Question

If \(f(t)\) is the inverse Laplace transform of \( F(s) = \frac{s+1+s^{-2}}{s^2-1} \), then \(f(t)\) is

• A. \(e^t + \sinh t + t\)
• B. \(e^t + \sinh t - t\)
• C. \(e^t - \sinh t + t\)
• D. \(e^t + \cosh t - t\)

Hint:

When dealing with complex rational functions for inverse Laplace transforms, partial fraction decomposition is the key method. Remember the standard transforms: \( \mathcal{L}^{-1}\{1/s^n\} = t^{n-1}/(n-1)! \), \( \mathcal{L}^{-1}\{1/(s-a)\} = e^{at} \).

Answer 1

The Correct Option is B

Step 1: Simplify the expression for F(s). 
The given expression appears unusual but can be simplified algebraically. \[ F(s) = \frac{s+1+s^{-2}}{s^2-1} = \frac{s+1+\frac{1}{s^2}}{s^2-1} = \frac{\frac{s^3+s^2+1}{s^2}}{s^2-1} = \frac{s^3+s^2+1}{s^2(s^2-1)} \]

Step 2: Decompose F(s) using partial fractions. 
We can write F(s) in the form: \[ \frac{s^3+s^2+1}{s^2(s-1)(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-1} + \frac{D}{s+1} \] 

To find the coefficients: 

- \( B = \left[ s^2 F(s) \right]_{s=0} = \left[ \frac{s^3+s^2+1}{s^2-1} \right]_{s=0} = \frac{1}{-1} = -1 \) 

- \( C = \left[ (s-1) F(s) \right]_{s=1} = \left[ \frac{s^3+s^2+1}{s^2(s+1)} \right]_{s=1} = \frac{1+1+1}{1(2)} = \frac{3}{2} \) 

- \( D = \left[ (s+1) F(s) \right]_{s=-1} = \left[ \frac{s^3+s^2+1}{s^2(s-1)} \right]_{s=-1} = \frac{-1+1+1}{(-1)^2(-2)} = -\frac{1}{2} \) 

- To find A, compare coefficients of \(s^3\) on both sides: \(1 = A+C+D \Rightarrow 1 = A + \frac{3}{2} - \frac{1}{2} \Rightarrow 1 = A+1 \Rightarrow A = 0\).

Step 3: Write the final form of F(s) and find the inverse Laplace transform. 
\[ F(s) = -\frac{1}{s^2} + \frac{3/2}{s-1} - \frac{1/2}{s+1} \] 

Taking the inverse Laplace transform term by term: \[ f(t) = \mathcal{L}^{-1}\{-\frac{1}{s^2}\} + \mathcal{L}^{-1}\{\frac{3/2}{s-1}\} - \mathcal{L}^{-1}\{\frac{1/2}{s+1}\} \] \[ f(t) = -t + \frac{3}{2}e^t - \frac{1}{2}e^{-t} \]

Step 4: Express the result in terms of hyperbolic functions. 
Recall that \( \sinh t = \frac{e^t - e^{-t}}{2} \) and \( e^t = \cosh t + \sinh t \). \[ f(t) = -t + e^t + \frac{1}{2}e^t - \frac{1}{2}e^{-t} = -t + e^t + \left(\frac{e^t - e^{-t}}{2}\right) = e^t + \sinh t - t \]