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Differential and Integral Equation




                    Notes          which on substitution becomes

                                            Z  m   Z  n
                                                        1.
                                            X      X
                                          The complete integral is
                                          Z = aX + bY + c

                                   where
                                           m
                                              n
                                          a  + b  = 1
                                              m n    (m n l )/(m n )
                                   and    Z         .z
                                             m n a
                                              1   1
                                          X    (x  sin2 ).
                                                       x
                                             2    2
                                             1    1
                                          Y    (y  sin2 ).
                                                       y
                                             2    2

                                                                           2
                                          Example 4: Solve: (y   x) (qy   px) = (p   q) .
                                   Solution:
                                   Put x + y = X, xy = Y

                                              z   z  X    z  Y
                                          p        .       .
                                              x   X  x   Y   x
                                             z    z
                                                    y
                                              .1    . ;
                                             X    Y
                                              z   z  X   z  Y
                                          q        .       .
                                              y   X  y   Y   y
                                             z    z
                                              .1    . .
                                                    x
                                             X    Y
                                   The given equation by this substitution becomes

                                                  z    z      z     z
                                          (y x )     x    y      y    x
                                                  X    Y      X    Y

                                                              2
                                              z    z   z     z
                                                 y        x    .
                                             X     Y   X    Y
                                                    2            2
                                                  z            z
                                          (y x ) 2     (y x ) 2
                                                  X            Y
                                                   2
                                           z     z
                                   or
                                           X    Y




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