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P. 94

We  make  characteristic  equation:   3    2 2      0   and
                                     , 0       1.
                                   1      2   3
                                                                   ~
                                                                                       x
                                     Common decision of LHDE  is  y   C    C (  2    C 3 x) e .
                                                                         1
                                     We  search     the  partial  decision  of    LNDR  as
                                  y*   Acos  x   Bsin  x,  because  right  part  of  the  equation  is
                                  f  (x )   4 (sin x   cos  ) x   (special  type  of  right  part  (6.17)):
                                  f ( x )  e  x  P ( x cos)   x   Q ( x sin)     x max(,  s ,  s )    s   and  as
                                             s 1           s 2               1  2
                                 if   s  , 0    , 0    1 thus a number    i   i  is not the root of
                                 characteristic equation.
                                     Farther, finding  * yy  ,  *  ,  y  *   we will put in given LNDE
                                 and, applying the method of indefinite coefficients (see previous
                                 Example 2.), will find coefficients:  A   B    2.
                                     Thus, the partial decision of LNDE will be

                                      y*   2  cos  x 2  sin  x.

                                     Common decision of LNDE is
                                                                  x
                                          y 
                                      y   ~  y*   C   C (    C  x) e   2 cos  x 2  sin  x.
                                                   1     2    3
                                     Then we will find:

                                                                x
                                             
                                            y   (C   C  1 (   x ))e   ( 2  sin x   cos  ) x ,
                                                  2    3
                                                                 x
                                               
                                            y   (C   C  2 (   x ))e   2 (cos x   sin  ) x .
                                                   2    3

                                     Taking       advantage      of      initial    conditions
                                     ) 0 ( y    , 2  ) 0 (  y    , 2  ) 0 (   y     1,  we  will  obtain  the  system  of
                                 equations
                                           C 1   C  2   2   ,2
                                          
                                              C 2   C 3   2   ,2
                                          
                                           C 2   2C 3   2     ,1

                                 from where C     , 1 C      , 1 C    1 .
                                              1      2       3
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