give the calculation step by step as like as full answer in 2 steps and explanation block in every steps.
And final answer summary
Definition of Parameters
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And final answer summary
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VIPSolutions · Accepted Answer

**Step 1: Apply the First Shifting Theorem (Time Delay Property)**

The given Laplace transform is:

$$
\mathcal{L}^{-1} \left[ \frac{e^{-3s}}{s^2 (s-1)} \right]
$$

The **First Shifting Theorem** states:
$$
\mathcal{L}^{-1} \left[ e^{-as} F(s) \right] = f(t-a) u(t-a)
$$
where $ u(t-a) $ is the unit step function.

Here, $ a = 3 $ and $ F(s) = \frac{1}{s^2 (s-1)} $.

**Explanation:**  
The exponential term $ e^{-3s} $ in the Laplace domain introduces a delay of 3 units in the time domain. The solution will have the form $ f(t-3) u(t-3) $ where $ f(t) $ is the inverse Laplace of $ \frac{1}{s^2 (s-1)} $.

---

**Step 2: Find the Inverse Laplace of $ \frac{1}{s^2 (s-1)} $ Using Partial Fractions**

Express $ \frac{1}{s^2 (s-1)} $ as partial fractions:
$$
\frac{1}{s^2 (s-1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-1}
$$

Multiply both sides by $ s^2 (s-1) $:
$$
1 = A s (s-1) + B (s-1) + C s^2
$$

Expand and collect like terms:
$$
1 = A (s^2 - s) + B (s-1) + C s^2 = (A + C)s^2 + (-A + B)s - B
$$

Equate coefficients:
- $ s^2 $ term: $ A + C = 0 $
- $ s $ term: $ -A + B = 0 $
- Constant term: $ -B = 1 $

Solve:
- $ -B = 1 \implies B = -1 $
- $ -A + (-1) = 0 \implies A = -1 $
- $ -1 + C = 0 \implies C = 1 $

So,
$$
\frac{1}{s^2 (s-1)} = -\frac{1}{s} - \frac{1}{s^2} + \frac{1}{s-1}
$$

Now, take the inverse Laplace:
$$
\mathcal{L}^{-1} \left[-\frac{1}{s}\right] = -1
$$
$$
\mathcal{L}^{-1} \left[-\frac{1}{s^2}\right] = -t
$$
$$
\mathcal{L}^{-1} \left[\frac{1}{s-1}\right] = e^{t}
$$

Combine:
$$
f(t) = -1 - t + e^{t}
$$

**Explanation:**  
Partial fraction decomposition allows the expression to be split into simpler terms, each of which has a known inverse Laplace transform.

---

**Final Answer (With Delay Applied):**

$$
\boxed{
\mathcal{L}^{-1} \left[ \frac{e^{-3s}}{s^2 (s-1)} \right] =
\left[ -1 - (t-3) + e^{t-3} \right] u(t-3)
}
$$

**Summary:**  
The inverse Laplace transform of $ \frac{e^{-3s}}{s^2 (s-1)} $ is $ \left[ -1 - (t-3) + e^{t-3} \right] u(t-3) $, where $ u(t-3) $ is the unit step function. The calculation uses the time-shifting property and partial fractions for the inverse Laplace.

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