1. **Problem Statement:** Find the inverse Laplace transform $\mathcal{L}^{-1}\{F(s)\}$ for the given functions using the Laplace transform table. --- ### a) $F(s) = \frac{10s + 23}{s^2 + 7s + 12}$ 2. **Factor the denominator:** $$s^2 + 7s + 12 = (s + 3)(s + 4)$$ 3. **Partial fraction decomposition:** Assume $$\frac{10s + 23}{(s + 3)(s + 4)} = \frac{A}{s + 3} + \frac{B}{s + 4}$$ Multiply both sides by $(s + 3)(s + 4)$: $$10s + 23 = A(s + 4) + B(s + 3)$$ 4. **Solve for A and B:** Set $s = -3$: $$10(-3) + 23 = A(-3 + 4) + B(0) \Rightarrow -30 + 23 = A(1) \Rightarrow A = -7$$ Set $s = -4$: $$10(-4) + 23 = A(0) + B(-4 + 3) \Rightarrow -40 + 23 = B(-1) \Rightarrow B = 17$$ 5. **Rewrite $F(s)$:** $$F(s) = \frac{-7}{s + 3} + \frac{17}{s + 4}$$ 6. **Use Laplace inverse formulas:** Recall: $$\mathcal{L}^{-1}\left\{\frac{1}{s + a}\right\} = e^{-at}$$ 7. **Apply inverse Laplace transform:** $$f(t) = -7 e^{-3t} + 17 e^{-4t}$$ --- ### b) $F(s) = \frac{2s + 12}{s^2 + 6s + 13}$ 8. **Complete the square in denominator:** $$s^2 + 6s + 13 = (s + 3)^2 + 4$$ 9. **Rewrite numerator to match Laplace forms:** $$2s + 12 = 2(s + 3) + 6$$ 10. **Express $F(s)$ as:** $$F(s) = \frac{2(s + 3)}{(s + 3)^2 + 2^2} + \frac{6}{(s + 3)^2 + 2^2}$$ 11. **Recall Laplace inverse formulas:** $$\mathcal{L}^{-1}\left\{\frac{s + a}{(s + a)^2 + b^2}\right\} = e^{-at} \cos(bt)$$ $$\mathcal{L}^{-1}\left\{\frac{b}{(s + a)^2 + b^2}\right\} = e^{-at} \sin(bt)$$ 12. **Apply inverse Laplace transform:** $$f(t) = 2 e^{-3t} \cos(2t) + 3 e^{-3t} \sin(2t)$$ --- **Final answers:** - a) $f(t) = -7 e^{-3t} + 17 e^{-4t}$ - b) $f(t) = 2 e^{-3t} \cos(2t) + 3 e^{-3t} \sin(2t)$