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 \[L^{-1}\left(F\left(s\right)\right)=L^{-1}\left[\frac{1}{4\left(s-1\right)}-\frac{1}{3s}+\frac{1}{12\left(s+3\right)}\right]\] \[L^{-1}\left(F\left(s\right)\right)=\frac{1}{4}L^{-1}\left[\frac{1}{\left(s-1\right)}\right]-\frac{1}{3}L^{-1}\left[\frac{1}{s}\right]+\frac{1}{12}L^{-1}\left[\frac{1}{\left(s+3\right)}\right]\] We know that inverse Laplace transeform formulas \[L^{-1}\left[\frac{1}{s}\right]=1,\ L^{-1}\left[\frac{1}{s-a}\right]=e^{at}\ L^{-1}\left[\frac{1}{s+a}\right]=e^{-at}\]…
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