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Differential and Integral Equation




                    Notes
                                             u v   v u      u v   v u    u v   v u
                                   or     p              q                           0
                                             z y   z y      x z   x z    x y   x y
                                                   u
                                                     v
                                             v
                                            u
                                           ( , )   ( , )  ( , )
                                                          u
                                                            v
                                         p       q                                                        ...(13)
                                             z
                                                            y
                                            y
                                           ( , )   ( , )  ( , )
                                                          x
                                                   z
                                                     x
                                   which is partial differential equation of the type (11). It should be noted that equation (13) is a
                                   linear partial differential equation i.e. the powers of p and q are both unity. Whereas the partial
                                   differentiation equation (11) need not be linear. To see that consider the equation
                                                            2
                                                    2
                                               (x   a)  + (y   b)  + z 2  = 1                             ...(14)
                                   Differentiating (14) with respect to x and y separately, we have
                                                          2(x   a) + 2zp = 0,  2(y   b) + 2zq = 0
                                   Substituting the values of (x   a) and (y   b) in equation (14) we have
                                                       2 2
                                                                    2
                                                                      2
                                                                         2
                                                            2
                                                  2 2
                                                  z p  + z q  + z  = 1 or z (p  + q  + 1) = 1.            ...(15)
                                   So powers of p and q are not one.
                                          Example 2: Eliminate the constants a and b from
                                                                         2
                                                              2z = (ax + y)  + b                           ...(1)
                                   Solution: Differentiate with respect to x we have
                                                              z
                                                            2    = 2p  2 (ax y )
                                                                        a
                                                              x
                                   Differentiating (1) with respect to y we have
                                                              z
                                                            2    = 2q  2(ax y )
                                                              y
                                   or                         p = a(ax + y)                                ...(2)
                                                              q = (ax + y)                                 ...(3)
                                                          px + qy = ax(ax + y) + y(ax + y)

                                                                         2
                                                                 = (ax + y)  = q 2
                                   or                     px + qy = q  2
                                   is the answer.


                                          Example 3: Eliminate the arbitrary function f from the equation

                                                                      xy
                                                               z =  f                                      ...(4)
                                                                      z
                                   Differentiating with respect to x and y respectively we have

                                                           z          y  xy
                                                              p =   f      p                              ...(15)
                                                           x          z  z 2





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