1. **State the problem:** Find the inverse Laplace transform of $$F(s) = \frac{(s - 2) e^{-2s}}{s^2 - 6s + 13}$$. 2. **Recall the shifting theorem:** The factor $$e^{-as}$$ in the Laplace domain corresponds to a time delay of $$a$$ in the time domain. Specifically, if $$\mathcal{L}\{f(t)\} = F(s)$$, then $$\mathcal{L}\{u(t - a) f(t - a)\} = e^{-as} F(s)$$ where $$u(t - a)$$ is the Heaviside step function. 3. **Focus on the non-shifted part:** Consider $$G(s) = \frac{s - 2}{s^2 - 6s + 13}$$. 4. **Complete the square in the denominator:** $$s^2 - 6s + 13 = (s - 3)^2 + 4$$ 5. **Rewrite numerator to match denominator shift:** $$s - 2 = (s - 3) + 1$$ 6. **Split the fraction:** $$G(s) = \frac{(s - 3) + 1}{(s - 3)^2 + 4} = \frac{s - 3}{(s - 3)^2 + 4} + \frac{1}{(s - 3)^2 + 4}$$ 7. **Recall Laplace transform pairs:** - $$\mathcal{L}\{e^{at} \cos(bt)\} = \frac{s - a}{(s - a)^2 + b^2}$$ - $$\mathcal{L}\{e^{at} \sin(bt)\} = \frac{b}{(s - a)^2 + b^2}$$ 8. **Match terms:** - For $$\frac{s - 3}{(s - 3)^2 + 4}$$, corresponds to $$e^{3t} \cos(2t)$$ - For $$\frac{1}{(s - 3)^2 + 4}$$, note numerator is 1 but for sine term numerator should be 2, so rewrite: $$\frac{1}{(s - 3)^2 + 4} = \frac{1}{2} \cdot \frac{2}{(s - 3)^2 + 4}$$ which corresponds to $$\frac{1}{2} e^{3t} \sin(2t)$$ 9. **Combine inverse transforms:** $$g(t) = e^{3t} \cos(2t) + \frac{1}{2} e^{3t} \sin(2t)$$ 10. **Apply the time shift:** Since the original function has $$e^{-2s}$$, the inverse Laplace transform is delayed by 2 units: $$f(t) = u(t - 2) g(t - 2) = u(t - 2) \left[ e^{3(t - 2)} \cos(2(t - 2)) + \frac{1}{2} e^{3(t - 2)} \sin(2(t - 2)) \right]$$ **Final answer:** $$\boxed{f(t) = u(t - 2) \left[ e^{3(t - 2)} \cos(2(t - 2)) + \frac{1}{2} e^{3(t - 2)} \sin(2(t - 2)) \right]}$$