Solve the Differential equation $left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin...












1












$begingroup$


Solve the equation



$$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$



My try:



Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$



so have modified the equation as:



$$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$



$$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$



$$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$



any way to proceed here?






algebra-precalculus ordinary-differential-equations derivatives indefinite-integrals







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$endgroup$

















    1












    $begingroup$


    Solve the equation



    $$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$



    My try:



    Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$



    so have modified the equation as:



    $$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$



    $$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$



    $$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$



    any way to proceed here?






    algebra-precalculus ordinary-differential-equations derivatives indefinite-integrals







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Solve the equation



      $$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$



      My try:



      Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$



      so have modified the equation as:



      $$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$



      $$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$



      $$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$



      any way to proceed here?






      algebra-precalculus ordinary-differential-equations derivatives indefinite-integrals







      share|cite|improve this question









      $endgroup$




      Solve the equation



      $$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$



      My try:



      Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$



      so have modified the equation as:



      $$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$



      $$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$



      $$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$



      any way to proceed here?






      algebra-precalculus ordinary-differential-equations derivatives indefinite-integrals




      algebra-precalculus ordinary-differential-equations derivatives indefinite-integrals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 8:26









      Umesh shankarUmesh shankar

      2,68031219




      2,68031219






















          1 Answer
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          $begingroup$

          If you multiply the expression on the left-hand side by $y$, you get the exact differential,
          $$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$



          So the solution is



          $$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            If you multiply the expression on the left-hand side by $y$, you get the exact differential,
            $$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$



            So the solution is



            $$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              If you multiply the expression on the left-hand side by $y$, you get the exact differential,
              $$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$



              So the solution is



              $$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                If you multiply the expression on the left-hand side by $y$, you get the exact differential,
                $$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$



                So the solution is



                $$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$






                share|cite|improve this answer









                $endgroup$



                If you multiply the expression on the left-hand side by $y$, you get the exact differential,
                $$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$



                So the solution is



                $$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '18 at 8:36









                Kenny WongKenny Wong

                18.7k21439




                18.7k21439






























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