1. **Problem Statement:** Evaluate the Laplace transforms of the given functions. 2. **Recall the Laplace transform definition:** $$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt$$ 3. **Useful formulas and properties:** - $\mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s - a)^{n+1}}$ - $\mathcal{L}\{e^{at} \sin(bt)\} = \frac{b}{(s - a)^2 + b^2}$ - $\mathcal{L}\{e^{at} \cos(bt)\} = \frac{s - a}{(s - a)^2 + b^2}$ - $\sinh(bt) = \frac{e^{bt} - e^{-bt}}{2}$ and $\cosh(bt) = \frac{e^{bt} + e^{-bt}}{2}$ - Linearity: $\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}$ --- 4. **Evaluate each:** (i) $\mathcal{L}\{t^3 e^{-3t}\}$ Using formula for $t^n e^{at}$ with $n=3$, $a=-3$: $$= \frac{3!}{(s - (-3))^{4}} = \frac{6}{(s + 3)^4}$$ (ii) $\mathcal{L}\{5 e^{3t} \sin 4t\}$ Using $\mathcal{L}\{e^{at} \sin(bt)\} = \frac{b}{(s - a)^2 + b^2}$: $$= 5 \times \frac{4}{(s - 3)^2 + 16} = \frac{20}{(s - 3)^2 + 16}$$ (iii) $\mathcal{L}\{(t + 2)^2 e^t\}$ Expand $(t+2)^2 = t^2 + 4t + 4$: $$= \mathcal{L}\{t^2 e^t\} + 4 \mathcal{L}\{t e^t\} + 4 \mathcal{L}\{e^t\}$$ Use formulas: - $\mathcal{L}\{t^2 e^{t}\} = \frac{2!}{(s - 1)^3} = \frac{2}{(s - 1)^3}$ - $\mathcal{L}\{t e^{t}\} = \frac{1!}{(s - 1)^2} = \frac{1}{(s - 1)^2}$ - $\mathcal{L}\{e^{t}\} = \frac{1}{s - 1}$ So: $$= \frac{2}{(s - 1)^3} + \frac{4}{(s - 1)^2} + \frac{4}{s - 1}$$ (iv) $\mathcal{L}\{e^{-t}(3 \sinh 2t - 5 \cosh 2t)\}$ Rewrite using linearity: $$= 3 \mathcal{L}\{e^{-t} \sinh 2t\} - 5 \mathcal{L}\{e^{-t} \cosh 2t\}$$ Recall $\sinh 2t = \frac{e^{2t} - e^{-2t}}{2}$ and $\cosh 2t = \frac{e^{2t} + e^{-2t}}{2}$ So: $$e^{-t} \sinh 2t = \frac{e^{t} - e^{-3t}}{2}, \quad e^{-t} \cosh 2t = \frac{e^{t} + e^{-3t}}{2}$$ Apply linearity: $$= 3 \times \frac{\mathcal{L}\{e^{t}\} - \mathcal{L}\{e^{-3t}\}}{2} - 5 \times \frac{\mathcal{L}\{e^{t}\} + \mathcal{L}\{e^{-3t}\}}{2}$$ Use $\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$: $$= \frac{3}{2} \left( \frac{1}{s - 1} - \frac{1}{s + 3} \right) - \frac{5}{2} \left( \frac{1}{s - 1} + \frac{1}{s + 3} \right)$$ Simplify: $$= \frac{3}{2(s - 1)} - \frac{3}{2(s + 3)} - \frac{5}{2(s - 1)} - \frac{5}{2(s + 3)} = -\frac{1}{s - 1} - \frac{4}{s + 3}$$ (v) $\mathcal{L}\{e^{-4t} \cosh 2t\}$ Use $\cosh 2t = \frac{e^{2t} + e^{-2t}}{2}$: $$= \frac{1}{2} \left( \mathcal{L}\{e^{-4t + 2t}\} + \mathcal{L}\{e^{-4t - 2t}\} \right) = \frac{1}{2} \left( \mathcal{L}\{e^{-2t}\} + \mathcal{L}\{e^{-6t}\} \right)$$ Apply formula: $$= \frac{1}{2} \left( \frac{1}{s + 2} + \frac{1}{s + 6} \right)$$ (vi) $\mathcal{L}\{e^{2t}(3 \sin 4t - 4 \cos 4t)\}$ Use linearity: $$= 3 \mathcal{L}\{e^{2t} \sin 4t\} - 4 \mathcal{L}\{e^{2t} \cos 4t\}$$ Use formulas: $$= 3 \times \frac{4}{(s - 2)^2 + 16} - 4 \times \frac{s - 2}{(s - 2)^2 + 16} = \frac{12 - 4(s - 2)}{(s - 2)^2 + 16} = \frac{20 - 4s}{(s - 2)^2 + 16}$$ (vii) $\mathcal{L}\{t e^{2t} (\sinh t + \cosh t)\}$ Recall $\sinh t + \cosh t = e^t$: $$= \mathcal{L}\{t e^{2t} e^t\} = \mathcal{L}\{t e^{3t}\}$$ Use formula for $t e^{at}$: $$= \frac{1}{(s - 3)^2}$$ --- **Final answers:** (i) $\frac{6}{(s + 3)^4}$ (ii) $\frac{20}{(s - 3)^2 + 16}$ (iii) $\frac{2}{(s - 1)^3} + \frac{4}{(s - 1)^2} + \frac{4}{s - 1}$ (iv) $-\frac{1}{s - 1} - \frac{4}{s + 3}$ (v) $\frac{1}{2} \left( \frac{1}{s + 2} + \frac{1}{s + 6} \right)$ (vi) $\frac{20 - 4s}{(s - 2)^2 + 16}$ (vii) $\frac{1}{(s - 3)^2}$