Explanation of differing solutions for an exact ODE












2












$begingroup$


I was solving the following ODE



$$ y cos(xy) + x cos(xy) y' = 0$$



My initial intuition to solving it is to observe that it can be factored as:



$$ cos(xy) (y + xy') = 0$$



So over $x,y$ that don't satisfy $cos(xy) = 0$ we have



$$ y + xy' = 0$$



And this is an exact eqation which has solutions of the form



$$ xy = C$$



Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$



So therefore the solutions are of the form $ sin(xy) = C$



clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.






ordinary-differential-equations pde







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    2












    $begingroup$


    I was solving the following ODE



    $$ y cos(xy) + x cos(xy) y' = 0$$



    My initial intuition to solving it is to observe that it can be factored as:



    $$ cos(xy) (y + xy') = 0$$



    So over $x,y$ that don't satisfy $cos(xy) = 0$ we have



    $$ y + xy' = 0$$



    And this is an exact eqation which has solutions of the form



    $$ xy = C$$



    Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$



    So therefore the solutions are of the form $ sin(xy) = C$



    clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.






    ordinary-differential-equations pde







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I was solving the following ODE



      $$ y cos(xy) + x cos(xy) y' = 0$$



      My initial intuition to solving it is to observe that it can be factored as:



      $$ cos(xy) (y + xy') = 0$$



      So over $x,y$ that don't satisfy $cos(xy) = 0$ we have



      $$ y + xy' = 0$$



      And this is an exact eqation which has solutions of the form



      $$ xy = C$$



      Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$



      So therefore the solutions are of the form $ sin(xy) = C$



      clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.






      ordinary-differential-equations pde







      share|cite|improve this question









      $endgroup$




      I was solving the following ODE



      $$ y cos(xy) + x cos(xy) y' = 0$$



      My initial intuition to solving it is to observe that it can be factored as:



      $$ cos(xy) (y + xy') = 0$$



      So over $x,y$ that don't satisfy $cos(xy) = 0$ we have



      $$ y + xy' = 0$$



      And this is an exact eqation which has solutions of the form



      $$ xy = C$$



      Now the video series that the ODE originated from takes a different approach. It notices that $$frac{partial}{partial y} [y cos(xy)] = cos(xy)- xy sin(xy) , frac{partial}{partial x} [xcos(xy)] = cos(xy) - xy sin(xy) $$



      So therefore the solutions are of the form $ sin(xy) = C$



      clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $cos(xy) ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.






      ordinary-differential-equations pde




      ordinary-differential-equations pde






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      asked Jan 27 at 22:11









      frogeyedpeasfrogeyedpeas

      7,63172054




      7,63172054






















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          $begingroup$

          Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh damn, since we can just take $arcsin$ of both sides.
            $endgroup$
            – frogeyedpeas
            Jan 27 at 22:38











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh damn, since we can just take $arcsin$ of both sides.
            $endgroup$
            – frogeyedpeas
            Jan 27 at 22:38
















          2












          $begingroup$

          Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh damn, since we can just take $arcsin$ of both sides.
            $endgroup$
            – frogeyedpeas
            Jan 27 at 22:38














          2












          2








          2





          $begingroup$

          Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.






          share|cite|improve this answer









          $endgroup$



          Indeed, $xy=C$ and $sin(xy)=C$ characterize the same family of curves. For example $sin(xy)=0$ is the curves $xy=kpi$, $kinmathbb{Z}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 27 at 22:19









          Eclipse SunEclipse Sun

          7,8401438




          7,8401438












          • $begingroup$
            oh damn, since we can just take $arcsin$ of both sides.
            $endgroup$
            – frogeyedpeas
            Jan 27 at 22:38


















          • $begingroup$
            oh damn, since we can just take $arcsin$ of both sides.
            $endgroup$
            – frogeyedpeas
            Jan 27 at 22:38
















          $begingroup$
          oh damn, since we can just take $arcsin$ of both sides.
          $endgroup$
          – frogeyedpeas
          Jan 27 at 22:38




          $begingroup$
          oh damn, since we can just take $arcsin$ of both sides.
          $endgroup$
          – frogeyedpeas
          Jan 27 at 22:38


















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