Page 88 - 4549
P. 88

b) We search the partial decision of heterogeneous Equation
                                 in a kind:
                                                        x
                                       *
                                      y   C  ( )x   C  ( )x e ,
                                            1      2

                                 and functions  C  ( )x   and   C  ( )x  we search from the system of
                                                1            1
                                 equations
                                                                x
                                                C ( ) 1x    C ( )x e   0
                                                 1        2
                                                                x    1
                                                C 1 ( ) 0x    C 2  ( )x e   x  ,
                                                                   1 e
                                 from where
                                       C         1
                                        2 ( )x   e x (1 e  x  )
                                      
                                      
                                       C ( )x    1  .
                                         1    1 e  x

                                     Then
                                                           dx
                                               C  ( )x         x   ln(1 e  x  ) ,
                                                1             x
                                                         1 e
                                                  C  ( )x   e  x    x   ln(1 e  x  ) .
                                                   2

                                     Constant  integration  we  scorn  (we  search  the  partial
                                 decision).
                                     Consequently:
                                       *
                                      y     ln(1 e  x ) 1 xe   x    e x  ln(1 e  x  ).
                                            x

                                     c) the common decision of equation is:
                                           *
                                                                                               x
                                      y   y     x  ln(1 e  ) x  ) 1 xe   x    e  x  ln(1 e  x  ) C    C e 
                                              y   
                                                                                        1    2
                                        C   1 x   e x  (C   x ) (1 e    x  ) ln(1 e  x ) .
                                         1             2



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