1-s-2-2s-5-2-find-inverse-laplace-

Question Number 129268 by BHOOPENDRA last updated on 14/Jan/21

\frac{1}{{(s^{2} + 2 s + 5)}^{2}} f i n d i n v e r s e l a p l a c e

Answered by mnjuly1970 last updated on 14/Jan/21

F (s) = \frac{1}{{({(s + 1)}^{2} + 2^{2})}^{2}}

w e k n o w : L^{- 1} (\frac{1}{{(s + 1)}^{2} + 2^{2}}) = \frac{1}{2} e^{- t} s i n (2 t)

∴ L^{- 1} [{(\frac{1}{{(s + 1)}^{2} + 2^{2}})}^{2}] = \frac{1}{4} \int_{0}^{t} (e^{- λ} s i n (2 λ)) . (e^{- (t - λ)} s i n (2 (t - λ)) d λ

= \frac{1}{4} \int_{0}^{t} e^{- t} s i n (2 λ) . s i n (2 t - 2 λ) d λ

= \frac{1}{8} \int_{0}^{t} e^{- t} [c o s (2 t - 2 λ - 2 λ) - c o s (2 t - 2 λ + 2 λ)] d λ

= \frac{1}{8} {\int_{0}^{t} e^{- t} c o s (4 λ - 2 t) d λ - \int_{0}^{t} e^{- t} c o s (2 t) d λ

= \dots

Commented by BHOOPENDRA last updated on 14/Jan/21

w h a t λ r e p r e s e n t h e r e ?