... 2) -2 t = e u ( t − 2) + e-4t u ( t − 2) = e-4 e-2(t - 2) u ( t − 2) + e-8 e-4(t - 2) u ( t − 2) cosh( t ) = L [ e -4 e -2 (t -2 ) u ( t − 2)] = e -4 e -2 s ⋅ L [ e -2 u ( t )] e -( 2s+ 4) L [ e -4 ... 4) L [ e -4 e -2 (t -2 ) u ( t − 2)] = s+2 Similarly, L [ e -8 e - 4(t - 2) 4e u ( t − 2) ] = -( 2s+ 8) s+4 Therefore, e -( 2s+ 4) e -( 2s+8) e -( 2s+ 6) [ (4 e + e -2 ) s + (16 e + e -2 )] + = F(s)... 15, Solution 31 (a) F(s) = 10s A B C = + + (s + 1)(s + 2)(s + 3) s + s + s + - 10 = -5 - 20 B = F(s) (s + 2) s= -2 = = 20 -1 - 30 C = F(s) (s + 3) s= -3 = = -1 5 A = F(s) (s + 1) s= -1 = F(s) = -5