1. We need to find the Laplace transform of each given function. The Laplace transform of a function $f(t)$ is defined as $$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt.$$ 2. For (i) $3e^{-2t}$, use the formula $\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$ for $s > a$. Here, $a = -2$, so: $$\mathcal{L}\{3e^{-2t}\} = 3 \times \frac{1}{s + 2} = \frac{3}{s + 2}.$$ 3. For (ii) $4t^{3} - e^{-t}$, use linearity and known transforms: - $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$ for $n=3$ gives $\frac{6}{s^4}$. - $\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$ with $a = -1$ gives $\frac{1}{s + 1}$. So, $$\mathcal{L}\{4t^{3} - e^{-t}\} = 4 \times \frac{6}{s^4} - \frac{1}{s + 1} = \frac{24}{s^4} - \frac{1}{s + 1}.$$ 4. For (iii) $7 \sin 2t - 3 \cos 2t$, use: - $\mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2}$, - $\mathcal{L}\{\cos at\} = \frac{s}{s^2 + a^2}$. Here, $a=2$, so: $$\mathcal{L}\{7 \sin 2t - 3 \cos 2t\} = 7 \times \frac{2}{s^2 + 4} - 3 \times \frac{s}{s^2 + 4} = \frac{14 - 3s}{s^2 + 4}.$$ 5. For (iv) $(t^{2} + 1)^{2} = t^4 + 2t^2 + 1$, use linearity and transforms: - $\mathcal{L}\{t^4\} = \frac{4!}{s^5} = \frac{24}{s^5}$, - $\mathcal{L}\{t^2\} = \frac{2!}{s^3} = \frac{2}{s^3}$, - $\mathcal{L}\{1\} = \frac{1}{s}$. So, $$\mathcal{L}\{(t^{2} + 1)^{2}\} = \frac{24}{s^5} + 2 \times \frac{2}{s^3} + \frac{1}{s} = \frac{24}{s^5} + \frac{4}{s^3} + \frac{1}{s}.$$ 6. For (v) $(4e^{2t} - 2)^{3}$, expand using binomial theorem: $$(4e^{2t} - 2)^3 = 64 e^{6t} - 96 e^{4t} + 48 e^{2t} - 8.$$ Apply Laplace transform term-wise: - $\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$ for $s > a$. So, $$\mathcal{L}\{(4e^{2t} - 2)^3\} = 64 \times \frac{1}{s - 6} - 96 \times \frac{1}{s - 4} + 48 \times \frac{1}{s - 2} - 8 \times \frac{1}{s} = \frac{64}{s - 6} - \frac{96}{s - 4} + \frac{48}{s - 2} - \frac{8}{s}.$$ Final answers: (i) $\frac{3}{s + 2}$ (ii) $\frac{24}{s^4} - \frac{1}{s + 1}$ (iii) $\frac{14 - 3s}{s^2 + 4}$ (iv) $\frac{24}{s^5} + \frac{4}{s^3} + \frac{1}{s}$ (v) $\frac{64}{s - 6} - \frac{96}{s - 4} + \frac{48}{s - 2} - \frac{8}{s}$