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




          This will be the case when all the roots, m , m , m ,... m  of the auxiliary equation are real, distinct  Notes
                                           1  2  3   n
          and different.
          Auxiliary Equation having Equal Roots

          If  the auxiliary equation has two equal roots, say m  and m , the solution of the given equation
                                                    1     2
                      n
                     d y   d n  1 y
                   P     P       ... P y  = 0
                    0  n  1  n  1    n
                     dx    dx
          will be                     y = (c   c  )e  m 1 x  c e m 2 x  ... c e  m n x
                                            1  2      3        n
          or                          y = ce m 1 x  c e m 2 x  ... c e m n x
                                                 3        n
          where                   c  + c  = c.
                                   1  2
          This is not the general solution of (i), because it contains (n   1) arbitrary constants while the
          order of the equation is n. To obtain the general solution of (i) in this case, we proceed as follows:

                                         dy     2
                                                                                 2
          Consider  the  repeated  factor  as   m 1  y  0. This  can  be  written  as  (D m 1 ) y  0,
                                         dx
                    d
          where D =   .
                   dx
          Put                  (D   m ) y =  ;
                                    1
          then                 (D   m )   = 0.
                                    1
                                    d
          Therefore                     = m 1
                                    dx
                                    d
          or                            = m  dx
                                            1

          Integrating, we have log    = m  x
                               c    1
                               2

          Hence                         = c e m 1 x .
                                            2
          or                   (D   m ) y = c e m 1 x
                                    1       2
                                dy
          or                       m y = c e m 1 x
                                     1
                                            2
                                dx
          This is a linear differential equation and we will have
                                  ye  m 1 x  = c 1  c e m 1 x . e  m 1 x  dx
                                                2
                                        = c 1  c x
                                               2
                                      y = (c 1  c  2  ) x e  m 1 x .







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