Let's solve **Problem 12**: Given: \[ F(s) = \frac{3}{(s+1)^2} + \frac{1}{s^3} \] We are asked to find the inverse Laplace transform, \( \mathcal{L}^{-1}(F(s)) \). --- ### Step 1: Use Laplace Transform Table - \( \mathcal{L}^{-1}\left(\frac{n!}{s^{n+1}}\right) = t^n \) - \( \mathcal{L}^{-1}\left(\frac{1}{(s+a)^n}\right) = \frac{t^{n-1}}{(n-1)!} e^{-a t} \) --- ### Step 2: Find Inverse Laplace for Each Term #### **First term:** \( \frac{3}{(s+1)^2} \) - \( \mathcal{L}^{-1}\left(\frac{1}{(s+a)^2}\right) = t e^{-a t} \) - So, \( \mathcal{L}^{-1}\left(\frac{3}{(s+1)^2}\right) = 3 t e^{-t} \) #### **Second term:** \( \frac{1}{s^3} \) - \( \mathcal{L}^{-1}\left(\frac{1}{s^{n+1}}\right) = \frac{t^n}{n!} \); here \( n=2 \) - So, \( \mathcal{L}^{-1}\left(\frac{1}{s^3}\right) = \frac{t^2}{2} \) --- ### Step 3: Combine Results \[ \boxed{ f(t) = 3 t e^{-t} + \frac{t^2}{2} } \] --- **Final Answer:** \[ \mathcal{L}^{-1}(F(s)) = 3 t e^{-t} + \frac{t^2}{2} \]