1. **State the problem:** Find the inverse Laplace transform of $$\frac{20}{(s^2 + 4)(s^2 + 2.5s + 2)}$$. 2. **Rewrite the denominator:** The first quadratic is $$s^2 + 4$$ which corresponds to $$s^2 + 2^2$$. The second quadratic is $$s^2 + 2.5s + 2$$. Complete the square: $$s^2 + 2.5s + 2 = \left(s + \frac{2.5}{2}\right)^2 - \left(\frac{2.5}{2}\right)^2 + 2 = (s + 1.25)^2 - 1.5625 + 2 = (s + 1.25)^2 + 0.4375$$ 3. **Express the denominator as:** $$ (s^2 + 2^2) \left[(s + 1.25)^2 + (\sqrt{0.4375})^2\right] $$ where $$\sqrt{0.4375} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \approx 0.6614$$. 4. **Use partial fraction decomposition:** Assume $$\frac{20}{(s^2 + 4)((s + 1.25)^2 + 0.4375)} = \frac{As + B}{s^2 + 4} + \frac{Cs + D}{(s + 1.25)^2 + 0.4375}$$ Multiply both sides by the denominator: $$20 = (As + B)((s + 1.25)^2 + 0.4375) + (Cs + D)(s^2 + 4)$$ 5. **Expand and collect terms:** Expand $$((s + 1.25)^2 + 0.4375) = s^2 + 2.5s + 1.5625 + 0.4375 = s^2 + 2.5s + 2$$. So: $$20 = (As + B)(s^2 + 2.5s + 2) + (Cs + D)(s^2 + 4)$$ Expand: $$20 = As^3 + 2.5As^2 + 2As + Bs^2 + 2.5Bs + 2B + Cs^3 + 4Cs + Ds^2 + 4D$$ Group by powers of $$s$$: $$20 = (A + C)s^3 + (2.5A + B + D)s^2 + (2A + 2.5B + 4C)s + (2B + 4D)$$ 6. **Set up system of equations by matching coefficients:** - Coefficient of $$s^3$$: $$A + C = 0$$ - Coefficient of $$s^2$$: $$2.5A + B + D = 0$$ - Coefficient of $$s$$: $$2A + 2.5B + 4C = 0$$ - Constant term: $$2B + 4D = 20$$ 7. **Solve the system:** From $$A + C = 0$$, we get $$C = -A$$. Substitute $$C = -A$$ into the $$s$$ coefficient equation: $$2A + 2.5B + 4(-A) = 0 \Rightarrow 2A + 2.5B - 4A = 0 \Rightarrow -2A + 2.5B = 0 \Rightarrow 2.5B = 2A \Rightarrow B = \frac{2A}{2.5} = 0.8A$$ Substitute $$B = 0.8A$$ and $$C = -A$$ into the $$s^2$$ coefficient equation: $$2.5A + 0.8A + D = 0 \Rightarrow 3.3A + D = 0 \Rightarrow D = -3.3A$$ Substitute $$B = 0.8A$$ and $$D = -3.3A$$ into the constant term equation: $$2(0.8A) + 4(-3.3A) = 20 \Rightarrow 1.6A - 13.2A = 20 \Rightarrow -11.6A = 20 \Rightarrow A = -\frac{20}{11.6} = -\frac{50}{29}$$ Calculate other coefficients: $$B = 0.8A = 0.8 \times -\frac{50}{29} = -\frac{40}{29}$$ $$C = -A = \frac{50}{29}$$ $$D = -3.3A = -3.3 \times -\frac{50}{29} = \frac{165}{29}$$ 8. **Rewrite the partial fractions:** $$\frac{20}{(s^2 + 4)((s + 1.25)^2 + 0.4375)} = \frac{-\frac{50}{29}s - \frac{40}{29}}{s^2 + 4} + \frac{\frac{50}{29}s + \frac{165}{29}}{(s + 1.25)^2 + 0.4375}$$ 9. **Find inverse Laplace transforms:** Recall: - $$\mathcal{L}^{-1}\left\{\frac{s}{s^2 + a^2}\right\} = \cos(at)$$ - $$\mathcal{L}^{-1}\left\{\frac{a}{s^2 + a^2}\right\} = \sin(at)$$ - $$\mathcal{L}^{-1}\left\{\frac{s + b}{(s + b)^2 + c^2}\right\} = e^{-bt} \cos(ct)$$ - $$\mathcal{L}^{-1}\left\{\frac{c}{(s + b)^2 + c^2}\right\} = e^{-bt} \sin(ct)$$ Rewrite numerator of second term to match form: $$\frac{50}{29}s + \frac{165}{29} = \frac{50}{29}(s + 1.25) + \frac{165}{29} - \frac{50}{29} \times 1.25 = \frac{50}{29}(s + 1.25) + \frac{165}{29} - \frac{62.5}{29} = \frac{50}{29}(s + 1.25) + \frac{102.5}{29}$$ 10. **Apply inverse Laplace:** - First term: $$\mathcal{L}^{-1}\left\{\frac{-\frac{50}{29}s - \frac{40}{29}}{s^2 + 4}\right\} = -\frac{50}{29} \cos(2t) - \frac{40}{29} \frac{\sin(2t)}{2} = -\frac{50}{29} \cos(2t) - \frac{20}{29} \sin(2t)$$ - Second term: $$\mathcal{L}^{-1}\left\{\frac{\frac{50}{29}(s + 1.25) + \frac{102.5}{29}}{(s + 1.25)^2 + (\frac{\sqrt{7}}{4})^2}\right\} = \frac{50}{29} e^{-1.25 t} \cos\left(\frac{\sqrt{7}}{4} t\right) + \frac{102.5}{29} \frac{4}{\sqrt{7}} e^{-1.25 t} \sin\left(\frac{\sqrt{7}}{4} t\right)$$ Simplify the sine coefficient: $$\frac{102.5}{29} \times \frac{4}{\sqrt{7}} = \frac{410}{29 \sqrt{7}}$$ 11. **Final answer:** $$f(t) = -\frac{50}{29} \cos(2t) - \frac{20}{29} \sin(2t) + \frac{50}{29} e^{-1.25 t} \cos\left(\frac{\sqrt{7}}{4} t\right) + \frac{410}{29 \sqrt{7}} e^{-1.25 t} \sin\left(\frac{\sqrt{7}}{4} t\right)$$