Characteristic curves of a pde












0












$begingroup$


I am new to these sorts of questions, and the method of characteristics. I've been asked to consider the equation:



$ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$



I need to find the characteristic curves in the explicit form, and obtain the general solution of the PDE and check that this solution satisfies the PDE by differentiation. I ended up with the general solution being:



$ frac{x^{2}}{2} + frac{1}{2y^{2}} $



Would appreciate any help.






pde characteristics







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    0












    $begingroup$


    I am new to these sorts of questions, and the method of characteristics. I've been asked to consider the equation:



    $ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$



    I need to find the characteristic curves in the explicit form, and obtain the general solution of the PDE and check that this solution satisfies the PDE by differentiation. I ended up with the general solution being:



    $ frac{x^{2}}{2} + frac{1}{2y^{2}} $



    Would appreciate any help.






    pde characteristics







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am new to these sorts of questions, and the method of characteristics. I've been asked to consider the equation:



      $ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$



      I need to find the characteristic curves in the explicit form, and obtain the general solution of the PDE and check that this solution satisfies the PDE by differentiation. I ended up with the general solution being:



      $ frac{x^{2}}{2} + frac{1}{2y^{2}} $



      Would appreciate any help.






      pde characteristics







      share|cite|improve this question









      $endgroup$




      I am new to these sorts of questions, and the method of characteristics. I've been asked to consider the equation:



      $ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$



      I need to find the characteristic curves in the explicit form, and obtain the general solution of the PDE and check that this solution satisfies the PDE by differentiation. I ended up with the general solution being:



      $ frac{x^{2}}{2} + frac{1}{2y^{2}} $



      Would appreciate any help.






      pde characteristics




      pde characteristics






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      asked Feb 2 at 10:08









      User19098User19098

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          $begingroup$

          $$ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$$



          $u= frac{x^{2}}{2} + frac{1}{2y^{2}} quad$ is not the general solution.



          This is only a particular solution.



          You apparently got the characteristic equation
          $$x^2+frac{1}{y^2}=c$$
          So, the general solution is :
          $$u(x,y)=Fleft(x^2+frac{1}{y^2}right)$$
          where $F$ is an arbitrary function, to be determined according to some boundary condition.






          share|cite|improve this answer









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            $begingroup$

            $$ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$$



            $u= frac{x^{2}}{2} + frac{1}{2y^{2}} quad$ is not the general solution.



            This is only a particular solution.



            You apparently got the characteristic equation
            $$x^2+frac{1}{y^2}=c$$
            So, the general solution is :
            $$u(x,y)=Fleft(x^2+frac{1}{y^2}right)$$
            where $F$ is an arbitrary function, to be determined according to some boundary condition.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              $$ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$$



              $u= frac{x^{2}}{2} + frac{1}{2y^{2}} quad$ is not the general solution.



              This is only a particular solution.



              You apparently got the characteristic equation
              $$x^2+frac{1}{y^2}=c$$
              So, the general solution is :
              $$u(x,y)=Fleft(x^2+frac{1}{y^2}right)$$
              where $F$ is an arbitrary function, to be determined according to some boundary condition.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                $$ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$$



                $u= frac{x^{2}}{2} + frac{1}{2y^{2}} quad$ is not the general solution.



                This is only a particular solution.



                You apparently got the characteristic equation
                $$x^2+frac{1}{y^2}=c$$
                So, the general solution is :
                $$u(x,y)=Fleft(x^2+frac{1}{y^2}right)$$
                where $F$ is an arbitrary function, to be determined according to some boundary condition.






                share|cite|improve this answer









                $endgroup$



                $$ frac{∂u}{∂x} + xy^{3}frac{∂u}{∂y} = 0$$



                $u= frac{x^{2}}{2} + frac{1}{2y^{2}} quad$ is not the general solution.



                This is only a particular solution.



                You apparently got the characteristic equation
                $$x^2+frac{1}{y^2}=c$$
                So, the general solution is :
                $$u(x,y)=Fleft(x^2+frac{1}{y^2}right)$$
                where $F$ is an arbitrary function, to be determined according to some boundary condition.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 2 at 10:21









                JJacquelinJJacquelin

                45.6k21857




                45.6k21857






























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