Can't plot DSolve's solution to Riccati differential equation











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DSolve gives a strange solution for the Riccati differential equation $ y' = (y^2) - 2 x^2 y + (x^4) + 2 x + 4 $



Opres = DSolve[y'[x] == y[x]^2-2x^2*y[x]+x^4+2x+4, y[x], x]


$left{left{y(x)to frac{1}{c_1 e^{4 i x}-frac{i}{4}}+x^2-2 iright}right}$



When I try plot this solution



Opresgraf = 
Plot[Evaluate[y[x] /. Opres /. C[1] -> Range[-3, 3]], {x, -4.7, 4.7},
PlotRange -> 4.7]


I get a blank graph.



My question is: how can I get a solution with DSolve (not with NDSolve, because in my student research project I need DSolve) and plot that solution, the most important is to plot that general solution with DSolve.










share|improve this question




















  • 1




    You can't plot a complex expression. You need to either plot its real value, its imaginary value or its modulus. Also you have a typo in the input, you need y[x] not y in the ODE itself.
    – Nasser
    Nov 27 at 18:24






  • 1




    Is Range[-3.3] supposed to be Range[-3,3]?
    – That Gravity Guy
    Nov 27 at 18:26















up vote
3
down vote

favorite












DSolve gives a strange solution for the Riccati differential equation $ y' = (y^2) - 2 x^2 y + (x^4) + 2 x + 4 $



Opres = DSolve[y'[x] == y[x]^2-2x^2*y[x]+x^4+2x+4, y[x], x]


$left{left{y(x)to frac{1}{c_1 e^{4 i x}-frac{i}{4}}+x^2-2 iright}right}$



When I try plot this solution



Opresgraf = 
Plot[Evaluate[y[x] /. Opres /. C[1] -> Range[-3, 3]], {x, -4.7, 4.7},
PlotRange -> 4.7]


I get a blank graph.



My question is: how can I get a solution with DSolve (not with NDSolve, because in my student research project I need DSolve) and plot that solution, the most important is to plot that general solution with DSolve.










share|improve this question




















  • 1




    You can't plot a complex expression. You need to either plot its real value, its imaginary value or its modulus. Also you have a typo in the input, you need y[x] not y in the ODE itself.
    – Nasser
    Nov 27 at 18:24






  • 1




    Is Range[-3.3] supposed to be Range[-3,3]?
    – That Gravity Guy
    Nov 27 at 18:26













up vote
3
down vote

favorite









up vote
3
down vote

favorite











DSolve gives a strange solution for the Riccati differential equation $ y' = (y^2) - 2 x^2 y + (x^4) + 2 x + 4 $



Opres = DSolve[y'[x] == y[x]^2-2x^2*y[x]+x^4+2x+4, y[x], x]


$left{left{y(x)to frac{1}{c_1 e^{4 i x}-frac{i}{4}}+x^2-2 iright}right}$



When I try plot this solution



Opresgraf = 
Plot[Evaluate[y[x] /. Opres /. C[1] -> Range[-3, 3]], {x, -4.7, 4.7},
PlotRange -> 4.7]


I get a blank graph.



My question is: how can I get a solution with DSolve (not with NDSolve, because in my student research project I need DSolve) and plot that solution, the most important is to plot that general solution with DSolve.










share|improve this question















DSolve gives a strange solution for the Riccati differential equation $ y' = (y^2) - 2 x^2 y + (x^4) + 2 x + 4 $



Opres = DSolve[y'[x] == y[x]^2-2x^2*y[x]+x^4+2x+4, y[x], x]


$left{left{y(x)to frac{1}{c_1 e^{4 i x}-frac{i}{4}}+x^2-2 iright}right}$



When I try plot this solution



Opresgraf = 
Plot[Evaluate[y[x] /. Opres /. C[1] -> Range[-3, 3]], {x, -4.7, 4.7},
PlotRange -> 4.7]


I get a blank graph.



My question is: how can I get a solution with DSolve (not with NDSolve, because in my student research project I need DSolve) and plot that solution, the most important is to plot that general solution with DSolve.







differential-equations






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edited Nov 27 at 23:01









kglr

174k8196401




174k8196401










asked Nov 27 at 18:10









Милош Вучковић

596




596








  • 1




    You can't plot a complex expression. You need to either plot its real value, its imaginary value or its modulus. Also you have a typo in the input, you need y[x] not y in the ODE itself.
    – Nasser
    Nov 27 at 18:24






  • 1




    Is Range[-3.3] supposed to be Range[-3,3]?
    – That Gravity Guy
    Nov 27 at 18:26














  • 1




    You can't plot a complex expression. You need to either plot its real value, its imaginary value or its modulus. Also you have a typo in the input, you need y[x] not y in the ODE itself.
    – Nasser
    Nov 27 at 18:24






  • 1




    Is Range[-3.3] supposed to be Range[-3,3]?
    – That Gravity Guy
    Nov 27 at 18:26








1




1




You can't plot a complex expression. You need to either plot its real value, its imaginary value or its modulus. Also you have a typo in the input, you need y[x] not y in the ODE itself.
– Nasser
Nov 27 at 18:24




You can't plot a complex expression. You need to either plot its real value, its imaginary value or its modulus. Also you have a typo in the input, you need y[x] not y in the ODE itself.
– Nasser
Nov 27 at 18:24




1




1




Is Range[-3.3] supposed to be Range[-3,3]?
– That Gravity Guy
Nov 27 at 18:26




Is Range[-3.3] supposed to be Range[-3,3]?
– That Gravity Guy
Nov 27 at 18:26










4 Answers
4






active

oldest

votes

















up vote
7
down vote













perhaps



Plot[Evaluate[ReIm@y[x] /. (Opres /. C[1] -> Range[-3, 3])], {x, -4.7, 4.7}, 
PlotRange -> 4.7]


enter image description here






share|improve this answer




























    up vote
    4
    down vote













    With a single graph you can only plot those solution that are imaginary or real.



    There are 2 real ones:



    sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
    zeroIm = FullSimplify[ComplexExpand[Im[y[x] /. sol]]] == 0 // Solve[#, C[1]] &



    $style{text-decoration:line-through}{left{left{C[1]to -frac{1}{4}right},left{C[1]to frac{1}{4}right}right}}$




    I forgot to consider complex values of C[1]:



    sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
    zeroIm = Numerator[FullSimplify[ComplexExpand[Im[y[x] /. sol], C[1]]]] == 0

    (* -2 + 32 Abs[C[1]]^2 == 0 *)


    which is the equation of a circle of real solutions:



    Manipulate[Plot[Evaluate[y[x] /. sol /. C[1] -> Sqrt[1/16] (Cos[t] + I Sin[t])],
    {x, -4.7, 4.7}, Exclusions -> All], {t, 0, 2 π}]







    share|improve this answer






























      up vote
      3
      down vote













      Try this



      Opres = DSolve[y'[x] == y[x]^2-2x^2 *y[x]+x^4+2x+4, y[x], x][[1]];
      Plot[{Re[y[x]/.Opres/.C[1]->Range[3.3]],Im[y[x]/.Opres/.C[1]->Range[3.3]]}, {x,-4.7,4.7}]





      share|improve this answer




























        up vote
        1
        down vote













        The general solution is not real valued. Try setting an initial condition:



        FullSimplify[
        DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 1},
        y[x], x]
        ]


        yielding



        {{y[x] -> -2 I + (4 + 8 I)/((2 - I) + (2 + I) E^(4 I x)) + x^2}}


        which is not real valued (almost everywhere). However, for a different initial condition



        FullSimplify[
        DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 0},
        y[x], x]
        ]

        {{y[x] -> x^2 + 2 Tan[2 x]}}


        the solution is real valued.



        We can use a symbolic initial condition



        FullSimplify[
        DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == c},
        y[x], x]
        ]

        {{y[x] -> -2 I + (8 - 4 I c)/(-2 I - c + (-2 I + c) E^(4 I x)) + x^2}}


        and see that this complex valued behaviour is generic, but can be hidden with particular choices of the initial condition, c. Note that we can give the initial condition at a different value of the independent variable, and get different behaviour altogether. In fact, providing an initial condition at x=1 gives a real valued generic solution.



        FullSimplify[
        DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[1] == c},
        y[x], x]
        ]

        {{ y[x] -> ( 2 (-1 + c + x^2) Cos[2 - 2 x] + (-4 + (-1 + c) x^2) Sin[2 - 2 x] )/
        ( 2 Cos[2 - 2 x] + (-1 + c) Sin[2 - 2 x] ) }}

        Plot[Table[y[x] /. %[[1]], {c, -2, 2}], {x, -2, 2}]


        Plot of several particular solutions.






        share|improve this answer





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          7
          down vote













          perhaps



          Plot[Evaluate[ReIm@y[x] /. (Opres /. C[1] -> Range[-3, 3])], {x, -4.7, 4.7}, 
          PlotRange -> 4.7]


          enter image description here






          share|improve this answer

























            up vote
            7
            down vote













            perhaps



            Plot[Evaluate[ReIm@y[x] /. (Opres /. C[1] -> Range[-3, 3])], {x, -4.7, 4.7}, 
            PlotRange -> 4.7]


            enter image description here






            share|improve this answer























              up vote
              7
              down vote










              up vote
              7
              down vote









              perhaps



              Plot[Evaluate[ReIm@y[x] /. (Opres /. C[1] -> Range[-3, 3])], {x, -4.7, 4.7}, 
              PlotRange -> 4.7]


              enter image description here






              share|improve this answer












              perhaps



              Plot[Evaluate[ReIm@y[x] /. (Opres /. C[1] -> Range[-3, 3])], {x, -4.7, 4.7}, 
              PlotRange -> 4.7]


              enter image description here







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 27 at 18:27









              kglr

              174k8196401




              174k8196401






















                  up vote
                  4
                  down vote













                  With a single graph you can only plot those solution that are imaginary or real.



                  There are 2 real ones:



                  sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                  zeroIm = FullSimplify[ComplexExpand[Im[y[x] /. sol]]] == 0 // Solve[#, C[1]] &



                  $style{text-decoration:line-through}{left{left{C[1]to -frac{1}{4}right},left{C[1]to frac{1}{4}right}right}}$




                  I forgot to consider complex values of C[1]:



                  sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                  zeroIm = Numerator[FullSimplify[ComplexExpand[Im[y[x] /. sol], C[1]]]] == 0

                  (* -2 + 32 Abs[C[1]]^2 == 0 *)


                  which is the equation of a circle of real solutions:



                  Manipulate[Plot[Evaluate[y[x] /. sol /. C[1] -> Sqrt[1/16] (Cos[t] + I Sin[t])],
                  {x, -4.7, 4.7}, Exclusions -> All], {t, 0, 2 π}]







                  share|improve this answer



























                    up vote
                    4
                    down vote













                    With a single graph you can only plot those solution that are imaginary or real.



                    There are 2 real ones:



                    sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                    zeroIm = FullSimplify[ComplexExpand[Im[y[x] /. sol]]] == 0 // Solve[#, C[1]] &



                    $style{text-decoration:line-through}{left{left{C[1]to -frac{1}{4}right},left{C[1]to frac{1}{4}right}right}}$




                    I forgot to consider complex values of C[1]:



                    sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                    zeroIm = Numerator[FullSimplify[ComplexExpand[Im[y[x] /. sol], C[1]]]] == 0

                    (* -2 + 32 Abs[C[1]]^2 == 0 *)


                    which is the equation of a circle of real solutions:



                    Manipulate[Plot[Evaluate[y[x] /. sol /. C[1] -> Sqrt[1/16] (Cos[t] + I Sin[t])],
                    {x, -4.7, 4.7}, Exclusions -> All], {t, 0, 2 π}]







                    share|improve this answer

























                      up vote
                      4
                      down vote










                      up vote
                      4
                      down vote









                      With a single graph you can only plot those solution that are imaginary or real.



                      There are 2 real ones:



                      sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                      zeroIm = FullSimplify[ComplexExpand[Im[y[x] /. sol]]] == 0 // Solve[#, C[1]] &



                      $style{text-decoration:line-through}{left{left{C[1]to -frac{1}{4}right},left{C[1]to frac{1}{4}right}right}}$




                      I forgot to consider complex values of C[1]:



                      sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                      zeroIm = Numerator[FullSimplify[ComplexExpand[Im[y[x] /. sol], C[1]]]] == 0

                      (* -2 + 32 Abs[C[1]]^2 == 0 *)


                      which is the equation of a circle of real solutions:



                      Manipulate[Plot[Evaluate[y[x] /. sol /. C[1] -> Sqrt[1/16] (Cos[t] + I Sin[t])],
                      {x, -4.7, 4.7}, Exclusions -> All], {t, 0, 2 π}]







                      share|improve this answer














                      With a single graph you can only plot those solution that are imaginary or real.



                      There are 2 real ones:



                      sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                      zeroIm = FullSimplify[ComplexExpand[Im[y[x] /. sol]]] == 0 // Solve[#, C[1]] &



                      $style{text-decoration:line-through}{left{left{C[1]to -frac{1}{4}right},left{C[1]to frac{1}{4}right}right}}$




                      I forgot to consider complex values of C[1]:



                      sol = First[DSolve[y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[x], x]];
                      zeroIm = Numerator[FullSimplify[ComplexExpand[Im[y[x] /. sol], C[1]]]] == 0

                      (* -2 + 32 Abs[C[1]]^2 == 0 *)


                      which is the equation of a circle of real solutions:



                      Manipulate[Plot[Evaluate[y[x] /. sol /. C[1] -> Sqrt[1/16] (Cos[t] + I Sin[t])],
                      {x, -4.7, 4.7}, Exclusions -> All], {t, 0, 2 π}]








                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Nov 28 at 11:21

























                      answered Nov 27 at 18:33









                      Coolwater

                      14.4k32452




                      14.4k32452






















                          up vote
                          3
                          down vote













                          Try this



                          Opres = DSolve[y'[x] == y[x]^2-2x^2 *y[x]+x^4+2x+4, y[x], x][[1]];
                          Plot[{Re[y[x]/.Opres/.C[1]->Range[3.3]],Im[y[x]/.Opres/.C[1]->Range[3.3]]}, {x,-4.7,4.7}]





                          share|improve this answer

























                            up vote
                            3
                            down vote













                            Try this



                            Opres = DSolve[y'[x] == y[x]^2-2x^2 *y[x]+x^4+2x+4, y[x], x][[1]];
                            Plot[{Re[y[x]/.Opres/.C[1]->Range[3.3]],Im[y[x]/.Opres/.C[1]->Range[3.3]]}, {x,-4.7,4.7}]





                            share|improve this answer























                              up vote
                              3
                              down vote










                              up vote
                              3
                              down vote









                              Try this



                              Opres = DSolve[y'[x] == y[x]^2-2x^2 *y[x]+x^4+2x+4, y[x], x][[1]];
                              Plot[{Re[y[x]/.Opres/.C[1]->Range[3.3]],Im[y[x]/.Opres/.C[1]->Range[3.3]]}, {x,-4.7,4.7}]





                              share|improve this answer












                              Try this



                              Opres = DSolve[y'[x] == y[x]^2-2x^2 *y[x]+x^4+2x+4, y[x], x][[1]];
                              Plot[{Re[y[x]/.Opres/.C[1]->Range[3.3]],Im[y[x]/.Opres/.C[1]->Range[3.3]]}, {x,-4.7,4.7}]






                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered Nov 27 at 18:24









                              Bill

                              5,42059




                              5,42059






















                                  up vote
                                  1
                                  down vote













                                  The general solution is not real valued. Try setting an initial condition:



                                  FullSimplify[
                                  DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 1},
                                  y[x], x]
                                  ]


                                  yielding



                                  {{y[x] -> -2 I + (4 + 8 I)/((2 - I) + (2 + I) E^(4 I x)) + x^2}}


                                  which is not real valued (almost everywhere). However, for a different initial condition



                                  FullSimplify[
                                  DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 0},
                                  y[x], x]
                                  ]

                                  {{y[x] -> x^2 + 2 Tan[2 x]}}


                                  the solution is real valued.



                                  We can use a symbolic initial condition



                                  FullSimplify[
                                  DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == c},
                                  y[x], x]
                                  ]

                                  {{y[x] -> -2 I + (8 - 4 I c)/(-2 I - c + (-2 I + c) E^(4 I x)) + x^2}}


                                  and see that this complex valued behaviour is generic, but can be hidden with particular choices of the initial condition, c. Note that we can give the initial condition at a different value of the independent variable, and get different behaviour altogether. In fact, providing an initial condition at x=1 gives a real valued generic solution.



                                  FullSimplify[
                                  DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[1] == c},
                                  y[x], x]
                                  ]

                                  {{ y[x] -> ( 2 (-1 + c + x^2) Cos[2 - 2 x] + (-4 + (-1 + c) x^2) Sin[2 - 2 x] )/
                                  ( 2 Cos[2 - 2 x] + (-1 + c) Sin[2 - 2 x] ) }}

                                  Plot[Table[y[x] /. %[[1]], {c, -2, 2}], {x, -2, 2}]


                                  Plot of several particular solutions.






                                  share|improve this answer

























                                    up vote
                                    1
                                    down vote













                                    The general solution is not real valued. Try setting an initial condition:



                                    FullSimplify[
                                    DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 1},
                                    y[x], x]
                                    ]


                                    yielding



                                    {{y[x] -> -2 I + (4 + 8 I)/((2 - I) + (2 + I) E^(4 I x)) + x^2}}


                                    which is not real valued (almost everywhere). However, for a different initial condition



                                    FullSimplify[
                                    DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 0},
                                    y[x], x]
                                    ]

                                    {{y[x] -> x^2 + 2 Tan[2 x]}}


                                    the solution is real valued.



                                    We can use a symbolic initial condition



                                    FullSimplify[
                                    DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == c},
                                    y[x], x]
                                    ]

                                    {{y[x] -> -2 I + (8 - 4 I c)/(-2 I - c + (-2 I + c) E^(4 I x)) + x^2}}


                                    and see that this complex valued behaviour is generic, but can be hidden with particular choices of the initial condition, c. Note that we can give the initial condition at a different value of the independent variable, and get different behaviour altogether. In fact, providing an initial condition at x=1 gives a real valued generic solution.



                                    FullSimplify[
                                    DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[1] == c},
                                    y[x], x]
                                    ]

                                    {{ y[x] -> ( 2 (-1 + c + x^2) Cos[2 - 2 x] + (-4 + (-1 + c) x^2) Sin[2 - 2 x] )/
                                    ( 2 Cos[2 - 2 x] + (-1 + c) Sin[2 - 2 x] ) }}

                                    Plot[Table[y[x] /. %[[1]], {c, -2, 2}], {x, -2, 2}]


                                    Plot of several particular solutions.






                                    share|improve this answer























                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      The general solution is not real valued. Try setting an initial condition:



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 1},
                                      y[x], x]
                                      ]


                                      yielding



                                      {{y[x] -> -2 I + (4 + 8 I)/((2 - I) + (2 + I) E^(4 I x)) + x^2}}


                                      which is not real valued (almost everywhere). However, for a different initial condition



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 0},
                                      y[x], x]
                                      ]

                                      {{y[x] -> x^2 + 2 Tan[2 x]}}


                                      the solution is real valued.



                                      We can use a symbolic initial condition



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == c},
                                      y[x], x]
                                      ]

                                      {{y[x] -> -2 I + (8 - 4 I c)/(-2 I - c + (-2 I + c) E^(4 I x)) + x^2}}


                                      and see that this complex valued behaviour is generic, but can be hidden with particular choices of the initial condition, c. Note that we can give the initial condition at a different value of the independent variable, and get different behaviour altogether. In fact, providing an initial condition at x=1 gives a real valued generic solution.



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[1] == c},
                                      y[x], x]
                                      ]

                                      {{ y[x] -> ( 2 (-1 + c + x^2) Cos[2 - 2 x] + (-4 + (-1 + c) x^2) Sin[2 - 2 x] )/
                                      ( 2 Cos[2 - 2 x] + (-1 + c) Sin[2 - 2 x] ) }}

                                      Plot[Table[y[x] /. %[[1]], {c, -2, 2}], {x, -2, 2}]


                                      Plot of several particular solutions.






                                      share|improve this answer












                                      The general solution is not real valued. Try setting an initial condition:



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 1},
                                      y[x], x]
                                      ]


                                      yielding



                                      {{y[x] -> -2 I + (4 + 8 I)/((2 - I) + (2 + I) E^(4 I x)) + x^2}}


                                      which is not real valued (almost everywhere). However, for a different initial condition



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == 0},
                                      y[x], x]
                                      ]

                                      {{y[x] -> x^2 + 2 Tan[2 x]}}


                                      the solution is real valued.



                                      We can use a symbolic initial condition



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[0] == c},
                                      y[x], x]
                                      ]

                                      {{y[x] -> -2 I + (8 - 4 I c)/(-2 I - c + (-2 I + c) E^(4 I x)) + x^2}}


                                      and see that this complex valued behaviour is generic, but can be hidden with particular choices of the initial condition, c. Note that we can give the initial condition at a different value of the independent variable, and get different behaviour altogether. In fact, providing an initial condition at x=1 gives a real valued generic solution.



                                      FullSimplify[
                                      DSolve[{y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y[1] == c},
                                      y[x], x]
                                      ]

                                      {{ y[x] -> ( 2 (-1 + c + x^2) Cos[2 - 2 x] + (-4 + (-1 + c) x^2) Sin[2 - 2 x] )/
                                      ( 2 Cos[2 - 2 x] + (-1 + c) Sin[2 - 2 x] ) }}

                                      Plot[Table[y[x] /. %[[1]], {c, -2, 2}], {x, -2, 2}]


                                      Plot of several particular solutions.







                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered Nov 28 at 4:48









                                      Eric Towers

                                      2,236613




                                      2,236613






























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