Page 139 - Schaum's Outline of Differential Equations
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122                   SECOND-ORDER  LINEAR  DIFFERENTIAL  EQUATIONS              [CHAP.  14




               This  problem  is, therefore,  an example  of forced  undamped  motion. The  solution to the  associated  homogeneous
               equation  is
                                            x h = GI  cos 4t + c 2 sin 4t

               A particular solution is found by the method  of undetermined coefficients (the modification described  in Chapter  11
               is necessary  here):  x  =-\ cos 4t. The solution to (1) is then



               Applying the initial conditions,  x(0) = -  y  and i(0) = 0, we obtain




                  Note  that  \x\  —>  °° as  t  —>  °°. This  phenomenon  is  called pure resonance.  It  is  due  to  the  forcing function  F(t)
               having the  same  circular frequency  as that of the associated  free  undamped  system.


         14.11.  Write the steady-state motion found in Problem  14.9 in the form  specified by Eq.  (14.13).
                  The  steady-state  displacement  is given by  (2) of Problem  14.9 as



               Its  circular frequency is  to =  1. Here






               and

               The coefficient of the cosine  term in the steady-state  displacement  is negative,  so k=  1, and Eq. (14.13)  becomes




         14.12.  An RCL circuit connected  in series has R = 180 ohms, C = 1/280 farad, L = 20 henries, and an applied
               voltage E(i)  = 10 sin t. Assuming no initial charge on the capacitor, but an initial current of  1 ampere at
               t = 0 when the voltage is first  applied, find  the subsequent charge on the capacitor.
                  Substituting the  given quantities into Eq. (14.5),  we  obtain



               This  equation  is  identical  in  form  to  (1)  of  Problem  14.9; hence,  the  solution  must  be  identical  in  form  to  the
               solution of that equation.  Thus,



               Applying the initial conditions  q(0)  = 0 and  q(0)  = 1, we obtain  Cj =  110/500 and c 2 = -101/500.  Hence,




                  As in Problem  14.9, the solution is the sum of transient and steady-state  terms.

                                                                2
         14.13.  An RCL circuit connected in series has R = 10 ohms, C = 10  farad,  L = jhenry,  and an applied voltage
               E = 12 volts. Assuming no initial current and no initial charge at t = 0 when the voltage is first  applied,
               find  the subsequent current in the system.
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