Page 17 - 4549
P. 17

where  f   –  continuous  function  and  terms  are  executed
                                 a    ; 0  b    0    .  Indeed,  then  we  do  replacement,
                                                                   t    a
                                                   
                                                                
                                 ax   by   c   t ,    t   a   y b ,  y     and  equation  (1.20)
                                                                     b
                                                   
                                 acquires  a  kind  t   a   f    t ,  that  is  get  equation  with  the
                                 separated variables.   

                                     Example  1.4  To  find  the  common  decision  of  equation
                                   
                                  y   sin x   . y
                                                                                 dy    dt
                                        We will do substitution  x   y   t , then 1     and
                                                                                 dx    dx
                                 the  given  equation  will  be  written    down  in  a  kind
                                  dt
                                       sin t  1.
                                  dx
                                     We will consider two cases.

                                     1)   Lets    sin t  1   . 0 Then   we   separate   variables
                                        dt                              dt
                                                               
                                 dx            0  and integrate  dx        .C
                                      sin t  1                       sin t  1

                                     We will find more difficult integral separately by universal
                                 trigonometric substitution:

                                                       t          2zdz
                                                z   tg  ,  sin t     ,
                                          dt          2          1 z 2            2dz
                                                                                           
                                       sin t  1      2dz                        2   2z     
                                                dt        , t    2arctg  z  1 z    2    1
                                                     1 z  2                        1   z  
                                              dz            dz        2              2
                                        2            2              C 2          C 2 .
                                          z  2    2 z  1   z   1  2  z    1    t
                                                                                  tg   1
                                                                                    2


                                                               15
   12   13   14   15   16   17   18   19   20   21   22