1. Find the Laplace Transforms of the following functions: (a) $f(t) = 2t^3 + 4 \cos 5t$ - The Laplace transform of $t^n$ is $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$. - The Laplace transform of $\cos(at)$ is $\mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}$. Calculate: $$\mathcal{L}\{2t^3\} = 2 \cdot \frac{3!}{s^{4}} = 2 \cdot \frac{6}{s^{4}} = \frac{12}{s^{4}}$$ $$\mathcal{L}\{4 \cos 5t\} = 4 \cdot \frac{s}{s^2 + 25} = \frac{4s}{s^2 + 25}$$ So, $$\mathcal{L}\{f(t)\} = \frac{12}{s^{4}} + \frac{4s}{s^2 + 25}$$ (b) $f(t) = t \cosh 4t$ - Recall $\cosh(at) = \frac{e^{at} + e^{-at}}{2}$. - The Laplace transform of $t f(t)$ is $-\frac{d}{ds} \mathcal{L}\{f(t)\}$. - The Laplace transform of $\cosh(at)$ is $\frac{s}{s^2 - a^2}$. Calculate $\mathcal{L}\{\cosh 4t\} = \frac{s}{s^2 - 16}$. Then, $$\mathcal{L}\{t \cosh 4t\} = -\frac{d}{ds} \left( \frac{s}{s^2 - 16} \right)$$ Differentiate: $$\frac{d}{ds} \left( \frac{s}{s^2 - 16} \right) = \frac{(s^2 - 16) \cdot 1 - s \cdot 2s}{(s^2 - 16)^2} = \frac{s^2 - 16 - 2s^2}{(s^2 - 16)^2} = \frac{-s^2 - 16}{(s^2 - 16)^2}$$ So, $$\mathcal{L}\{t \cosh 4t\} = - \left( \frac{-s^2 - 16}{(s^2 - 16)^2} \right) = \frac{s^2 + 16}{(s^2 - 16)^2}$$ (c) $f(t) = t^3 e^{-5t}$ - The Laplace transform of $t^n e^{at}$ is $\frac{n!}{(s - a)^{n+1}}$. Calculate: $$\mathcal{L}\{t^3 e^{-5t}\} = \frac{3!}{(s + 5)^4} = \frac{6}{(s + 5)^4}$$ (d) $f(t) = \frac{\sinh 2t}{t}$ - Recall $\sinh(at) = \sum_{n=0}^\infty \frac{a^{2n+1} t^{2n+1}}{(2n+1)!}$. - Dividing by $t$ gives $\frac{\sinh 2t}{t} = \sum_{n=0}^\infty \frac{2^{2n+1} t^{2n}}{(2n+1)!}$. The Laplace transform is linear, so: $$\mathcal{L}\left\{ \frac{\sinh 2t}{t} \right\} = \sum_{n=0}^\infty \frac{2^{2n+1}}{(2n+1)!} \mathcal{L}\{t^{2n}\} = \sum_{n=0}^\infty \frac{2^{2n+1}}{(2n+1)!} \cdot \frac{(2n)!}{s^{2n+1}}$$ This series converges to: $$\mathcal{L}\left\{ \frac{\sinh 2t}{t} \right\} = \frac{1}{s} \arctan \left( \frac{2}{s} \right)$$ 2. Find the inverse Laplace Transform: (a) $F(s) = \frac{1}{5s + 9}$ Rewrite: $$F(s) = \frac{1}{5(s + \frac{9}{5})} = \frac{1}{5} \cdot \frac{1}{s + \frac{9}{5}}$$ Inverse transform: $$\mathcal{L}^{-1}\left\{ \frac{1}{s + a} \right\} = e^{-at}$$ So, $$f(t) = \frac{1}{5} e^{-\frac{9}{5} t}$$ (b) $F(s) = \frac{s}{s^2 + 6s + 13}$ Complete the square: $$s^2 + 6s + 13 = (s + 3)^2 + 4$$ Rewrite: $$F(s) = \frac{s + 3 - 3}{(s + 3)^2 + 2^2} = \frac{s + 3}{(s + 3)^2 + 4} - \frac{3}{(s + 3)^2 + 4}$$ Inverse transforms: $$\mathcal{L}^{-1}\left\{ \frac{s + a}{(s + a)^2 + b^2} \right\} = e^{-at} \cos bt$$ $$\mathcal{L}^{-1}\left\{ \frac{b}{(s + a)^2 + b^2} \right\} = e^{-at} \sin bt$$ So, $$f(t) = e^{-3t} \cos 2t - \frac{3}{2} e^{-3t} \sin 2t$$ (c) $F(s) = \frac{5s^2 - 4s - 7}{(s - 3)(s^2 + 4)}$ Use partial fractions: $$\frac{5s^2 - 4s - 7}{(s - 3)(s^2 + 4)} = \frac{A}{s - 3} + \frac{Bs + C}{s^2 + 4}$$ Multiply both sides by denominator: $$5s^2 - 4s - 7 = A(s^2 + 4) + (Bs + C)(s - 3)$$ Expand: $$5s^2 - 4s - 7 = A s^2 + 4A + B s^2 - 3 B s + C s - 3 C$$ Group terms: $$5s^2 - 4s - 7 = (A + B) s^2 + (-3B + C) s + (4A - 3C)$$ Equate coefficients: - $s^2$: $5 = A + B$ - $s$: $-4 = -3B + C$ - Constant: $-7 = 4A - 3C$ Solve system: From $5 = A + B$, $B = 5 - A$. Substitute into $-4 = -3B + C$: $$-4 = -3(5 - A) + C = -15 + 3A + C \Rightarrow C = 11 - 3A$$ Substitute $A$ and $C$ into $-7 = 4A - 3C$: $$-7 = 4A - 3(11 - 3A) = 4A - 33 + 9A = 13A - 33$$ Solve for $A$: $$13A = 26 \Rightarrow A = 2$$ Then, $$B = 5 - 2 = 3$$ $$C = 11 - 3(2) = 11 - 6 = 5$$ So, $$F(s) = \frac{2}{s - 3} + \frac{3s + 5}{s^2 + 4}$$ Inverse Laplace: $$\mathcal{L}^{-1}\left\{ \frac{1}{s - a} \right\} = e^{at}$$ $$\mathcal{L}^{-1}\left\{ \frac{s}{s^2 + b^2} \right\} = \cos bt$$ $$\mathcal{L}^{-1}\left\{ \frac{b}{s^2 + b^2} \right\} = \sin bt$$ Rewrite numerator: $$3s + 5 = 3s + 6 - 1 = 3s + 6 - 1$$ Split: $$\frac{3s + 5}{s^2 + 4} = 3 \cdot \frac{s}{s^2 + 4} + \frac{5}{s^2 + 4} = 3 \cdot \frac{s}{s^2 + 4} + \frac{6 - 1}{s^2 + 4} = 3 \cdot \frac{s}{s^2 + 4} + 3 \cdot \frac{2}{s^2 + 4} - \frac{1}{s^2 + 4}$$ But better to write: $$\frac{3s + 5}{s^2 + 4} = 3 \cdot \frac{s}{s^2 + 4} + \frac{5}{s^2 + 4}$$ Since $\frac{5}{s^2 + 4} = \frac{5}{2} \cdot \frac{2}{s^2 + 4}$, Inverse transform: $$3 \cos 2t + \frac{5}{2} \sin 2t$$ Final solution: $$f(t) = 2 e^{3t} + 3 \cos 2t + \frac{5}{2} \sin 2t$$ 3. Solve the differential equation: $$\frac{dx}{dt} + 4x = e^{-4t} \alpha, \quad x(0) = 2$$ Step 1: Take Laplace transform of both sides: $$s X(s) - x(0) + 4 X(s) = \alpha \cdot \frac{1}{s + 4}$$ Substitute $x(0) = 2$: $$s X(s) - 2 + 4 X(s) = \frac{\alpha}{s + 4}$$ Step 2: Solve for $X(s)$: $$X(s)(s + 4) = 2 + \frac{\alpha}{s + 4}$$ Multiply both sides by $s + 4$: $$X(s)(s + 4)^2 = 2(s + 4) + \alpha$$ So, $$X(s) = \frac{2(s + 4) + \alpha}{(s + 4)^2} = \frac{2s + 8 + \alpha}{(s + 4)^2}$$ Step 3: Split into two terms: $$X(s) = \frac{2s}{(s + 4)^2} + \frac{8 + \alpha}{(s + 4)^2}$$ Rewrite numerator of first term: $$2s = 2(s + 4) - 8$$ So, $$X(s) = \frac{2(s + 4) - 8}{(s + 4)^2} + \frac{8 + \alpha}{(s + 4)^2} = \frac{2(s + 4)}{(s + 4)^2} - \frac{8}{(s + 4)^2} + \frac{8 + \alpha}{(s + 4)^2}$$ Simplify: $$X(s) = \frac{2}{s + 4} + \frac{\alpha}{(s + 4)^2}$$ Step 4: Take inverse Laplace transform: $$\mathcal{L}^{-1}\left\{ \frac{1}{s + a} \right\} = e^{-at}$$ $$\mathcal{L}^{-1}\left\{ \frac{1}{(s + a)^2} \right\} = t e^{-at}$$ So, $$x(t) = 2 e^{-4t} + \alpha t e^{-4t}$$ This is the solution satisfying the initial condition.