Let's find the **Laplace Transform** of each function \( f(t) \). Recall that the Laplace transform is defined as: \[ \mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t) dt \] where \( u(t) \) is the unit step function (so all functions are 0 for \( t < 0 \)). We'll use standard Laplace transform tables for efficiency. --- ### (e) \( f(t) = t^2 e^{-at} u(t) \) **Laplace transform of \( t^n e^{-at} \):** \[ \mathcal{L}\{t^n e^{-at}\} = \frac{n!}{(s+a)^{n+1}} \] For \( n = 2 \): \[ \mathcal{L}\{t^2 e^{-at}\} = \frac{2!}{(s+a)^{3}} = \frac{2}{(s+a)^3} \] --- ### (f) \( f(t) = (\sin \omega t)(\cos \omega t) u(t) \) First, use the identity: \[ \sin \omega t \cos \omega t = \frac{1}{2} \sin 2\omega t \] So, \[ f(t) = \frac{1}{2} \sin 2\omega t \cdot u(t) \] Laplace transform of \( \sin a t \): \[ \mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2} \] Here, \( a = 2\omega \): \[ F(s) = \frac{1}{2} \frac{2\omega}{s^2 + (2\omega)^2} = \frac{\omega}{s^2 + 4\omega^2} \] --- ### (g) \( f(t) = 0.03(1 - \cos 2t) u(t) \) Break it up: - Laplace of 1: \( \frac{1}{s} \) - Laplace of \( \cos 2t \): \( \frac{s}{s^2 + 4} \) So, \[ F(s) = 0.03\left( \frac{1}{s} - \frac{s}{s^2 + 4} \right) \] --- ### (h) \( f(t) = t e^{-t} \sin 5t \cdot u(t) \) Use the Laplace transform: \[ \mathcal{L}\{t e^{-\alpha t} \sin \beta t\} = \frac{2\beta (s+\alpha) - \beta^3}{[(s+\alpha)^2 + \beta^2]^2} \] But let's use the property: \[ \mathcal{L}\{t f(t)\} = -\frac{d}{ds}[F(s)] \] First, \( \mathcal{L}\{e^{-t} \sin 5t\} \): Let’s recall: \[ \mathcal{L}\{e^{-at} \sin(\omega t)\} = \frac{\omega}{(s+a)^2 + \omega^2} \] Here, \( a = 1 \), \( \omega = 5 \): \[ F(s) = \frac{5}{(s+1)^2 + 25} \] Now, Laplace of \( t e^{-t} \sin 5t \): \[ \mathcal{L}\{t e^{-t} \sin 5t\} = -\frac{d}{ds} \left( \frac{5}{(s+1)^2 + 25} \right) \] Let \( u = (s+1)^2 + 25 \): \[ \frac{d}{ds} \left( \frac{5}{u} \right) = 5 \cdot \frac{-1}{u^2} \cdot \frac{d}{ds}[u] = -5 \cdot \frac{2(s+1)}{u^2} \] So, \[ F(s) = -\frac{d}{ds} \left( \frac{5}{(s+1)^2 + 25} \right) = \frac{10(s+1)}{\left[ (s+1)^2 + 25 \right]^2} \] --- ## **Final Answers:** ### (e) \[ \boxed{F(s) = \frac{2}{(s+a)^3}} \] ### (f) \[ \boxed{F(s) = \frac{\omega}{s^2 + 4\omega^2}} \] ### (g) \[ \boxed{F(s) = 0.03\left( \frac{1}{s} - \frac{s}{s^2 + 4} \right)} \] ### (h) \[ \boxed{F(s) = \frac{10(s+1)}{\left[(s+1)^2 + 25\right]^2}} \] If you need any further detailed steps for any part, let me know!

Certainly! Here are 50-word theoretical explanations for each problem: **(e)** The Laplace transform of \( t^n e^{-at} \) involves factorials and powers of \( s + a \). Specifically, for integer \( n \), it simplifies to \( \frac{n!}{(s+a)^{n+1}} \), capturing the polynomial time dependence and exponential decay. **(f)** Using the identity \( \sin \omega t \cos \omega t = \frac{1}{2} \sin 2\omega t \), the transform simplifies. The Laplace of sine functions is standard, resulting in a rational function with \( s^2 + a^2 \) in the denominator, representing oscillatory behavior in the frequency domain. **(g)** The Laplace transform of a constant minus a cosine function separates into the difference of their transforms. The transform of 1 is \( 1/s \), and of cosine is \( s/(s^2 + \omega^2) \). This approach simplifies complex functions involving step functions and trigonometric components. **(h)** The Laplace transform of \( t e^{-\alpha t} \sin \beta t \) uses differentiation of the basic transform \( \frac{\beta}{(s+\alpha)^2 + \beta^2} \) with respect to \( s \). This method leverages the differentiation property to handle multiplication by \( t \), yielding a rational function representing damping and oscillation.