Differential equations: $f(x,y) dx + g(x,y) dy = 0$

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$$(2xy + 3y^2)dx + (x^2 + 6xy - 2y)dy = 0$$
$$y(1) = -1/2$$



How do you solve this? I have just started learning Differential equations and I have some trouble.



Is this equivalent with this?




$$2y + (6x - 2) = 0$$
$$y(1) = -1/2$$





differential-equations




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migrated from mathematica.stackexchange.com Dec 12 '12 at 22:42


This question came from our site for users of Wolfram Mathematica.










  • 1




    See exact equations.
    – Artem
    Dec 12 '12 at 22:45






  • 1




    Notice that $(2xy+3y^2)dx+(x^2+6xy-2y)dy$ is the divergent of $F(x,y)=x^2y+3xy^2-y^2+c$, c is a constant.
    – Marra
    Dec 12 '12 at 22:52











  • @GustavoMarra I think you mean it is $dbf r bullet nabla F$.
    – Eric Angle
    Dec 12 '12 at 23:14










  • ooops, that is correct, sorry!
    – Marra
    Dec 12 '12 at 23:16






  • 1




    See here.
    – Mhenni Benghorbal
    Dec 13 '12 at 0:31














up vote
1
down vote

favorite












$$(2xy + 3y^2)dx + (x^2 + 6xy - 2y)dy = 0$$
$$y(1) = -1/2$$



How do you solve this? I have just started learning Differential equations and I have some trouble.



Is this equivalent with this?




$$2y + (6x - 2) = 0$$
$$y(1) = -1/2$$





differential-equations




share|cite|improve this question














migrated from mathematica.stackexchange.com Dec 12 '12 at 22:42


This question came from our site for users of Wolfram Mathematica.










  • 1




    See exact equations.
    – Artem
    Dec 12 '12 at 22:45






  • 1




    Notice that $(2xy+3y^2)dx+(x^2+6xy-2y)dy$ is the divergent of $F(x,y)=x^2y+3xy^2-y^2+c$, c is a constant.
    – Marra
    Dec 12 '12 at 22:52











  • @GustavoMarra I think you mean it is $dbf r bullet nabla F$.
    – Eric Angle
    Dec 12 '12 at 23:14










  • ooops, that is correct, sorry!
    – Marra
    Dec 12 '12 at 23:16






  • 1




    See here.
    – Mhenni Benghorbal
    Dec 13 '12 at 0:31












up vote
1
down vote

favorite









up vote
1
down vote

favorite











$$(2xy + 3y^2)dx + (x^2 + 6xy - 2y)dy = 0$$
$$y(1) = -1/2$$



How do you solve this? I have just started learning Differential equations and I have some trouble.



Is this equivalent with this?




$$2y + (6x - 2) = 0$$
$$y(1) = -1/2$$





differential-equations




share|cite|improve this question














$$(2xy + 3y^2)dx + (x^2 + 6xy - 2y)dy = 0$$
$$y(1) = -1/2$$



How do you solve this? I have just started learning Differential equations and I have some trouble.



Is this equivalent with this?




$$2y + (6x - 2) = 0$$
$$y(1) = -1/2$$






differential-equations





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 at 8:00









Abcd

2,5481926




2,5481926










asked Dec 12 '12 at 21:55









Grimmer Wolfgang

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82




migrated from mathematica.stackexchange.com Dec 12 '12 at 22:42


This question came from our site for users of Wolfram Mathematica.






migrated from mathematica.stackexchange.com Dec 12 '12 at 22:42


This question came from our site for users of Wolfram Mathematica.









  • 1




    See exact equations.
    – Artem
    Dec 12 '12 at 22:45






  • 1




    Notice that $(2xy+3y^2)dx+(x^2+6xy-2y)dy$ is the divergent of $F(x,y)=x^2y+3xy^2-y^2+c$, c is a constant.
    – Marra
    Dec 12 '12 at 22:52











  • @GustavoMarra I think you mean it is $dbf r bullet nabla F$.
    – Eric Angle
    Dec 12 '12 at 23:14










  • ooops, that is correct, sorry!
    – Marra
    Dec 12 '12 at 23:16






  • 1




    See here.
    – Mhenni Benghorbal
    Dec 13 '12 at 0:31












  • 1




    See exact equations.
    – Artem
    Dec 12 '12 at 22:45






  • 1




    Notice that $(2xy+3y^2)dx+(x^2+6xy-2y)dy$ is the divergent of $F(x,y)=x^2y+3xy^2-y^2+c$, c is a constant.
    – Marra
    Dec 12 '12 at 22:52











  • @GustavoMarra I think you mean it is $dbf r bullet nabla F$.
    – Eric Angle
    Dec 12 '12 at 23:14










  • ooops, that is correct, sorry!
    – Marra
    Dec 12 '12 at 23:16






  • 1




    See here.
    – Mhenni Benghorbal
    Dec 13 '12 at 0:31







1




1




See exact equations.
– Artem
Dec 12 '12 at 22:45




See exact equations.
– Artem
Dec 12 '12 at 22:45




1




1




Notice that $(2xy+3y^2)dx+(x^2+6xy-2y)dy$ is the divergent of $F(x,y)=x^2y+3xy^2-y^2+c$, c is a constant.
– Marra
Dec 12 '12 at 22:52





Notice that $(2xy+3y^2)dx+(x^2+6xy-2y)dy$ is the divergent of $F(x,y)=x^2y+3xy^2-y^2+c$, c is a constant.
– Marra
Dec 12 '12 at 22:52













@GustavoMarra I think you mean it is $dbf r bullet nabla F$.
– Eric Angle
Dec 12 '12 at 23:14




@GustavoMarra I think you mean it is $dbf r bullet nabla F$.
– Eric Angle
Dec 12 '12 at 23:14












ooops, that is correct, sorry!
– Marra
Dec 12 '12 at 23:16




ooops, that is correct, sorry!
– Marra
Dec 12 '12 at 23:16




1




1




See here.
– Mhenni Benghorbal
Dec 13 '12 at 0:31




See here.
– Mhenni Benghorbal
Dec 13 '12 at 0:31










1 Answer
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You can rewrite your equation as
$$
df = fracpartial fpartial x dx + fracpartial fpartial y dy = 0,
$$
where $df$ is the total differential of a function $fleft(x,yright)$ that you should determine. Then $df = 0$ implies $fleft(x,yright)$ is a constant. Determine this constant with the condition on $yleft(1right)$.






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    up vote
    4
    down vote



    accepted










    You can rewrite your equation as
    $$
    df = fracpartial fpartial x dx + fracpartial fpartial y dy = 0,
    $$
    where $df$ is the total differential of a function $fleft(x,yright)$ that you should determine. Then $df = 0$ implies $fleft(x,yright)$ is a constant. Determine this constant with the condition on $yleft(1right)$.






    share|cite|improve this answer
























      up vote
      4
      down vote



      accepted










      You can rewrite your equation as
      $$
      df = fracpartial fpartial x dx + fracpartial fpartial y dy = 0,
      $$
      where $df$ is the total differential of a function $fleft(x,yright)$ that you should determine. Then $df = 0$ implies $fleft(x,yright)$ is a constant. Determine this constant with the condition on $yleft(1right)$.






      share|cite|improve this answer






















        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        You can rewrite your equation as
        $$
        df = fracpartial fpartial x dx + fracpartial fpartial y dy = 0,
        $$
        where $df$ is the total differential of a function $fleft(x,yright)$ that you should determine. Then $df = 0$ implies $fleft(x,yright)$ is a constant. Determine this constant with the condition on $yleft(1right)$.






        share|cite|improve this answer












        You can rewrite your equation as
        $$
        df = fracpartial fpartial x dx + fracpartial fpartial y dy = 0,
        $$
        where $df$ is the total differential of a function $fleft(x,yright)$ that you should determine. Then $df = 0$ implies $fleft(x,yright)$ is a constant. Determine this constant with the condition on $yleft(1right)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '12 at 22:51









        Eric Angle

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