L^-1((s-5)/((s-1)(s^2)+4))

To find the inverse Laplace transform of the given expression, we can use partial fraction decomposition.<br /><br />The given expression is:<br /><br />\( \frac{s-5}{(s-1)(s^2+4)} \)<br /><br />We can decompose this fraction into simpler fractions:<br /><br />\( \frac{s-5}{(s-1)(s^2+4)} = \frac{A}{s-1} + \frac{Bs+C}{s^2+4} \)<br /><br />Multiplying both sides by the denominator \((s-1)(s^2+4)\), we get:<br /><br />\( s-5 = A(s^2+4) + (Bs+C)(s-1) \)<br /><br />Expanding and equating the coefficients of \(s\), we have:<br /><br />\( s-5 = As^2 + 4A + Bs^2 - Bs + Cs - C \)<br /><br />Grouping the terms, we get:<br /><br />\( s-5 = (A+B)s^2 + (4A-C)s + (4A-Bs-C) \)<br /><br />Equating the coefficients of \(s^2\), \(s\), and the constant term, we have:<br /><br />\( A+B = 0 \)<br /><br />\( 4A-C = 1 \)<br /><br />\( 4A-Bs-C = -5 \)<br /><br />Solving these equations, we find:<br /><br />\( A = \frac{1}{5} \)<br /><br />\( B = -\frac{1}{5} \)<br /><br />\( C = -\frac{4}{5} \)<br /><br />Substituting these values back into the partial fractions, we get:<br /><br />\( \frac{s-5}{(s-1)(s^2+4)} = \frac{1/5}{s-1} + \frac{-s/5 - 4/5}{s^2+4} \)<br /><br />Now, we can take the inverse Laplace transform of each term separately:<br /><br />\( L^{-1}\left(\frac{1/5}{s-1}\right) = \frac{1}{5}e^t \)<br /><br />\( L^{-1}\left(\frac{-s/5 - 4/5}{s^2+4}\right) = -\frac{1}{5}\cos(2t) - \frac{2}{5}\sin(2t) \)<br /><br />Therefore, the inverse Laplace transform of the given expression is:<br /><br />\( L^{-1}\left(\frac{s-5}{(s-1)(s^2+4)}\right) = \frac{1}{5}e^t - \frac{1}{5}\cos(2t) - \frac{2}{5}\sin(2t) \)