Question : how do you find inverse Laplace transform of 1/(s^2+2s-3)? Solution : Given \[f(s) =\frac{1}{(s^2+2s-3)}\] It’s factorization is given as \[ f(s)=\frac{1}{(s+3)(s-1)} \] \[=> f(s)=\frac{1}{4}\left(\frac{1}{(s-1)}- \frac{1}{(s+3)}\right) \] Now using inverse Laplace transform both sides \[L^-1\left(f(s)\right) = \frac{1}{4}\left(L^-1\left(\frac{1}{s-1}\right)-L^-1\left(\frac{1}{s+3}\right)\right)\] By inverse…
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