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Unit 7: General Properties of Solutions of Linear Differential Equations of Order n




                                                                                                Notes
                      1
                              ,
          To evaluate    sinax  where  ( )f D  P D n  P D n  1  ...P  .
                       D
                     f ( )                 0     1        n
          Case I. When f(D) contains even powers of D
          Let                      ( f D 2  ) = P  (D 2 n  P  (D 2 n  1    P
                                                       )
                                               )
                                            0      1            . n
                                 2
                                            2
          We notice that        D  sin ax =  a  sin ax.
                                             2 2
                                 4
                                D  sin ax = ( a )  sin ax
                                             2 3
                                 6
                                D  sin ax = ( a )  sin ax
                                                        .................................................
                                                        .................................................
                                             2 n
                                2 n
                              (D )  sin ax = ( a )  sin ax
          Therefore            ( f D 2 )sin ax = P  (D 2n  P D 2n  2    P  )sin ax
                                            0      1         n
          or                   ( f D 2 )sin ax
                                        = P D 2n  sinax P D  2n  2  sin ax   P  sin ax
                                            0         1               n
                                        = P  ( a 2 n     ( a 2 n  1  sin ax  ... P  sin ax
                                                ) sin ax P
                                                            )
                                            0           1                 n
                                        =   ( f  a 2 )sin ax .
                                     1
          Operating on both sides with   2  ,  we have
                                    ( f D  )

                          1      2           1   ( f  a  2 )sin ax
                              ( f D  )sin ax =  2
                                             D
                          ( f D 2 )        f  ( )
                                                   1
                                               2
          or                      sin ax =  ( f  a  ).  2  sin ax .
                                                  ( f D  )
          Dividing both sides by  (f  a  2 ),  we have

                               1             1  sin ax .
                              ( f D 2  ) sin ax =  ( f  a 2 )
          Case II. When f(D) contains odd powers of D.

          Let it be put in the form  f  (D 2 ) Df  (D 2 );  then
                               1       2
                               1                 1
                                  sin ax  =            2  sin ax
                                D
                              f  ( )       f  1 (D  ) D f 2 (D  )
                                                  1
                                        =      2        2  sin ax
                                            ( f  a  ) D f 2 ( a  )

                                             1
                                        =        sin ax  say
                                           m nD



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