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I've to solve this differential equation but I don't know how to do it. Someone can help me?

$\frac{dy}{dx} = \frac{1}{x} \Bigg( -1 - \frac{2a}{\sqrt{a^2 - 3x L f^2 e^{3y}}} \Bigg)$

where a, L and f are costant.

I've tried to solve it in Mathematica with DSolve in this way:

DSolve[1/x (-1 - (2 a/Sqrt[a^2 - 3 x L f^2 Exp[3 y[x]]])) == y'[x], 
 y[x], x]

Thanks everyone.

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  • $\begingroup$ Is this question about the Mathematica software? If not, this question is better suited for math.stackexchange. If you are asking about the Mathematica software, could you include what you have tried? $\endgroup$ – eyorble Mar 30 at 16:19
  • $\begingroup$ yes it's about Mathematica software and I've tried with DSolve $\endgroup$ – Laria Mar 30 at 16:23
  • $\begingroup$ Please include what you have already tried with DSolve in your question $\endgroup$ – MarcoB Mar 30 at 16:24
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    $\begingroup$ Your DSolve works in 12.2 without providing values to the constants, and gives multiple solutions of the form y[x]->Log[Root[...]] $\endgroup$ – flinty Mar 30 at 16:32
  • $\begingroup$ a = 1; f = 1; L = 1; DSolve[ 1/x (-1 - (2 a/Sqrt[a^2 - 3 x L f^2 Exp[3 y[x]]])) == y'[x], y[x], x] produces a useless output. Try ParametricNDSolve. $\endgroup$ – user64494 Mar 30 at 16:34
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Mathematica seems to have problems with Sqrt. "Squaring" gives

ode = ( -(2 a/Sqrt[a^2 - 3 x L f^2 Exp[3 y[x]]]))^2 == (x y'[x] + 1)^2     
(*(4 a^2)/(a^2 - 3 E^(3 y[x]) f^2 L x) == (1 + x Derivative[1][y][x])^2*)

DSolveis now able to solve the ode

DSolve[ode, y[x], x][[1]]
(*Solve[(3 (6 Sqrt[a^2 x^2 (a^2 - 3 E^(3 y[x]) f^2 L x)]
      ArcTan[Sqrt[-a^2 + 3 E^(3 y[x]) f^2 L x]/(3 a)] - 
    2 Sqrt[a^2 x^2 (a^2 - 3 E^(3 y[x]) f^2 L x)]
      ArcTan[Sqrt[-a^2 + 3 E^(3 y[x]) f^2 L x]/a] + 
    a x Sqrt[-a^2 + 
      3 E^(3 y[x]) f^2 L x] (5 Log[x] + 
       3 Log[8 a^2 + 3 E^(3 y[x]) f^2 L x])))/(16 a x Sqrt[-a^2 + 
   3 E^(3 y[x]) f^2 L x]) - (3 y[x])/16 == C[1], y[x]]*)

As the Maple result, Mathematica evaluates an implicit equation in y[x]!

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  • $\begingroup$ You obtain a non-equivalent ODE by squaring. $\endgroup$ – user64494 Mar 31 at 15:35
  • $\begingroup$ This ode includes all solution branches of the original ode! $\endgroup$ – Ulrich Neumann Mar 31 at 17:30
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As mentioned by flinty in the comment, v12.2 can solve the problem directly, with a Solve::ifun warning generated. If you need the implicit solution as given by Maple, you can:

Trace[
  DSolve[1/x (-1 - (2 a/Sqrt[a^2 - 3 x L f^2 Exp[3 y[x]]])) == y'[x], y[x], x], 
  Solve[_, y[x]], TraceInternal -> True] // Flatten
(* 
  {HoldForm[
    Solve[(3/8)*(3*Log[a - Sqrt[a^2 - 3*E^(3*y[x])*f^2*L*x]] + 
        2*Log[a + Sqrt[a^2 - 3*E^(3*y[x])*f^2*L*x]] + 
                3*Log[3*a + Sqrt[a^2 - 3*E^(3*y[x])*f^2*L*x]]) - 3*y[x] == C[1], y[x]]]}
 *)
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