Question

(3) [1 point] Section 7.3 # 25: Use formula (6) and the fact that ỵ (os bt} (s) to help determine (a) t cos bt) and (b) (t2 cos bt) E(t"f(t)) (s) = (-1)_rdm(s). (Formula 6) ds"

Answer 1

(i) L {t cos bt} = ? comparing it with L {t^n f (t)} we get n = 1 & f (t) = cos bt so L {t cos bt} = (- 1)^1 d/ds (s/s^2 + b^2) = -d/ds (s/s^2 + b^2) = -[(s^2 + b^2) 1/(s^2 + b^2)^2 - s (2s)] therefore L {t cos bt} = s^2 - b^2/(s^2 + b^2)^2 (ii) now L {t^2 cos bt} = ? here n = 2, f (t) = cos bt so , L {t^2 cos bt} = (- 1)^2 d^2/ds^2 (s/s^2 + b^2) = d/ds [d/ds (s/s^2 + b^2)] = d/ds (b^2 - s^2/(s^2 + b^2)^2) (because using (i)) = (s^2 + b^2)^2 (- 2s) - (b^2 - s^2) 2 (s^2 + b^2)2s/(s^2 + b^2)^4 = (s^2 + b^2) [(s^2 + b^2) (- 2s) - 4s (b^2 - s^2)]/(s^2 + b^2)^4 =