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3
                                               4
                                        *
                                 0     y   Ax   Bx   Cx 2  ,
                                                  3
                                 0     (y    4Ax   3Bx   2Cx ,
                                                         2
                                         *
                                          ) 
                                 -3    (y   12Ax   6Bx   2C ,
                                                   2
                                         *
                                          ) 
                                 0     (y    24Ax   6B ,
                                         *
                                          ) 
                                 1     (y  * IV    24A ,
                                          )
                                           
                                                      2
                                  y  *IV    3y  *     36Ax  18Bx   6C   24A   9x  2 .

                                     Equating coefficients at the identical degrees х in left and
                                 right  parts  of  the  last  identity,  we  have  the  system  of  linear
                                 algebraic Equations for determination А, В, С:

                                  2
                                 х   -36А=9,
                                 х 1   -18В=0,
                                  0
                                 х   -6С+24А=0,
                                                                       1  2
                                                                   2
                                                              *
                                 So  А=-1/4, B=0, C=-1. So,  y   x  ( x    ) 1 . Consequently,
                                                                       4
                                           1  2
                                   *
                                        2
                                  y   x  ( x    ) 1  is partial decision of given LNDE.
                                           4

                                     A  function  is  the  common  decision  of  equation
                                          *
                                                                           4
                                  y   ~  y   c   c  x   c  e  3x    c  e   3x    1  x   x 2 . 
                                     y 
                                               1   2    3       4
                                                                        4

                                     Example 6.3 To find partial decision of equation
                                      y    2y    y   4 (sin x   cos  ) x ,  which  satisfies  initial
                                                 
                                 conditions  (y  ) 0   , 2  ) 0 (  y    , 2  ) 0 (   y     1.

                                       We have the LNDE third order with constant coefficients
                                 with right part of the special kind. Proper to him LHDE:

                                             y  2  y     y    0.

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