Question

Give the inverse Laplace transform of \(F(s) = \frac{4}{s^2} + \frac{2}{s} - \frac{2e^{-2s}}{s^2} - \frac{2e^{-2s}}{(s-2)^2}\) a) \(f(x) = \begin{cases} 8x + 2, & 0 \le x < 2\\ 6x + 2 - 2(x - 2)e^{-4+2x}, & 2 \le x \end{cases}\) \(f(x) = \begin{cases} 4x + 2, & 0 \le x < 2\\ 2x + 6 - 2(x - 2)e^{-4+2x}, & 2 \le x \end{cases}\) \(f(x) = \begin{cases} 8x + 2, & 0 \le x < 2\\ 6x + 2 - 2e^{4+2x}, & 2 \le x \end{cases}\) \(f(x) = \begin{cases} 4x + 2, & 0 \le x < 2\\ 2x + 6 - 2xe^{-4+2x}, & 2 \le x \end{cases}\) \(f(x) = \begin{cases} 4x + 2, & 0 \le x < 2\\ 2x - 2 - 2xe^{2x}, & 2 \le x \end{cases}\) None of the above.

          Give the inverse Laplace transform of
\(F(s) = \frac{4}{s^2} + \frac{2}{s} - \frac{2e^{-2s}}{s^2} - \frac{2e^{-2s}}{(s-2)^2}\)
a) \(f(x) = \begin{cases} 8x + 2, & 0 \le x < 2\\ 6x + 2 - 2(x - 2)e^{-4+2x}, & 2 \le x \end{cases}\)
\(f(x) = \begin{cases} 4x + 2, & 0 \le x < 2\\ 2x + 6 - 2(x - 2)e^{-4+2x}, & 2 \le x \end{cases}\)
\(f(x) = \begin{cases} 8x + 2, & 0 \le x < 2\\ 6x + 2 - 2e^{4+2x}, & 2 \le x \end{cases}\)
\(f(x) = \begin{cases} 4x + 2, & 0 \le x < 2\\ 2x + 6 - 2xe^{-4+2x}, & 2 \le x \end{cases}\)
\(f(x) = \begin{cases} 4x + 2, & 0 \le x < 2\\ 2x - 2 - 2xe^{2x}, & 2 \le x \end{cases}\)
None of the above.

Give the inverse Laplace transform of
F(s) = (4)/(s^2) + (2)/(s) - (2e^-2s)/(s^2) - (2e^-2s)/((s-2)^2)
a) f(x) = 8x + 2, 0 ≤ x < 2
6x + 2 - 2(x - 2)e^-4+2x, 2 ≤ x
f(x) = 4x + 2, 0 ≤ x < 2
2x + 6 - 2(x - 2)e^-4+2x, 2 ≤ x
f(x) = 8x + 2, 0 ≤ x < 2
6x + 2 - 2e^4+2x, 2 ≤ x
f(x) = 4x + 2, 0 ≤ x < 2
2x + 6 - 2xe^-4+2x, 2 ≤ x
f(x) = 4x + 2, 0 ≤ x < 2
2x - 2 - 2xe^2x, 2 ≤ x
None of the above.

Added by Mark D.

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