Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.7 The Second Shifting Theorem - Problems - Page 705: 24

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We are given that $F(s)=\frac{e^{-s}(2s-1)}{s^2+4s+5}$ Now,$f(t)= L^{-1}[\frac{e^{-s}(2s-1)}{s^2+4s+5}]\\ =L^{-1}[e^{-4s}(\frac{e^{-s}(s+2)}{(s+2)^2+1}-\frac{5e^{-s}}{(s+2)^2+1})]\\ =L^{-1}[e^{-s}][L(2e^{-2t}\cos t-5e^{-2t}\sin t)]\\ =2e^{-2(t-1)}\cos (t-1)u_1(t)-5e^{-2(t-1)}\sin (t-1)u_1(t)\\ =u_1(t)e^{-2(t-1)}[2\cos (t-1)-5\sin (t-1)]$