Page 29 - Pra U STPM 2022 Penggal 2 - Mathematics
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Mathematics Semester 2  STPM  Chapter 4 Differential Equations

                                         dy          y
                                      So     + x y = –   reduces to
                                               2
                                         dx          x
                                       1 dv     v     2 v     1  v
                                                              x 1 x 2
                                      1 x    dx  –   x 22  + x 1 x 2   = –
                                           1 dv   –   v   + vx  = –  v

                                           x  dx   x 2        x 2
                                                1 dv   + vx  = 0

                                                x  dx

                                      \   dv   = –vx  which is the form where the variables are separable.
                                                 2
                                         dx
                                      Separating the variables and integrating with respect to x.
                                                 ∫
                                                    2
                                      \     dv   = –    x  dx
                                        ∫ v
                                      \ ln v = –  x 3 2Sdn Bhd. All Rights Reserved.
                                                   + A
                                                3
                                      Substituting v = yx
                                      \ ln yx = –  x 3  + A is the general solution.
                                                 3


                     Example 11


                                                       2
                                                   3
                                            dy    y  – x y
                  Solve the differential equation     =    .
                                            dx    x  + xy 2
                                                   3
                Penerbitan Pelangi
                                                                                                              4
                                              3
                                       dy    y  – x y
                  Solution:               =
                                              3
                                       dx    x  + xy 2
                                      Substituting  y  = vx,
                                                         3
                                             v + x  dv    =   v  – v      dy   = v + x  dv
                                                  dx    1 + v 2       dx       dx
                                                         3
                                                x  dv    =   v  – v   – v
                                                  dx    1 + v 2
                                                x  dv    = –   2v
                                                  dx     1 + v 2

                                                         ∫
                                            ∫   1 + v 2  dv =    –   dx
                                                            2
                                                v
                                                            x
                                           ln | v | +   1  v  = –2 ln | x | + C    C is an arbitrary constant.
                                                      2
                                                   2
                                              2
                                         ln | vx  | +   1  v  = C
                                                      2
                                                   2
                                                ln | yx | = –   y 2   + C
                                                          2x 2
                                                            y 2
                                                           –—
                                                    yx = Ae 2x  where A = e C

                                                             2
                                                                                                       137



         04 STPM Math(T) T2.indd   137                                                                 28/01/2022   5:44 PM
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