Question

inverse laplace transform, using completing the square and t-shifting

$(se^{-3s})/(s^2+4s+5)$

the book gives the below answer but i am not following where -2 (red circle) comes from, shouldnt that be just s as +2 -2=0?

Answer 1

s.e-35 5² +45+ S 2 a=3 8+45+5 stustS Apply invosse transform dale: if L' {f(s)} = f(x) then L1 {e-as f(s)} = H (t-a) f ( t - a) where (A) is heasiside step function: for seg ses Se 35 $60-mes, az3 L' (austs) = 1" ( statusas) - Contacts) -La (muns) sausis) = L ( thosht) - 24" (stusvesi) - L1 (3+234 + 1) - 2 21 (3+25+1) Apply inverse transform wale: It [ {f(d} = f(t) then [ {f(s-a)} = eat f(x) Therefore [ { 2 } = x 2t [l (8) [((start) - & 2* i ( 3 ) Use inverse Laplace transform table L (3 tab) = (as (at), L (ata) - din (at) so, II (str) = coset, L' ( 7 ) = sin(t)

Put these values we get & é et coolt - 2 é at sinkt) L( Sess = f-3) H(+-3)= X(+-5) [63(4-3)-2 sin(4-3) ] H(+-3) L stusts)