Let's solve each problem step-by-step. --- ## 1. **Laplace Transform of \( u_2(t) \)** \( u_2(t) \) is the **Heaviside (unit step) function** shifted to \( t = 2 \). \[ u_2(t) = \begin{cases} 0, & t < 2 \\ 1, & t \geq 2 \end{cases} \] **Laplace Transform Formula:** \[ \mathcal{L}\{u_c(t)\} = \frac{e^{-cs}}{s} \] **For \( c = 2 \):** \[ \boxed{\mathcal{L}\{u_2(t)\} = \frac{e^{-2s}}{s}} \] --- ## 2. **Laplace Transform of \( \delta_3(t) \)** \( \delta_3(t) \) is the **Dirac delta function** at \( t = 3 \). **Laplace Transform Formula:** \[ \mathcal{L}\{\delta_c(t)\} = e^{-cs} \] **For \( c = 3 \):** \[ \boxed{\mathcal{L}\{\delta_3(t)\} = e^{-3s}} \] --- ## 3. **Inverse Laplace Transform of \( e^{s} \frac{1}{s^2+1} \)** ### Step 1: Recognize the Laplace Shift Theorem \[ \mathcal{L}\{f(t-a)u_a(t)\} = e^{-as}F(s) \] Here, we have \( e^{s} \) instead of \( e^{-as} \), so \( a = -1 \). ### Step 2: Identify \( F(s) = \frac{1}{s^2+1} \) \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2+1}\right\} = \sin t \] ### Step 3: Apply the **time advance** property: \[ \mathcal{L}^{-1}\left\{e^{as}F(s)\right\} = f(t+a) \] So, \[ \mathcal{L}^{-1}\left\{e^{s}\frac{1}{s^2+1}\right\} = \sin(t+1) \] **Final Answer:** \[ \boxed{\mathcal{L}^{-1}\left\{e^{s}\frac{1}{s^2+1}\right\} = \sin(t+1)} \] --- ## 4. **Inverse Laplace Transform of \( e^{-2s} \frac{s}{s^2+9} \)** ### Step 1: Recognize the Laplace Shift Theorem \[ \mathcal{L}^{-1}\left\{e^{-as} F(s)\right\} = f(t-a) u_a(t) \] Here, \( a = 2 \), \( F(s) = \frac{s}{s^2+9} \). ### Step 2: Inverse Laplace of \( F(s) \): \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+\omega^2}\right\} = \cos(\omega t) \] Here, \( \omega = 3 \), so \( \cos(3t) \). ### Step 3: Apply the shift: \[ f(t-2)u_2(t) = \cos(3(t-2))u_2(t) \] **Final Answer:** \[ \boxed{\mathcal{L}^{-1}\left\{e^{-2s} \frac{s}{s^2+9}\right\} = \cos(3(t-2))\,u_2(t)} \] --- ## **Summary of Answers** 1. \( \mathcal{L}\{u_2(t)\} = \boxed{\frac{e^{-2s}}{s}} \) 2. \( \mathcal{L}\{\delta_3(t)\} = \boxed{e^{-3s}} \) 3. \( \mathcal{L}^{-1}\left\{e^{s} \frac{1}{s^2+1}\right\} = \boxed{\sin(t+1)} \) 4. \( \mathcal{L}^{-1}\left\{e^{-2s} \frac{s}{s^2+9}\right\} = \boxed{\cos(3(t-2)) u_2(t)} \)