Tom N. answered 08/29/19
Strong proficiency in elementary and advanced mathematics
C(s)(s(s+2)(s+3) = R(s)(s+1) C(s)(s(s2 +5s +6)) = R(s)s +R(s) so now C(s)s3 +C(s)5s2 +C(s)6s = R(s)s +R(s). So the differential equation would become d3c(t)/dt3 + 5d2c(t)/dt2 +6dc(t)/dt =dr(t)/dt +r(t) all initial conditions are considered to be 0. Hence c'''(t) + 5c''(t) + 6c'(t) =r'(t) + r(t).