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Basic Mathematics-II




                    Notes                        1          1
                                                                y
                                                         x
                                   or,             x 2   xy   y 2     . C
                                                 2          2
                                   or,           x 2   y 2   2xy  2   2  C  . 
                                                                y
                                                             x
                                          Example: Solve (ey + 1)cosx dx + ey sinx dy = 0.
                                   Solution:
                                   The given equation is (ey + 1) cos x dx + ey sin x dy = 0               ...(1)
                                   This is of the form Mdx + Ndy = 0
                                   where   M = (ey + 1) cos x,  N = ey sin x

                                      M   y       N   y
                                              x
                                        e  cos ,    e  cosx
                                     y           x
                                         M   N
                                   since      ,  the equation (1) is exact.
                                        y   x
                                   General solution is given by
                                        Mdx    (terms of N  independent of  )dy   C
                                                                   x
                                   y  constant


                                              y
                                           (e  1) cosxdx   0dy   C
                                   i.e.,
                                       y  constant
                                   or, (ey + 1) sin x = C is the general solution.

                                                                  / 
                                          Example: Solve  1 e  / x y  dx e x y   1    x    dy    0
                                                               
                                                                         y
                                   Solution:

                                                                 / 
                                                          / x y  x y     x
                                   The given equation is   1 e   dx e   1     dy    0               ...(1)
                                                              
                                                                        y
                                          This is of the form  Mdx + Ndy = 0, where
                                                      ~
                                   M = 1 + ex/y,  N = ex/y (1x/y)
                                      M    / x y    x     N  / x y    x  
                                         e     ,     e     
                                      y        y 2    x    y 2  

                                         M   N
                                   since      ,  the given equation is exact.
                                        y   x
                                   The general solution is given by

                                                                    x
                                        Mdx     (terms of N  independent of  )dy    C
                                   y  constant
                                                / x y                / x y
                                         (1+ e  )dx      . o dy   C    y e   C
                                                                x
                                     y  constant
                                   This is the required general solution.



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