Page 30 - Pra U STPM 2022 Penggal 2 - Mathematics
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Mathematics Semester 2  STPM  Chapter 4 Differential Equations

                  Example 12


                                               dy   x + y + 5                dv   2(v + 1)
               If v = x + y, show that the equation     =   x + y – 3  can be reduced to    =   v – 3   , and hence solve
               the equation.                   dx                            dx
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                                     dy   x + y + 5
               Solution:                =
                                     dx   x + y – 3
                                    Substituting v = x + y or y = v – x,
                                                     d           v + 5
                                                        (v – x)  =
                                                     dx          v – 3
                                                         dv   – 1  =   v + 5
                                                        dx       v – 3
                                                            dv    =   v + 5  + 1
                                                           dx    v – 3
                                                                =   2(v + 1)
                                                                   v – 3
                                                                 ∫
                                                     ∫   v – 3  dv  = 2    dx
                                                      v + 1
                                                                 ∫
                                                       4
                                                ∫1    v + 1 2  dv  = 2  dx
                                                   1 –
                                                v – 4 ln | v + 1 |  = 2x + C
                                          x + y – 4 ln | x + y + 1 |  = 2x + C
                                                                                         1
                                                 ln | x + y + 1 |  =   1   (y – x) + D   D = –   C
                                                                 4                       4
                                                                   1
                                                      x + y + 1  = Ae —(y – x)   A = e D
                                                                   4
                                                         1
                                             x + y + 1 – Ae —(y – x)   = 0
                                                         4
          4
                  Example 13


                                                             3
               By writing u =   1  , reduce the equation 2  dy   –   y   = y  to   du  +   u  = –1.
                             y 2                  dx    x        dx    x
               Hence solve the differential equation.
               Solution:            Differentiating u =   1   with respect to x
                                                    y 2
                                    du    d         2  dy
                                              –2
                                        =     y  = –
                                    dx    dy       y 3  dx
                                                        dy    y
                                                                   3
                                    Dividing the equation 2    –    = y  with y .
                                                                         3
                                                        dx    x
                                                    2     dy   –   1    = 1
                                                    y 3  dx  xy 2
                                                        du    u
                                                    \ –     –     = 1
                                                        dx    x
                                                        du    u
                                                     \      +     = –1
                                                        dx    x
                                                            1   dx
                                                         ∫
                                                               ln x
                                    The integrating factor is e     x   = e  = x.
               138






         04 STPM Math(T) T2.indd   138                                                                 28/01/2022   5:44 PM
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