Page 108 - Math Grade 12 (Advance)
P. 108

PROEUS DARA




                3.  ~m.m :tMUJfUtrfili  y' + ay  =  p(x) mtfifitrtttmnrruG88tta
                                                                      lf   lf
                         cr         ~   e»   u   ,  e1           .....
                2fTIU'JUU1:  tR''i:flSti515fn~~ttHbi:ti5Jlli y'+2y = 3e--'"'   (E)  '1

                                                       1
                                          1
                fl.  H'iGt~tl5!H9ils tU~fn~ )' + 2y = 0   (E )  '1
                        "'   u
                ~- ffi,'jHS~'i5B f  Rrulshtflru  f(x)  = e- xg(x)  '1  R'MSl  f'(x)  dlHSR''i§~ns
                                                     2
                         I                                                     I
                                          1
                  g(x) Sb  g (x)  '1  R'MSl  g (X)  ttl  f(x)  dlutiruls  (E)  '1
                            1
                R'.  9lFIJH1  g(x)  JG  f(x)  t~15j2J  f(x)  dlutirulstU~fn~ (E)  '1
                orfJw  :
                  fill
                                ""
                        ""  "
                                      1
                n.  ~ nut'i5liHS tU'i5 m~ Y + 2y  =  o   (E')
                        "'
                         d)'
                         - = -2y  il2J  dy  = -2dx
                         dx              y
                                      lnjyj  =  -2x+c
                                        y  = ±e   2x+c

                          -2x
                   y  = Ae
                                    1
                2.  R'MSl  f'(x)  Sb  g (x)
                                 -2x
                         f(x)  = e   g(x)
                          ,         -2x      -2x  1
                         f  (x)  = - 2e   g(x) + e   g  (x)




                       1               -3x
                      f  (x) + 2f(x) = 3e
                         -2x      -2 r  ,    -2x        -3 r
                      -2e   g(x) + e  · g  (x) + 2e   g(x)  = 3e  ·

                                          e  -2x  '(  )   J  -3x
                                             g  x  =  e
                                              3e-Jx    -x
                                       1
                                      g (x)  = --=2:;  = 3e   '1
                                              e
                          1
                R'.  ttflru  g (X)  =  3e-x  tSlg  g(x)  =  J 3e-xdx =  -3e-x + c
                                                        1
                   ttflru  f(x)  dlm~ruls (E)  mrurun  g (x)  = 3e-x  tSlg
                                   G\J
                          -2x
                  f(x)  = e   g(x)








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