Provide a correct and error-free solution. Ensure the method is correct, there are no calculation errors, and the solution is complete and well-framed properly and Don't give direct answers; it must b

Question

Provide a correct and error-free solution. Ensure the method is correct, there are no calculation errors, and the solution is complete and well-framed properly and Don't give direct answers; it must be clear and properly framed and aligned. Avoid using personal pronouns like 'we,' 'you,' and 'I' in the answer. Give the answer as a teacher who is giving the answer to a student and making them understand properly. Don't use Python, please. The solutions will be evaluated based on six key parameters:Relevancy, ensuring closely matches with the question; Completeness , including like calculations, diagrams, procedural steps, & explanatory statements in the solution; Accuracy, ensuring correctness of concepts and methods; Clarity , assessing readability and simplicity of language; Structure , focusing on organization and presentation; and Voice: Determining the response feels natural and human-like rather than robotic. Give a two-line best explanation or interpretation where there is a need for it. Give an overall short final answer last. The complete solution is lacking; please complete the whole answer till the end. Please adhere to and strictly follow all these guidelines anyhow. Tha

VIPSolutions · Accepted Answer

# Inverse Laplace Transform Solution

Given Problem:
$$
\text{ the inverse Laplace transform of:$$
$$
F(s) \frac{2s2 - 6s + 5}{s^3 - 6s^2 + 11s - 6}
$$

---

## **Step 1: Factor the Denominator**

First, factor the denominator $ s^3 - 6s^2 + 11s - 6 $.

To factor, check for rational roots using the Rational Root Theorem:

Possible roots: $ \pm1, \pm2, \pm3, \pm6 $.

Testing $ s = 1 $:

$$
1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 
$$

So, $ s = 1 $ is a root.

Divide $ s^3 - 6s^2 + 11s - 6 $ by $ (s - 1) $:

Using synthetic or long division, the quotient is $ s^2 - 5s + 6 $.

Factor $ s^2 - 5s + 6 $:

$$
s^2 - 5s + 6 = (s - 2)(s - 3)
$$

So, the complete factorization is:
$$
s^3 - 6s^2 + 11s - 6 = (s - 1)(s - 2)(s - 3)
$$

---

## **Step 2: Express as Partial Fractions**

Write:
$$
\frac{2s^2 - 6s + 5}{(s-1)(s-2)(s-3)} = \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3}
$$

Multiply both sides by the denominator:
$$
2s^2 - 6s + 5 = A(s-2)(s-3) + B(s-1)(s-3) + C(s-1)(s-2)
$$

---

## **Step 3: Solve for Coefficients $A, B, C$**

### Substituting Roots

#### For $ s = 1 $:
$$
2(1)^2 - 6(1) + 5 = A(1-2)(1-3) +  + 
$$
$$
2 - 6 + 5 = A(-1)(-2)
$$
$$
1 = 2A \implies A = \frac{1}{2}
$$

#### For $ s = 2 $:
$$
2(2)^2 - 6(2) + 5 =  + B(2-1)(2-3) + 
$$
$$
8 - 12 + 5 = B(1)(-1)
$$
$$
1 = -B \implies B = -1
$$

#### For $ s = 3 $:
$$
2(3)^2 - 6(3) + 5 =  +  + C(3-1)(3-2)
$$
$$
18 - 18 + 5 = C(2)(1)
$$
$$
5 = 2C \implies C = \frac{5}{2}
$$

---

## **Step 4: Substitute Coefficients**

$$
\frac{2s^2 - 6s + 5}{(s-1)(s-2)(s-3)} = \frac{1}{2}\frac{1}{s-1} - \frac{1}{s-2} + \frac{5}{2}\frac{1}{s-3}
$$

---

## **Step 5: Find the Inverse Laplace Transform**

Recall:
$$
\mathcal{L}^{-1}\left\{ \frac{1}{s-a} \right\} = e^{at}
$$

Apply this to each term:

- $ \frac{1}{2} \frac{1}{s-1} $ transforms to $ \frac{1}{2} e^{t} $
- $ -\frac{1}{s-2} $ transforms to $ -e^{2t} $
- $ \frac{5}{2} \frac{1}{s-3} $ transforms to $ \frac{5}{2} e^{3t} $

---

## **Step 6: Write the Final Solution**

$$
\mathcal{L}^{-1}\left\{ \frac{2s^2 - 6s + 5}{s^3 - 6s^2 + 11s - 6} \right\}
= \frac{1}{2} e^{t} - e^{2t} + \frac{5}{2} e^{3t}
$$

---

### **Interpretation**

The solution is a linear combination of exponentials, reflecting the system's response for three distinct poles at $ s=1, 2, 3 $. Each term in the solution corresponds to the time-domain response associated with one root of the denominator.

---

## **Final Answer**

$$
\boxed{
\mathcal{L}^{-1}\left\{ \frac{2s^2 - 6s + 5}{s^3 - 6s^2 + 11s - 6} \right\}
= \frac{1}{2} e^{t} - e^{2t} + \frac{5}{2} e^{3t}
}
$$

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