What-is-the-Laplace-transform-of-f-t-4t-2-5sin-3t-

Question Number 129048 by bramlexs22 last updated on 12/Jan/21

What is the Laplace transform

of f (t) = - {4 t}^{2} - 5 \sin 3 t

Answered by Dwaipayan Shikari last updated on 12/Jan/21

L (f (t)) = - 4 \int_{0}^{\infty} t^{2} e^{- s t} - 5 \int_{0}^{\infty} e^{- s t} s i n 3 t

= - \frac{4}{s^{3}} \int_{0}^{\infty} j^{2} e^{- j} d j - \frac{5}{2 i} \int_{0}^{\infty} e^{- (s - 3 i)} - e^{- (s + 3 i)} d t

= - \frac{8}{s^{3}} - \frac{5}{2 i} (\frac{1}{s - 3 i} - \frac{1}{s + 3 i}) = - \frac{8}{s^{3}} - \frac{15}{s^{2} + 9}