how to Find Inverse Laplace transform of Y(s)=1/((s^2+7s+9))

Question : Find Inverse Laplace transform of Y(s)=1/((s^2+7s+9)) Solution : To Find Inverse Laplace transform of Y(s)=1/((s^2+7s+9)) Given function \[Y\left(s\right)=\frac{1}{(s^2+7s+9)}\] It can be written as \[Y\left(s\right)=\frac{1}{\left(s+\frac{7}{2}\right)^2-\frac{13}{4}}\] \[Y\left(s\right)=\frac{1}{\left(s+\frac{7}{2}\right)^2-\left(\sqrt{\frac{13}{2}}\right)^2}\] \[Y\left(s\right)=\frac{1}{\left(s+\frac{7}{2}-\sqrt{\frac{13}{2}}\right)\left(s+\frac{7}{2}+\sqrt{\frac{13}{2}}\right)}\] So the…

Question : Find Inverse Laplace transform of 1/(s(s^2+2s-3))

Question : Find Inverse Laplace transform of 1/(s(s^2+2s-3)) Solution : To Find Inverse Laplace transform of F(s)=1/s(s-1)(s+3) Partial fraction of F(s) is given as \[F\left(s\right)=\frac{1}{4\left(s-1\right)}-\frac{1}{3s}+\frac{1}{12\left(s+3\right)}\]Applying inverse Laplace transeform \[L^{-1}\left(F\left(s\right)\right)=L^{-1}\left[\frac{1}{4\left(s-1\right)}-\frac{1}{3s}+\frac{1}{12\left(s+3\right)}\right]\] \[L^{-1}\left(F\left(s\right)\right)=\frac{1}{4}L^{-1}\left[\frac{1}{\left(s-1\right)}\right]-\frac{1}{3}L^{-1}\left[\frac{1}{s}\right]+\frac{1}{12}L^{-1}\left[\frac{1}{\left(s+3\right)}\right]\]…