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Differential and Integral Equation




                    Notes          Differentiating with respect to a and b,
                                                                           1
                                                              0 = x      . b  ,
                                                                          2 a

                                                                         a
                                                              0 = y       ,
                                                                      2 b

                                                               b          a
                                                                     x  and  y
                                                               a          b
                                   Eliminating a and b, the singular integral is
                                          xy = 1.


                                                                        2
                                                                           2
                                          Example 2: Solve z   px   qy = c (1 + p  + q ).
                                   Solution:
                                   The complete integral is
                                                                                2
                                                                                    2
                                                               z = ax + by + c (1 + a  + b )               ... (1)
                                   Differentiating with respect to a and b,
                                                                          ca
                                                              0 = x             ,                          ... (2)
                                                                       (1 a 2  b 2 )

                                                                          bc
                                                              0 = y             .                          ... (3)
                                                                       (1 a 2  b 2  )

                                                                   c  2 (a 2  b 2 )
                                                           2
                                                          x  + y 2  =  2  2  .
                                                                    1 a  b
                                                                       c  2 (a 2  b  2 )
                                                                    2
                                                           2
                                                        2
                                                       c    x    y 2  = c  2  2
                                                                       1 a   b
                                                                      c 2
                                                                 =     2  2  .
                                                                   1 a   b
                                                                       c 2
                                                           2
                                                        1 + a  + b 2  =  2  2  2  .
                                                                   c   x  y
                                   Putting in (2), (3),

                                                                     x  (1 a 2  b 2  )  x
                                                               a =                  2   2  2
                                                                          c        (c  x  y  )

                                                                         y
                                   and                        b =     2   2  2  .
                                                                     (c  x  y  )







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