0
$\begingroup$

In DIFFERENTIAL EQUATION SOLVING WITH DSOLVE by D. Kapadia, there is a Chini differential equation :

sol = DSolve[x'[t] == 5 x[t]^4 + 3 x[t]^(-4/3), x[t], t]

The answer implies RootSum. The question is how to use and plot the solution. I have tried 

Plot[x[t] /. sol /. {C[1] -> 2}, {t, 0, 10}]

which doesn't works and used Evaluate and/or Normal and/or ToRadical@Normal both on x[t] and on sol. But nothing works. Moreover, this case is not documented in the documentation "Plotting the Solution".

Need help; Thanks

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Hold up… if you try N[x[10] /. sol /. {C[1] -> 2}], what do you get? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jun 26, 2016 at 6:31
  • $\begingroup$ If I do that I have that the sol is neither a list of replacement rule... $\endgroup$
    – cyrille.piatecki
    Commented Jun 26, 2016 at 6:42
  • 1
    $\begingroup$ …and now you know why it doesn't work within Plot[]. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jun 26, 2016 at 7:09
  • $\begingroup$ Ok but that does not resolve my problem. I allready have found this but I do not know how to proceed. It's strange to find a solution in a manual of how to and not to be able to apply it. $\endgroup$
    – cyrille.piatecki
    Commented Jun 26, 2016 at 7:29
  • $\begingroup$ I don't have Mathematica on hand at the moment. Can you paste in the result sol in InputForm? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jun 26, 2016 at 8:24

2 Answers 2

Reset to default
4
$\begingroup$

An alternative approach is to solve for t[x] instead of x[t].

t[x] /. First@DSolve[t'[x] == 1/(5 x^4 + 3 x^(-4/3)), t[x], x] /. C[1] -> 0;
ParametricPlot[{Chop@%, x}, {x, 0, 3}, AxesLabel -> {t + C[1], x}, 
    AspectRatio -> GoldenRatio]

enter image description here

Because the ODE determines t[x] only up to an arbitrary constant, the curve above can be shifted by an arbitrary amount to the left or right.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I wonder why Chop@% work and not Chop@t $\endgroup$
    – cyrille.piatecki
    Commented Jun 29, 2016 at 10:21
  • $\begingroup$ @cyrille.piatecki t actually is not defined by my two lines of code. To define it, use t = s[x] /. First@DSolve[s'[x] == 1/(5 x^4 + 3 x^(-4/3)), s[x], x] /. C[1] -> 0; ParametricPlot[{Chop@t, x}, {x, 0, 3}, AxesLabel -> {"t+C[1]", x}, AspectRatio -> GoldenRatio]. Note that I use s[x] as the dummy variable in the first equation here to avoid possible issues. $\endgroup$
    – bbgodfrey
    Commented Jun 29, 2016 at 16:52
3
$\begingroup$

Maybe like this:

sol = DSolve[x'[t] == 5 x[t]^4 + 3 x[t]^(-4/3), x[t], t]
eq = sol[[1, 1, 2, 0, 1]][x] == (sol[[1, 1, 2, 1]] /. C[1] -> 2)
ContourPlot[sol[[1, 1, 2, 0, 1]][x] == (sol[[1, 1, 2, 1]] /. C[1] -> 2),
{t, -3, 3}, {x, -3, 3}, Axes -> True, Frame -> False, AxesLabel -> {t, x[t]}]

enter image description here

EDITED:

If You exectue this code:

 Internal`InheritedBlock[{Solve}, Unprotect[Solve];
 Solve[x___] := 
 Block[{$guard = True}, Print["Solve called : ", HoldForm[Solve[x]]];
 Solve[x]] /; ! TrueQ[$guard];
 DSolve[{x'[t] == 5 x[t]^4 + 3 x[t]^(-4/3)}, x[t], t]];

enter image description here

You can see DSolve cant find solution.Then solution of Solve convert to InverseFuntion. I'm only extract this solution using Part function.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you Mariusz it works perfectly even if I do not understand the philosophy behind. The mainchange is from x to x[t], in Dsolve and it change all. $\endgroup$
    – cyrille.piatecki
    Commented Jun 26, 2016 at 10:47
  • $\begingroup$ I,m only using Part function.That all. $\endgroup$
    – Mariusz Iwaniuk
    Commented Jun 26, 2016 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.