Page 46 - 4549
P. 46

0
                                     Let  us  t  ,  from    an  initial  condition    (2.39)    we  find
                                       С
                                 10      ,    that  is  C   1000 .  Therefore  the  law  of  change  of
                                      100
                                 amount of salt   x   in kilograms, that is in a reservoir, depending
                                 on the passed time  in minutes  (rice. 2.1б)  is set by a formula

                                                        1000
                                                             x   .                                         (2.40)
                                                       (10 t  ) 2

                                     We  will  notice  that  from  a  formula  (2.40),  knowing  the
                                 amount of salt, that remained in a reservoir (last it is easily to
                                 set, measuring the volume of reservoir and concentration of salt
                                 in him), it is possible to define the amount of time, which passed
                                 from  the beginning of process. On this  idea  the calculation of
                                 age of seas and oceans is founded.   

                                     Exercises:

                                     1.    For  each  of  the  following equations  to  set  argued  his
                                 type and method of decision :

                                   ) a  e y  (1 x  2 )dy   2 (1x   e y  )dx   0 ;
                                               y           y
                                 б ) (x   y  cos )dx   x  cos  dy   0 ;
                                               x           x
                                               3
                                                       3
                                         2
                                 в ) (3xy   2 )x dx   y dy   0 ;
                                 г ) cos x y  ' sin x y      1 ;
                                       2
                                 д ) ( y   ) x dx   2xydy   0 ;
                                          y    3  4
                                  ) е  ' y     x y  ;
                                          x
                                 є ) (1 y  2 )xdx   (1 x  2 )dy   0 ;
                                 ж  ) (x   y ) ' (y   x   ) y   0 .

                                     2.  To find the common decision of equation :
                                                      2
                                       2
                                 а ) (x   2xy )dx   (x   y  2 )dy   0
                                                               44
   41   42   43   44   45   46   47   48   49   50   51