Question

The L(te^-3t cos⁡(2t)cos⁡(3t)) is given by  \(k\left [\frac{25-(s+3)^2}{((s+3)^2+25)^2} + \frac{(1-(s+3)^2)}{((s+3)^2+1)^2}\right ]\). Find the value of k.

(a) 0

(b) 1

(c) \(\frac{1}{2}\)

(d) \(\frac{-1}{2}\)

This question was posed to me during an interview.

My question is from Table of General Properties of Laplace Transform in division Laplace Transform of Engineering Mathematics

Answer 1

Right option is (d) \(\frac{-1}{2}\)

The best explanation: In the given question,

L(e^-3t cos⁡(2t)cos⁡(3t))

L(cos⁡(2t) cos⁡(3t))

=\(\frac{1}{2} L(cos⁡(5t)+cos⁡(t))\)

=\(\frac{1}{2} \left (\frac{s}{(s^2+25)}\right )+\frac{1}{2} \left (\frac{s}{(s^2+1)}\right )\)

By effect of multiplication by t

L(t×cos⁡(2t) cos⁡(3t))=\((-1)×\frac{1}{2}×\frac{d}{ds} \frac{1}{2} \left (\frac{s}{(s^2+25)}\right )+\frac{1}{2} \left (\frac{s}{(s^2+1)}\right )\)

=\(\frac{-1}{2} \left [\frac{25-s^2}{(s^2+25)^2} + \frac{(1-s^2)}{(s^2+1)^2}\right]\)

By the effect of first shifting,

L(te^-3t cos⁡(2t)cos⁡(3t))=\(\frac{-1}{2} \left [\frac{25-(s+3)^2}{((s+3)^2+25)^2} + \frac{1-(s+3)^2}{((s+3)^2+1)^2}\right ]\)

k=\(\frac{-1}{2}\).