How to find a special integrating factor which is not in form of $x^m y^n$ but is both function of $x$ and $y$?

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I was watching this youtube video.



Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?



suppose you have this differential equation $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$ how would you get the integrating factor of the above differential equation?



although it is $ycos y$




line-integrals




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  • I tried to make what you wrote more readable, though you might want to check. In particular, your differential equation does not look like an equation
    – Henry
    Aug 15 at 9:13










  • @Henry what's wrong in it ? Please tell me. I'll try to improve.
    – user187604
    Aug 15 at 9:15










  • @Henry anyway thanks for editing
    – user187604
    Aug 15 at 9:17










  • equations often have an equals sign
    – Henry
    Aug 15 at 9:23














up vote
0
down vote

favorite












I was watching this youtube video.



Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?



suppose you have this differential equation $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$ how would you get the integrating factor of the above differential equation?



although it is $ycos y$




line-integrals




share|cite|improve this question






















  • I tried to make what you wrote more readable, though you might want to check. In particular, your differential equation does not look like an equation
    – Henry
    Aug 15 at 9:13










  • @Henry what's wrong in it ? Please tell me. I'll try to improve.
    – user187604
    Aug 15 at 9:15










  • @Henry anyway thanks for editing
    – user187604
    Aug 15 at 9:17










  • equations often have an equals sign
    – Henry
    Aug 15 at 9:23












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was watching this youtube video.



Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?



suppose you have this differential equation $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$ how would you get the integrating factor of the above differential equation?



although it is $ycos y$




line-integrals




share|cite|improve this question














I was watching this youtube video.



Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?



suppose you have this differential equation $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$ how would you get the integrating factor of the above differential equation?



although it is $ycos y$





line-integrals





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 15 at 10:25









James

2,002619




2,002619










asked Aug 15 at 9:06









user187604

2286




2286











  • I tried to make what you wrote more readable, though you might want to check. In particular, your differential equation does not look like an equation
    – Henry
    Aug 15 at 9:13










  • @Henry what's wrong in it ? Please tell me. I'll try to improve.
    – user187604
    Aug 15 at 9:15










  • @Henry anyway thanks for editing
    – user187604
    Aug 15 at 9:17










  • equations often have an equals sign
    – Henry
    Aug 15 at 9:23
















  • I tried to make what you wrote more readable, though you might want to check. In particular, your differential equation does not look like an equation
    – Henry
    Aug 15 at 9:13










  • @Henry what's wrong in it ? Please tell me. I'll try to improve.
    – user187604
    Aug 15 at 9:15










  • @Henry anyway thanks for editing
    – user187604
    Aug 15 at 9:17










  • equations often have an equals sign
    – Henry
    Aug 15 at 9:23















I tried to make what you wrote more readable, though you might want to check. In particular, your differential equation does not look like an equation
– Henry
Aug 15 at 9:13




I tried to make what you wrote more readable, though you might want to check. In particular, your differential equation does not look like an equation
– Henry
Aug 15 at 9:13












@Henry what's wrong in it ? Please tell me. I'll try to improve.
– user187604
Aug 15 at 9:15




@Henry what's wrong in it ? Please tell me. I'll try to improve.
– user187604
Aug 15 at 9:15












@Henry anyway thanks for editing
– user187604
Aug 15 at 9:17




@Henry anyway thanks for editing
– user187604
Aug 15 at 9:17












equations often have an equals sign
– Henry
Aug 15 at 9:23




equations often have an equals sign
– Henry
Aug 15 at 9:23










1 Answer
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$$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$
The integrating factor is $ycos(y)$ . After reducing to the common denominator, this is easy to guess. If you cannot see it at first place, try various forms of integrating factors. When trying an integrating factor of the form $f(y)$ you find the function $f(y)$ by identification.



Checking :



$$ycos(y)(2xy^2 sec y + 2xfractan yy) dx + ycos(y)(3x^2y sec y +3y sec y +fracx^2y) dy = 0$$



$$(2xy^3 + 2xsin(y)) dx + (3x^2y^2 +3y^2 +x^2sin(y)) dy = 0$$
$$dleft(x^2y^3+y^3+x^2sin(y)right)=0$$
$$x^2y^3+y^3+x^2sin(y)=c$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$
    The integrating factor is $ycos(y)$ . After reducing to the common denominator, this is easy to guess. If you cannot see it at first place, try various forms of integrating factors. When trying an integrating factor of the form $f(y)$ you find the function $f(y)$ by identification.



    Checking :



    $$ycos(y)(2xy^2 sec y + 2xfractan yy) dx + ycos(y)(3x^2y sec y +3y sec y +fracx^2y) dy = 0$$



    $$(2xy^3 + 2xsin(y)) dx + (3x^2y^2 +3y^2 +x^2sin(y)) dy = 0$$
    $$dleft(x^2y^3+y^3+x^2sin(y)right)=0$$
    $$x^2y^3+y^3+x^2sin(y)=c$$






    share|cite|improve this answer


























      up vote
      1
      down vote



      accepted










      $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$
      The integrating factor is $ycos(y)$ . After reducing to the common denominator, this is easy to guess. If you cannot see it at first place, try various forms of integrating factors. When trying an integrating factor of the form $f(y)$ you find the function $f(y)$ by identification.



      Checking :



      $$ycos(y)(2xy^2 sec y + 2xfractan yy) dx + ycos(y)(3x^2y sec y +3y sec y +fracx^2y) dy = 0$$



      $$(2xy^3 + 2xsin(y)) dx + (3x^2y^2 +3y^2 +x^2sin(y)) dy = 0$$
      $$dleft(x^2y^3+y^3+x^2sin(y)right)=0$$
      $$x^2y^3+y^3+x^2sin(y)=c$$






      share|cite|improve this answer
























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$
        The integrating factor is $ycos(y)$ . After reducing to the common denominator, this is easy to guess. If you cannot see it at first place, try various forms of integrating factors. When trying an integrating factor of the form $f(y)$ you find the function $f(y)$ by identification.



        Checking :



        $$ycos(y)(2xy^2 sec y + 2xfractan yy) dx + ycos(y)(3x^2y sec y +3y sec y +fracx^2y) dy = 0$$



        $$(2xy^3 + 2xsin(y)) dx + (3x^2y^2 +3y^2 +x^2sin(y)) dy = 0$$
        $$dleft(x^2y^3+y^3+x^2sin(y)right)=0$$
        $$x^2y^3+y^3+x^2sin(y)=c$$






        share|cite|improve this answer














        $$(2xy^2 sec y + 2xfractan yy) dx + (3x^2y sec y +3y sec y +fracx^2y) dy = 0$$
        The integrating factor is $ycos(y)$ . After reducing to the common denominator, this is easy to guess. If you cannot see it at first place, try various forms of integrating factors. When trying an integrating factor of the form $f(y)$ you find the function $f(y)$ by identification.



        Checking :



        $$ycos(y)(2xy^2 sec y + 2xfractan yy) dx + ycos(y)(3x^2y sec y +3y sec y +fracx^2y) dy = 0$$



        $$(2xy^3 + 2xsin(y)) dx + (3x^2y^2 +3y^2 +x^2sin(y)) dy = 0$$
        $$dleft(x^2y^3+y^3+x^2sin(y)right)=0$$
        $$x^2y^3+y^3+x^2sin(y)=c$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 15 at 10:42

























        answered Aug 15 at 10:35









        JJacquelin

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