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Unit 24: Integral Equations and Algebraic System of Linear Equations




                                                                                                Notes
                                            (x  1)t  if t  x
          with                    K(x, t) =                                      ...(vii)
                                             ( x t  1) if t  x
          Using this in (i),

                           1
                              x
                       x
                                   t
                                 G
                                t
                     G ( ) a K ( , ) ( )dt = 0
                           0
                                            2
                                                           t
                                           d y         (x  1) , t  x
                                                 x
                                                   t
          where                    G(x) =    2  , K ( , )
                                                           x
                                           dx          (t  1) , t  x
          Method 2:
          Integrating (i) from 0 to x
                        x        x
                           t
                                   t
                         y  ( )dt a y ( )dt  = 0
                        0        0
                                 x
                             x
                         y  ( )  a y ( )dt  = 0
                                   t
                           t
                             0
                                 0
                                 t
                                    t
                       x
          or         y  ( ) y  (0) a y ( )dt = 0
                                 0
          Again  integrating,
                                       x
                           x        x
                                              t
                         t
                                  t
                                             y
                        y ( )  y  (0)[ ]  a  (x t ) ( )dt = 0
                           0        0
                                       0
                                       x
                                             y
          or          y ( ) y (0) y  (0)x a  (x t ) ( )dt = 0
                        x
                                              t
                                       0
                                       x
          or               y ( ) y  (0)x a  (x t ) ( )dt = 0                     ...(viii)
                             x
                                              t
                                             y
                                       0
          Putting x = 1, this gives
                                       1
                            y (1) y  (0) a  (1 t ) ( )dt = 0
                                             y
                                              t
                                       0
          or as y(1) = 0, we have
                                                       1
                                              y  (0) = a  (1 t ) ( )dt
                                                            y
                                                              t
                                                       0




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