inverse laplace: 3s/(s^2 -2s + 7)

[TwistedNerve]

hey. i need help with the inverse laplace of something. im guessing theres an algebra trick and i just dont see it...

3s/(s^2 -2s + 7)

 

[galactus]

Re: inverse laplace

Inverse laplace of \(\displaystyle \frac{3s}{s^{2}-2s+7}\)?

Complete the square in the denominator.

\(\displaystyle \frac{3s}{(s-1)^{2}+6}\)

\(\displaystyle \frac{b}{(p+a)^{2}+b^{2}}=e^{-at}sin(bt)\)

\(\displaystyle \frac{p+a}{(p+a)^{2}+b^{2}}=e^{-at}cos(bt)\)

I get \(\displaystyle 3e^{t}cos(\sqrt{6}t)+\frac{\sqrt{6}e^{t}sin(\sqrt{6}t)}{2}=\frac{e^{t}(6cos(\sqrt{6}t)+\sqrt{6}sin(\sqrt{6}t))}{2}\)