Question

Find the Inverse Laplace Transform
(7s+5)/(s^2+s+7)

Answer 1

I tried doing some partial fractions completing the square
\[ \frac{7s+5}{(s+1/2)^2+27/4}\]

Answer 2

I tried separating it
\[\frac{7s}{(s+1/2)^2+27/4}+\frac{5}{(s+1/2)^2+27/4}\] and then adjusting to see if
\[\frac{7(s+1/2-1/2)}{(s+1/2)^2+27/4}+\frac{5}{(s+1/2)^2+27/4}\] to try to get it to something that looks like something in the table but I can't seem to make it work

Answer 3

are you sure it's 7 in denominator. Maybe it should be:
\(\huge\frac{7s+5}{s^2+s-2}\)

Answer 4

\[\frac{\color{red}7(s+1/2\color{red}{-1/2})}{(s+1/2)^2+27/4}+\frac{5}{(s+1/2)^2+27/4}\\\quad =\frac{7(s+1/2)}{(s+1/2)^2+27/4}+\frac{5\color{red}{-7(1/2)}}{(s+1/2)^2+27/4}\\\quad =\frac{7(s+1/2)}{(s+1/2)^2+27/4}+\frac{3/2}{(s+1/2)^2+27/4}\\\quad =7\frac{(s+1/2)}{(s+1/2)^2+27/4}+\frac{3/2}{3\sqrt 3/2}\cdot\frac{3\sqrt 3/2}{(s+1/2)^2+27/4}\]
the transform is \[e^{-t/2}[7\cos(3\sqrt 3/2)t+(\sqrt 3/3)\sin(3\sqrt 3/2)t]\]