Btukfyl

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.

perhaps

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

  PlotRange -> 4.7]

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}]

Code for GIF-animation:

n = 70;

pics = Table[Rasterize[#, "Image"] & @ Plot[

        Evaluate[y[x] /. sol /. C[1] -> Sqrt[1/16] (Cos[t] + I Sin[t])], {x, -4.7, 4.7},

        PlotRange -> {{-4.7, 4.7}, {-30, 46}}, ImageSize -> {500, 300}, AspectRatio -> Full,

        PlotRangePadding -> None, PlotRangeClipping -> True,

        ClippingStyle -> False], {t, 0, 2 π - #, #}] &[2 π/n];

Export["asd.gif", pics, "AnimationRepetitions" -> ∞,

        "ColorMapLength" -> 16, "DisplayDurations" -> ConstantArray[0.04, n]]

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}] 

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}]

Starting over in full generality, we can get an unintelligible plot.

genSol = FullSimplify[ ComplexExpand[

  y[x] /. DSolve[{

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

    y[d] == c}, 

    y[x], x][[1]]

]];

Plot[Flatten[

  Table[genSol, {c, -20, 20, 10}, {d, -2, 2, 1}

  ], 1], {x, -4, 4}]

Here we have applied the generic initial condition $y(d) = c$ (via y[d] == c) so that we have labels for the parts of a point of a particular solution. Since we only want real solutions, we want $c$, $d$, and $x$ to be real. Applying ComplexExpand treats all the variables as if they are real, yielding real solutions passing through the point $(c,d)$. (The largest effect here is that the complex exponentials are rewritten in terms of sine and cosine.)