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P. 97

 2    2   2   0, and     1   i .
                                                            2 , 1

                                                                   ~    x
                                     Common decision of LHDE is  y    e  (C 1  cos x   C  2  sin  ) x .
                                     We  search  the  partial  decision  LNDE  in  a  kind
                                         2
                                  y *  Ax   Bx   C .
                                                         2
                                 Right   part    f  (x )   x    corresponds   to   special   type
                                                x
                                               
                                  f ( x)   Q ( x) e  of right part and   s  , 2    0  are not the solves
                                          s
                                 of characteristic equation.
                                     We  shall  find  y  , * y  *   and  put  y *, y  , * y  *   in  given
                                 LNDE.
                                     How  agreed  already,  for  comfort  of  calculations  we  will
                                 write expressions for the function of  *y  and its derivative in
                                 separate  lines  and  to  the  left  after  a  vertical  hyphen  to  write
                                 down the proper coefficients, with which they enter to left part
                                 of given LNDE. Executing the increase of expressions on these
                                 coefficients and subsequent addition, we will get:

                                                   2
                                         2  y*   Ax   Bx   C
                                                
                                          2  y*   2 Ax   B
                                                    
                                         1      y*   2 A
                                                           2
                                      y  *    2y  *    2y *   2Ax   x  2 ( B   4A )   2 ( C   2B   2A )   x 2


                                     We equate coefficients at identical degrees of  x :
                                      x 2     2 A  1
                                      x 1  2 B  4 A  0

                                      x  0  2 C  2 B  2 A  0
                                                      1           1
                                     From where  A    , B    , 1 C    .
                                                      2           2

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