We are tasked with finding the inverse Laplace transform $L^{-1}$ of the given function:

$\frac{s^2 - 3s + 4}{s^3}.$

Step-by-step Solution:

1. Rewrite the given expression by dividing each term in the numerator by $s^3$:

   $\frac{s^2 - 3s + 4}{s^3} = \frac{s^2}{s^3} - \frac{3s}{s^3} + \frac{4}{s^3}.$

   Simplify each term:

   $\frac{s^2}{s^3} = \frac{1}{s}, \quad \frac{3s}{s^3} = \frac{3}{s^2}, \quad \frac{4}{s^3} = \frac{4}{s^3}.$

   Thus, the expression becomes:

   $\frac{s^2 - 3s + 4}{s^3} = \frac{1}{s} - \frac{3}{s^2} + \frac{4}{s^3}.$

2. Find the inverse Laplace transform term by term:

   Use the standard Laplace transform properties:

   • $L^{-1}\left(\frac{1}{s}\right) = 1$,
   • $L^{-1}\left(\frac{1}{s^2}\right) = t$,
   • $L^{-1}\left(\frac{1}{s^3}\right) = \frac{t^2}{2}$.

   Apply these to each term:

   • $L^{-1}\left(\frac{1}{s}\right) = 1$,
   • $L^{-1}\left(-\frac{3}{s^2}\right) = -3t$,
   • $L^{-1}\left(\frac{4}{s^3}\right) = \frac{4t^2}{2} = 2t^2$.

3. Combine the results:

   The inverse Laplace transform is:

   $L^{-1}\left[\frac{s^2 - 3s + 4}{s^3}\right] = 1 - 3t + 2t^2.$

Final Answer:

$L^{-1}\left[\frac{s^2 - 3s + 4}{s^3}\right] = 1 - 3t + 2t^2.$

Would you like a detailed explanation of Laplace transform properties or another example?

Follow-Up Questions:

1. What is the general formula for $L^{-1}\left(\frac{1}{s^n}\right)$?
2. How can Laplace transforms be used to solve differential equations?
3. What is the Laplace transform of a unit step function?
4. How do we handle Laplace transforms involving exponential functions (e.g., $e^{-as}$)?
5. What are the conditions for the existence of a Laplace transform?

Tip:

When solving Laplace transform problems, break the function into simpler components that match standard Laplace transform tables.