Question

What is the Laplace transform of t^3 + cos3t-e^-3t?

Answer 1

We have $F_{(t)}= {t^3 + \cos{(3t)}-e^{-3t}}$

We proceed to use the rules for Laplace transform and we proceed with each one of the terms:

$\mathscr{L}[F_{(t)}]_{(s)}=\mathscr{L}[{t^3 + \cos{(3t)}-e^{-3t}}]_{(s)} \\ \mathscr{L}[F_{(t)}]_{(s)}=\mathscr{L}[{t^3}]_{(s)}+\mathscr{s}[{\cos{(3t)}}]_{(s)}-\mathscr{L}[{e^{-3t}}]_{(s)} \\ \mathscr{L}[F_{(t)}]_{(s)}=\dfrac{3!}{s^{3+1}}+\dfrac{s}{s^2+3^2}-\dfrac{1}{s-{(-3)}} \\ \therefore \mathscr{L}[F_{(t)}]_{(s)}=\dfrac{6}{s^{4}}+\dfrac{s}{s^2+9}-\dfrac{1}{s+3}$

In conclusion, the Laplace transform is $\mathscr{L}[F_{(t)}]_{(s)}=\dfrac{6}{s^{4}}+\dfrac{s}{s^2+9}-\dfrac{1}{s+3}$.