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CHAP.  71                       STATE SPACE ANALYSIS                                 415



            7.48.  Using the state variables method, solve the second-order linear differential equation

                                             y"(t) + 5y'(t) + 6y(t) =x(t)                    (7.127)
                  with the initial conditions y(0) = 2, yl(0) = 1, and  x(t) = e-'u(t) (Prob. 3.38).
                      Let the state variables qJt) and q,(t) be

                                             qdt) =Y(I)       ~2( =~'(t)
                                                                  )
                                                                 f
                  Then the state space representation of Eq. (7.127) is given by  [Eq. (7.19)l
                                                  q(t) = Aq(t) + bx(t)

                                                 ~tt) =cq(t)
                         A=[-:  -:I                                       q'ol  = [q2(0)] = [i]
                  with                        b=[y]        c=[l  O]              qdo)


                  Thus, by Eq. (7.65)




                  with d = 0. Now, from the result from Prob. 7.39




                                                         -:I   +e-''[-:  -:I)[:]
                  and         c~q(0) [I  ~](e-~'[-:
                                      =








                  Thus,















            7.49.  Consider the network shown in Fig. 7-20. The initial voltages across the capacitors C,
                  and  C, are  f  V  and  1  V,  respectively.  Using  the  state  variable  method,  find  the
                  voltages  across  these  capacitors  for  t > 0. Assume  that  R, = R, = R, = 1  0 and
                            1
                  C, =C2= F.
                     Let the state variables q,(t) and q2( t) be
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