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we come to equation of the n-k -th order of relatively unknown
                                 function of z(x):

                                         ( zxF  ,  ,  ,  z  ,   z  ..., z  ( kn  )  )   , 0                                       (3.10)

                                  that is we bring an order down of equation on k units. If it is
                                 succeeded  to  find  the  common  decision  of  this  equation  in  a
                                 kind  z   (x ,c  ,c  ,...,c  ),  we will get differential equation of
                                               1  2    n k
                                 previous kind
                                          z   y (k )    (x ,c  ,c  ,...,c  ),                                   (3.11)
                                                       1  2    n k

                                 the decision of which is found k-times integration. In particular,
                                 if n=2, k=1 equation (3.11) –  first order.
                                     On  occasion  finding  the  decision  as  an  explicit  or  non-
                                 obvious function is difficult, but it is succeeded to find decision
                                 in a parametric form.
                                     We will consider examples, which illustrate the considered
                                 types of equations of higher order and methods of their solving.

                                     Example  3.3  To  find  the  common  decision  of  equation
                                  y     ctg  x      y    . 2

                                        The given  equation is equation of the considered  type,
                                 where n=3, k=2. We enter a new unknown function  z   y    and
                                 get linear equation of the first order  ctgz  x   z    , 2  which write
                                 down in a kind  z    z tg  x   2tg  . x

                                 Its common decision:

                                       
                                       tgxdx       tgxdx        ln  cos x    ln  cos x
                                 z   e    (   2tgx e  dx   c 1 )   e  2 (   tgx e  dx   c 1  )  
                                                 tgx                   sin  x  dx
                                      cos  x 2(     dx   c )   cos2  x     c cos  x  
                                                                                 1
                                                           1
                                                 cos x                  cos 2  x

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