How to approximate this nonlinear ODE?












0












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I need try to approximate this nonlinear ODE



$left( frac{d^2}{dx^2}y(x) right)left( 1+4(y(x))^2 right)+8xleft( frac{d}{dx}y(x) right)^2-4(y(x))left( frac{d}{dx}y(x) right)^2+2(9.81)(y(x))=0$



But, i could not find a method to achieve it. I appreciate your help in this problem.






ordinary-differential-equations numerical-methods







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    0












    $begingroup$


    I need try to approximate this nonlinear ODE



    $left( frac{d^2}{dx^2}y(x) right)left( 1+4(y(x))^2 right)+8xleft( frac{d}{dx}y(x) right)^2-4(y(x))left( frac{d}{dx}y(x) right)^2+2(9.81)(y(x))=0$



    But, i could not find a method to achieve it. I appreciate your help in this problem.






    ordinary-differential-equations numerical-methods







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I need try to approximate this nonlinear ODE



      $left( frac{d^2}{dx^2}y(x) right)left( 1+4(y(x))^2 right)+8xleft( frac{d}{dx}y(x) right)^2-4(y(x))left( frac{d}{dx}y(x) right)^2+2(9.81)(y(x))=0$



      But, i could not find a method to achieve it. I appreciate your help in this problem.






      ordinary-differential-equations numerical-methods







      share|cite|improve this question









      $endgroup$




      I need try to approximate this nonlinear ODE



      $left( frac{d^2}{dx^2}y(x) right)left( 1+4(y(x))^2 right)+8xleft( frac{d}{dx}y(x) right)^2-4(y(x))left( frac{d}{dx}y(x) right)^2+2(9.81)(y(x))=0$



      But, i could not find a method to achieve it. I appreciate your help in this problem.






      ordinary-differential-equations numerical-methods




      ordinary-differential-equations numerical-methods






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      share|cite|improve this question











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      asked Feb 3 at 3:08









      retro_varretro_var

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          $begingroup$

          Starting from the ODE $(1+4y^2)y''+4(2x-y)(y')^2+2gy = 0$ ($g := 9.81$), we define the vector-valued function $boldsymbol{y} := (y,y')^{top}$. We then obtain a first-order ODE for $boldsymbol{y}$:
          begin{equation}
          boldsymbol{y}' = left( begin{array}{c}
          y'\
          y''
          end{array}
          right) = left( begin{array}{c}
          y'\
          -frac{4(2x-y)(y')^2+2gy}{1+4y^2}
          end{array}
          right) =: boldsymbol{f}(x,boldsymbol{y}).
          end{equation}
          If you also know some initial value $boldsymbol{y}(x_0) = boldsymbol{y}_0$, you can then use a Runge-Kutta method, for example, to solve this system of ODEs numerically.



          The numerical solution $(x_i,boldsymbol{y}_i)$ will contain approximations for $y(x_i)$ and for $y'(x_i)$ at the points $x_i$, $i=1,2,dots$






          share|cite|improve this answer









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            $begingroup$

            Starting from the ODE $(1+4y^2)y''+4(2x-y)(y')^2+2gy = 0$ ($g := 9.81$), we define the vector-valued function $boldsymbol{y} := (y,y')^{top}$. We then obtain a first-order ODE for $boldsymbol{y}$:
            begin{equation}
            boldsymbol{y}' = left( begin{array}{c}
            y'\
            y''
            end{array}
            right) = left( begin{array}{c}
            y'\
            -frac{4(2x-y)(y')^2+2gy}{1+4y^2}
            end{array}
            right) =: boldsymbol{f}(x,boldsymbol{y}).
            end{equation}
            If you also know some initial value $boldsymbol{y}(x_0) = boldsymbol{y}_0$, you can then use a Runge-Kutta method, for example, to solve this system of ODEs numerically.



            The numerical solution $(x_i,boldsymbol{y}_i)$ will contain approximations for $y(x_i)$ and for $y'(x_i)$ at the points $x_i$, $i=1,2,dots$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Starting from the ODE $(1+4y^2)y''+4(2x-y)(y')^2+2gy = 0$ ($g := 9.81$), we define the vector-valued function $boldsymbol{y} := (y,y')^{top}$. We then obtain a first-order ODE for $boldsymbol{y}$:
              begin{equation}
              boldsymbol{y}' = left( begin{array}{c}
              y'\
              y''
              end{array}
              right) = left( begin{array}{c}
              y'\
              -frac{4(2x-y)(y')^2+2gy}{1+4y^2}
              end{array}
              right) =: boldsymbol{f}(x,boldsymbol{y}).
              end{equation}
              If you also know some initial value $boldsymbol{y}(x_0) = boldsymbol{y}_0$, you can then use a Runge-Kutta method, for example, to solve this system of ODEs numerically.



              The numerical solution $(x_i,boldsymbol{y}_i)$ will contain approximations for $y(x_i)$ and for $y'(x_i)$ at the points $x_i$, $i=1,2,dots$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Starting from the ODE $(1+4y^2)y''+4(2x-y)(y')^2+2gy = 0$ ($g := 9.81$), we define the vector-valued function $boldsymbol{y} := (y,y')^{top}$. We then obtain a first-order ODE for $boldsymbol{y}$:
                begin{equation}
                boldsymbol{y}' = left( begin{array}{c}
                y'\
                y''
                end{array}
                right) = left( begin{array}{c}
                y'\
                -frac{4(2x-y)(y')^2+2gy}{1+4y^2}
                end{array}
                right) =: boldsymbol{f}(x,boldsymbol{y}).
                end{equation}
                If you also know some initial value $boldsymbol{y}(x_0) = boldsymbol{y}_0$, you can then use a Runge-Kutta method, for example, to solve this system of ODEs numerically.



                The numerical solution $(x_i,boldsymbol{y}_i)$ will contain approximations for $y(x_i)$ and for $y'(x_i)$ at the points $x_i$, $i=1,2,dots$






                share|cite|improve this answer









                $endgroup$



                Starting from the ODE $(1+4y^2)y''+4(2x-y)(y')^2+2gy = 0$ ($g := 9.81$), we define the vector-valued function $boldsymbol{y} := (y,y')^{top}$. We then obtain a first-order ODE for $boldsymbol{y}$:
                begin{equation}
                boldsymbol{y}' = left( begin{array}{c}
                y'\
                y''
                end{array}
                right) = left( begin{array}{c}
                y'\
                -frac{4(2x-y)(y')^2+2gy}{1+4y^2}
                end{array}
                right) =: boldsymbol{f}(x,boldsymbol{y}).
                end{equation}
                If you also know some initial value $boldsymbol{y}(x_0) = boldsymbol{y}_0$, you can then use a Runge-Kutta method, for example, to solve this system of ODEs numerically.



                The numerical solution $(x_i,boldsymbol{y}_i)$ will contain approximations for $y(x_i)$ and for $y'(x_i)$ at the points $x_i$, $i=1,2,dots$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 3 at 7:22









                ChristophChristoph

                59616




                59616






























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