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# The Inverse Laplace Transform of a Fraction

DIFFEQ-TOXJVI

The inverse Laplace Transform of $F(s)=\cfrac{s^2-2s+1}{(s^2+2s+4)(s^2-4)}e^{-2s}$ is:

In the answers, $u(t)$ is the Heaviside step function and $u_2(t)=u(t-2)$ is the step function with the jump at $t=2$. In some books, the notation is $H(t-2)$ or $\theta(t-2)$.

A

$u_2[\frac{13}{24}\cos(\sqrt{3}(t-2))e^{-(t-2)} +\frac{\sqrt{3}}{8}\sin(\sqrt{3}(t-2))e^{-(t-2)}-\frac{9}{16}e^{-2(t-2)}+ \frac{1}{48}e^{2(t-2)}]$

B

$u_2[\frac{13}{24}\cos(\sqrt{3}(t-2))e^{-t} +\frac{\sqrt{3}}{8}\sin(\sqrt{3}(t-2))e^{-t}-\frac{9}{16}e^{-2(t-2)}+ \frac{1}{48}e^{2(t-2)}]$

C

$u_2[\frac{13}{24}\cos(\sqrt{3}(t-2))e^{-(t-2)} +\frac{3}{8}\sin(\sqrt{3}(t-2))e^{-(t-2)}-\frac{9}{16}e^{-2(t-2)}+ \frac{1}{48}e^{2(t-2)}]$

D

$u_2[\frac{13}{24}\cos(\sqrt{3}(t-2))e^{-t} +\frac{3}{8}\sin(\sqrt{3}(t-2))e^{-t}-\frac{9}{16}e^{-2(t-2)}+ \frac{1}{48}e^{2(t-2)}]$