Question

guys check this out: 1/(s^2+2s+10) in laplace transform (differential eq. subject) =D

Answer 1

\[\frac{1}{s^2+2s+10}=\frac{1}{(s+1)^2+3^2}=\]\[=\frac{1}{6i}\left(\frac{1}{s+1-3i}-\frac{1}{s+1+3i}\right)\]\[L^{-1}\left(\frac{1}{s^2+2s+10}\right)=\frac{1}{6i}(e^{(-1+3i)t}-e^{(-1-3i)t})=\]\[=\frac{1}{6i}e^{-t}(e^{3it}-e^{-3it})=\frac{1}{6i}e^{-t}\cdot 2i\sin{3t}=\]\[=\frac{1}{3}e^{-t}\sin{3t}\]