1. **Problem statement:** Find the inverse Laplace transform of the given functions: (a) $$F(s) = \frac{7}{(s+3)^3}$$ (b) $$F(s) = \frac{2s - 3}{s^2 + 2s + 10}$$ (c) $$F(s) = \frac{s - 2}{s^2 - 4s + 3}$$ --- 2. **Recall the inverse Laplace transform formulas and rules:** - For $$\mathcal{L}^{-1}\left\{\frac{n!}{(s-a)^{n+1}}\right\} = t^n e^{at}$$ - For quadratic denominators, complete the square and use transforms of the form $$\frac{s-a}{(s-a)^2 + b^2}$$ and $$\frac{b}{(s-a)^2 + b^2}$$ which correspond to $$e^{at}\cos(bt)$$ and $$e^{at}\sin(bt)$$ respectively. - Factor denominators when possible to use partial fractions. --- 3. **Solution for (a):** Given $$F(s) = \frac{7}{(s+3)^3}$$ Rewrite as $$7 \cdot \frac{1}{(s+3)^3}$$. Using the formula for inverse Laplace of $$\frac{n!}{(s-a)^{n+1}}$$ with $$n=2$$ and $$a=-3$$: $$\mathcal{L}^{-1}\left\{\frac{2!}{(s+3)^3}\right\} = t^2 e^{-3t}$$ Since numerator is 7 instead of 2!, multiply by $$\frac{7}{2!} = \frac{7}{2}$$: $$f(t) = \frac{7}{2} t^2 e^{-3t}$$ --- 4. **Solution for (b):** Given $$F(s) = \frac{2s - 3}{s^2 + 2s + 10}$$ Complete the square in denominator: $$s^2 + 2s + 10 = (s+1)^2 + 9$$ Rewrite numerator in terms of $$s+1$$: $$2s - 3 = 2(s+1) - 5$$ So, $$F(s) = \frac{2(s+1) - 5}{(s+1)^2 + 3^2} = 2 \cdot \frac{s+1}{(s+1)^2 + 3^2} - 5 \cdot \frac{1}{(s+1)^2 + 3^2}$$ Using inverse Laplace transforms: $$\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2 + b^2}\right\} = e^{at} \cos(bt)$$ $$\mathcal{L}^{-1}\left\{\frac{b}{(s-a)^2 + b^2}\right\} = e^{at} \sin(bt)$$ Rewrite second term numerator to match $$b$$: $$\frac{1}{(s+1)^2 + 3^2} = \frac{3}{3((s+1)^2 + 3^2)} = \frac{1}{3} \cdot \frac{3}{(s+1)^2 + 3^2}$$ So, $$F(s) = 2 \cdot \frac{s+1}{(s+1)^2 + 3^2} - \frac{5}{3} \cdot \frac{3}{(s+1)^2 + 3^2}$$ Inverse Laplace: $$f(t) = 2 e^{-t} \cos(3t) - \frac{5}{3} e^{-t} \sin(3t)$$ --- 5. **Solution for (c):** Given $$F(s) = \frac{s - 2}{s^2 - 4s + 3}$$ Factor denominator: $$s^2 - 4s + 3 = (s-1)(s-3)$$ Use partial fractions: $$\frac{s - 2}{(s-1)(s-3)} = \frac{A}{s-1} + \frac{B}{s-3}$$ Multiply both sides by denominator: $$s - 2 = A(s-3) + B(s-1)$$ Set $$s=3$$: $$3 - 2 = A(0) + B(2) \Rightarrow 1 = 2B \Rightarrow B = \frac{1}{2}$$ Set $$s=1$$: $$1 - 2 = A(-2) + B(0) \Rightarrow -1 = -2A \Rightarrow A = \frac{1}{2}$$ So, $$F(s) = \frac{1/2}{s-1} + \frac{1/2}{s-3}$$ Inverse Laplace: $$f(t) = \frac{1}{2} e^{t} + \frac{1}{2} e^{3t} = \frac{1}{2} (e^{t} + e^{3t})$$ --- **Final answers:** (a) $$f(t) = \frac{7}{2} t^2 e^{-3t}$$ (b) $$f(t) = 2 e^{-t} \cos(3t) - \frac{5}{3} e^{-t} \sin(3t)$$ (c) $$f(t) = \frac{1}{2} (e^{t} + e^{3t})$$