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Homogeneous differential equation solutions difficulties

Posted 13 days ago
6 Replies
4 Total Likes

Every one.
I appreciate if someone can help me.

I've been taking difficulties with solutions of homogeneous differential equations. I'm not sure with this solution.

I think that there is a problem with Vector Plot and equations family.

POSTED BY: Narciso Agudo
6 Replies

I am afraid I do not understand your problem, but what do you think about this?

Define a tangent-vector

tvec[u_, f_] := Module[{},
  abl = D[f, x];
  v1 = {u, f /. x -> u};
  v2 = {u + .1, (f /. x -> u) + .1 (abl /. x -> u)};
  {v1, v2}

and then plot these tangent-vectors with the solutions of your diff.eq.

{Arrow /@ Table[tvec[a, funcionesh[[2]]], {a, 0, 2, .25}],
Arrow /@ Table[tvec[a, funcionesh[[4]]], {a, 0, 2, .25}]}],
Axes -> True,
AxesOrigin -> {0, 0}

OK, that ist not yet really nice, but with a bit of playing around with the parameters it could be acceptable.

POSTED BY: Hans Dolhaine
Posted 11 days ago

Don't worry. It's o.k. By the way, I'm not working in this area.


POSTED BY: Narciso Agudo


  funcionesh = Table[y[x, c1], {c1, 0, 10, 1/5}];
  famh = Plot[funcionesh, {x, -5, 5}, PlotRange -> {-30, 0}, 
    AxesLabel -> {"x", "y"}, AxesStyle -> Black, PlotRange -> All]
  campoh = VectorPlot[{1, (x - y)/(x + y)}, {x, -5, 6}, {y, -30, 20}, 
    Axes -> True, AxesStyle -> Black, Axes -> Black]
  Show[{famh, campoh}]

Regards M.I.

POSTED BY: Mariusz Iwaniuk
Posted 12 days ago

Thanks for your help.

POSTED BY: Narciso Agudo

Just to put into words what is going on.

You defined the vector field in VectorPlot to be {(x-y)/(x+y), x}.

The vector field should be {1, (x-y)/(x+y)}.

General remark: A vector field of the form $(v_x, v_y)$ represents the system of differential equations $dx/dt = v_x$, $dy/dt = v_y$.

Thus a vector field of the form $(1, f(x, y))$ represents the system $dx/dt = 1$, $dy/dt = f(x, y))$. Since $dy/dx = \big(dy/dt\big) \big/ \big(dx/dt\big)$, this system is equivalent to $dy/dx = f(x,y)$.

Gratuitous remarks:

The integration parameter shows up as E^(2 C[1]), which can be negative if C[1] is complex. This is inconvenient, perhaps. We can replace C[1] -> Log[c1]/2 to get simply c1 as a parameter, which now give negative term inside Sqrt[] whenever c1 is negative.

A characteristic of homogeneous DEs is that the integral curves cross a radial line at the same angle. Since in recent versions of WL in VectorPlot, many vectors lie on radial lines at angles of 0º, 30º, 60º, etc., this is easy to show.

solhomo2 = solhomo /. C[1] -> Log[c1]/2  (* put parameter in a convenient form *)
(*  {{y[x] -> -x - Sqrt[c1 + 2 x^2]}, {y[x] -> -x + Sqrt[c1 + 2 x^2]}}  *)

funcionesh = Flatten@Table[y[x] /. solhomo2,
  {c1, 2 Abs[#] # &@Range[-3, 3]}]; (* c1 = ±2 times squares *)
  VectorPlot[{1, (x - y)/(x + y)}, {x, -5, 5}, {y, -5, 5},
    Prolog -> { (* omit axes, add radial lines *)
      Thin, Black,
        InfiniteLine[{0, 0}, {Cos[t], Sin[t]}], {t, 0, 5 Pi/6, Pi/6}],
        InfiniteLine[{0, 0}, {Cos[t], Sin[t]}], {t, Pi/12, Pi, Pi/6}]}
  Plot[funcionesh, {x, -10, 10}]

POSTED BY: Michael Rogers
Posted 12 days ago

Thanks for your help.

POSTED BY: Narciso Agudo
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