1. Problem 4a: Find the inverse Laplace transform of $$\frac{4s - 2}{s^2 - 6s + 18}$$. 2. First, complete the square in the denominator: $$s^2 - 6s + 18 = (s^2 - 6s + 9) + 9 = (s - 3)^2 + 9$$. 3. Rewrite the numerator to match the form $$A(s - 3) + B$$: $$4s - 2 = 4(s - 3) + 4 \times 3 - 2 = 4(s - 3) + 10$$. 4. Split the fraction: $$\frac{4s - 2}{(s - 3)^2 + 9} = \frac{4(s - 3)}{(s - 3)^2 + 9} + \frac{10}{(s - 3)^2 + 9}$$. 5. Recall Laplace transform pairs: - $$\mathcal{L}\{e^{at} \cos(bt)\} = \frac{s - a}{(s - a)^2 + b^2}$$ - $$\mathcal{L}\{e^{at} \sin(bt)\} = \frac{b}{(s - a)^2 + b^2}$$ 6. Using these, the inverse Laplace transform is: $$4 e^{3t} \cos(3t) + \frac{10}{3} e^{3t} \sin(3t)$$. --- 7. Problem 4b: Find the inverse Laplace transform of $$\frac{20}{(s^2 + 4)(s^2 + 2s + 2)}$$. 8. Complete the square in the second denominator: $$s^2 + 2s + 2 = (s + 1)^2 + 1$$. 9. Use partial fraction decomposition: $$\frac{20}{(s^2 + 4)((s + 1)^2 + 1)} = \frac{As + B}{s^2 + 4} + \frac{Cs + D}{(s + 1)^2 + 1}$$. 10. Multiply both sides by the denominator and equate coefficients: $$20 = (As + B)((s + 1)^2 + 1) + (Cs + D)(s^2 + 4)$$. 11. Expanding and equating coefficients, solve for A, B, C, D: - Coefficient of $$s^3$$: $$A + C = 0$$ - Coefficient of $$s^2$$: $$2A + B + D = 0$$ - Coefficient of $$s$$: $$A + 2B + 4C = 0$$ - Constant term: $$B + 4D = 20$$ 12. Solving the system: - From $$A + C = 0$$, $$C = -A$$ - Substitute $$C$$ into other equations and solve: - $$2A + B + D = 0$$ - $$A + 2B + 4(-A) = 0 \Rightarrow A + 2B - 4A = 0 \Rightarrow -3A + 2B = 0 \Rightarrow 2B = 3A \Rightarrow B = \frac{3A}{2}$$ - $$B + 4D = 20$$ 13. Substitute $$B = \frac{3A}{2}$$ into $$2A + B + D = 0$$: $$2A + \frac{3A}{2} + D = 0 \Rightarrow \frac{4A}{2} + \frac{3A}{2} + D = 0 \Rightarrow \frac{7A}{2} + D = 0 \Rightarrow D = -\frac{7A}{2}$$ 14. Substitute $$B$$ and $$D$$ into $$B + 4D = 20$$: $$\frac{3A}{2} + 4 \left(-\frac{7A}{2}\right) = 20 \Rightarrow \frac{3A}{2} - 14A = 20 \Rightarrow \frac{3A - 28A}{2} = 20 \Rightarrow \frac{-25A}{2} = 20 \Rightarrow -25A = 40 \Rightarrow A = -\frac{40}{25} = -\frac{8}{5}$$ 15. Then: $$B = \frac{3}{2} \times -\frac{8}{5} = -\frac{12}{5}$$ $$D = -\frac{7}{2} \times -\frac{8}{5} = \frac{28}{5}$$ $$C = -A = \frac{8}{5}$$ 16. So the decomposition is: $$\frac{-\frac{8}{5} s - \frac{12}{5}}{s^2 + 4} + \frac{\frac{8}{5} s + \frac{28}{5}}{(s + 1)^2 + 1}$$. 17. Write as: $$-\frac{8}{5} \frac{s + \frac{3}{2}}{s^2 + 4} + \frac{8}{5} \frac{s + \frac{7}{2}}{(s + 1)^2 + 1}$$ (adjusted for clarity). 18. Use Laplace inverse formulas: - $$\mathcal{L}^{-1}\left\{\frac{s}{s^2 + a^2}\right\} = \cos(at)$$ - $$\mathcal{L}^{-1}\left\{\frac{a}{s^2 + a^2}\right\} = \sin(at)$$ - $$\mathcal{L}^{-1}\left\{\frac{s + a}{(s + a)^2 + b^2}\right\} = e^{-at} \cos(bt)$$ - $$\mathcal{L}^{-1}\left\{\frac{b}{(s + a)^2 + b^2}\right\} = e^{-at} \sin(bt)$$ 19. Express numerators to match these forms and find inverse transforms: 20. Final inverse Laplace transform: $$-\frac{8}{5} \cos(2t) - \frac{12}{5} \frac{\sin(2t)}{2} + \frac{8}{5} e^{-t} \cos(t) + \frac{28}{5} e^{-t} \sin(t)$$ 21. Simplify: $$-\frac{8}{5} \cos(2t) - \frac{6}{5} \sin(2t) + \frac{8}{5} e^{-t} \cos(t) + \frac{28}{5} e^{-t} \sin(t)$$