Question

The inverse Laplace transform of F(s) = \(\frac{5}{s^2-9}\) is ______________

(a) \(\frac{1}{6}e^{3t} + \frac{5}{6e^{-3t}}\)

(b) \(\frac{1}{6}e^{3t} – \frac{5}{6e^{-3t}}\)

(c) \(\frac{5}{6}e^{3t} + \frac{5}{6e^{-3t}}\)

(d) \(\frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}\)

This question was addressed to me in an international level competition.

This intriguing question comes from Problems on Initial and Final Value Theorem topic in division Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

The correct choice is (d) \(\frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}\)

To explain I would say: F (t) = L^-1{F(s)}

= L^-1{\(\frac{5}{s^2-9}\)}

= L^-1\({\frac{A}{s-3} + \frac{B}{s+3}}\)

Hence, A = (s-3) \(\frac{5}{(s-3)(s+3)}|_{s=3} = \frac{5}{6}\)

And, B = (s+3) \(\frac{5}{(s-3)(s+3)}|_{s=-3} = -\frac{5}{6}\)

The inverse Laplace transform is \(\frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}\).