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Differential and Integral Equation




                    Notes          Thus y(x) is given by the integral equation
                                                                      1
                                                                              u
                                                                                     x
                                                                            y
                                                                         u
                                                                          x
                                                            y(x) =     K ( , ) ( )du  ( )                  ...(7)
                                                                      0
                                   Self Assessment
                                   2.  Express
                                        2y  ( ) 3 ( ) 2 ( )  4e  t  2cos t with  initial  conditions  y (0) 4, (0)  1,   into
                                                      y
                                           x
                                               y
                                                  x
                                                                                                  y
                                                       x
                                       integral equation.
                                       (Hint: Integrate the differential equation twice and use initial conditions.)
                                   24.4 Relation between Integral Equations and Algebraic System of
                                       Linear Equations
                                   Consider the general linear Fredholm integral of the second kind for a function  (x) of the type

                                                       1
                                                                     x
                                                  ( )   K ( , )dy =  f ( )  (0  x  1)                      ...(1)
                                                  x
                                                           y
                                                          x
                                                       0
                                   and the linear Fredholm equation of the first kind is given by
                                                   1
                                                                     x
                                                           y
                                                        y
                                                      x
                                                    K ( , ) ( )dy  =  f ( )  (0  x  1)                     ...(2)
                                                   0
                                   The problem of solving  (1) and  (2) can be considered as a generalization of  the problem  of
                                   solving a set of n linear algebraic equations in n unknown:
                                                          n
                                                            a x  = b .  (r = 1, 2, ...n)                   ...(3)
                                                            rs s
                                                                    r
                                                         s  1
                                   For this purpose we divide the interval (0  x  1) into n segments and define
                                                          K(x, y) = K    (r, s = 1, 2, ... n)
                                                                    rs
                                   and                       f(x) = f
                                                                   r
                                   Here, x, y are divided into strips as                                   ...(4)

                                                            r  1       r
                                                                 < x        (r = 1, 2, ... n)
                                                             n        n
                                                            s  1       s
                                                                 < y        (s = 1, 2, 3, ... n)
                                                             n        n
                                   Then equation (1) becomes

                                                                        n  s /n
                                                                                   y
                                                             x
                                                             ( ) =  f r        K rs  ( )dy   r  1  x  r          ...(5)
                                                                        s  1 (s  1)/n         n      n
                                   Equation (5) shows that if a function  ( )x exists it must be a step function, i.e.




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