) = s^2 \mathcal{L} (t \, \sin t)$

$2\mathcal{L} (\cos t) = s^2 \mathcal{L} (t \, \sin t) + \mathcal{L} (t \, \sin t)$

$s^2 \mathcal{L} (t \, \sin t) + \mathcal{L} (t \, \sin t) = 2\mathcal{L} (\cos t)$

$(s^2 + 1) \mathcal{L} (t \, \sin t) = 2\mathcal{L} (\cos t)$

$(s^2 + 1) \mathcal{L} (t \, \sin t) = 2 \left( \dfrac{s}{s^2 + 1^2} \right)$

$(s^2 + 1) \mathcal{L} (t \, \sin t) = \dfrac{2s}{s^2 + 1}$

$\mathcal{L} (t \, \sin t) = \dfrac{2s}{(s^2 + 1)(s^2 + 1)}$

$\mathcal{L} (t \, \sin t) = \dfrac{2s}{(s^2 + 1)^2}$ answer

Problem 04 | Laplace Transform of Derivatives