# Find the inverse Laplace transform of the function

Of $\displaystyle \frac{1}{4s^{2}-8s}$

im approaching the partial fraction $\displaystyle \frac{1}{4s\left(s-2\right)}=\frac{A}{4s}+\frac{B}{s-2},then,A=-\frac{1}{2},B=\frac{1}{8}$

memorizing the table i have $\displaystyle f(t)=-\frac{1}{8}+\frac{1}{8}e^{2t}$

however the answer seems to be $\displaystyle \frac{1}{4}e^{t}\sinh{t}$

how come?

-
The two answers agree: just expand $\text{sinh}t$ –  awllower Mar 19 '13 at 4:21
$$\frac{1}{4}e^{t}\sinh{(t)} = \frac{1}{4}e^{t}\left(\frac{e^t - e^{-t}}{2}\right)= \frac{e^{2t}}{8} -\frac{1}{8} \\$$