Solution of Nonhomogeneous D.E. using Variation of Parameters (continued)

 2 Use algebra to solve (1) and (2) namely                           u1’(x) sin 3x  +  u2’(x) cos 3x  =   0                 - - - - -    (1)                          3u1’(x) cos 3x  -  3u2’(x) sin 3x   =  2 sec  3x  - - - - - -    (2)     for   u1’(x)   and   u2’(x) .     The result is:                                u1’(x)   =  2/3 ,            u2’(x)   =   -2/3  tan 3x          Now  continuing            u1’(x)   =  2/3 ,            u2’(x)   =   -2/3  tan 3x     Integrate these expressions to find     u1(x)     and    u2(x) .  The result is:       u1(x)    =  2x/3  +  B1      and      u2(x)    =   (2/9) ln (cos 3x)  +  B2     Therefore the particular solution                              yp(x)   =   u1(x) y1(x)  +  u2(x) y2(x)     is                      yp(x)  =   (2x/3  + B1 ) sin 3x   +  (2/9) ln (cos 3x  + B2  ) cos 3x     and the general solution for y(x) is the sum of the complementary and particular solutions.  This is the final result (below).      y(x)  =  C1  sin 3x   +  C2  cos 3x   +  (2x/3) sin 3x   +   (2/9) ln (cos 3x) cos 3x\   (here C1  and  C2  are constants of integration)     where we have absorbed constants of integration  B1    and  B2   into   C1    and    C2