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?
differential-equations
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up vote
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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?
differential-equations
add a comment |Â
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?
differential-equations
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
differential-equations
asked Sep 7 at 8:48
user251130
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