Differential equation $(ax+by+c)dx+(ex+fy+g)dy=0$

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On Wikipedia:Homogeneous differential equation, it is said that a first order differential equation of the form ($a$, $b$, $c$, $e$, $f$, $g$ are all constants)
$$displaystyle (ax+by+c)dx+(ex+fy+g)dy=0,$$

where $af not= be$ can be transformed into a homogeneous type.



What if $af = be$? Is there a general method for solving this kind of equations?










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    On Wikipedia:Homogeneous differential equation, it is said that a first order differential equation of the form ($a$, $b$, $c$, $e$, $f$, $g$ are all constants)
    $$displaystyle (ax+by+c)dx+(ex+fy+g)dy=0,$$

    where $af not= be$ can be transformed into a homogeneous type.



    What if $af = be$? Is there a general method for solving this kind of equations?










    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      On Wikipedia:Homogeneous differential equation, it is said that a first order differential equation of the form ($a$, $b$, $c$, $e$, $f$, $g$ are all constants)
      $$displaystyle (ax+by+c)dx+(ex+fy+g)dy=0,$$

      where $af not= be$ can be transformed into a homogeneous type.



      What if $af = be$? Is there a general method for solving this kind of equations?










      share|cite|improve this question













      On Wikipedia:Homogeneous differential equation, it is said that a first order differential equation of the form ($a$, $b$, $c$, $e$, $f$, $g$ are all constants)
      $$displaystyle (ax+by+c)dx+(ex+fy+g)dy=0,$$

      where $af not= be$ can be transformed into a homogeneous type.



      What if $af = be$? Is there a general method for solving this kind of equations?







      differential-equations






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      asked Sep 7 at 8:48









      user251130

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