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I want to produce all possible solutions for this equation:

Clear["Global`*"]

eqn = {(y'[x]/y[x])^2 + 1/y[x]^2 - b/y[x]^4 - f/3 == 0};

DSolve[eqn, y[x], x]

I know the solution are:

   y[x] = Sqrt[3/(2 f)] Sqrt[1 - Cosh[2 x/Sqrt[3/f]] + 
Sqrt[f/Subscript[f, 0]] Sinh[2 x/Sqrt[3/f]]]   When f > Subscript[f, 0]

   y[x] = Sqrt[3/Subscript[f, 0]] Sqrt[1 + Exp[2 x/Sqrt[3/Subscript[f, 0]]]]   When 
   f= Subscript[f, 0] , y > Subscript[y, 0]

   y[x] = Sqrt[3/Subscript[f, 0]] Sqrt[1 - Exp[-2 x/Sqrt[3/Subscript[f, 0]]]]   When f = Subscript[f, 0], y < Subscript[y, 0]

   y[x] = Sqrt[3/(2 f)] Sqrt[1 + Sqrt[1 - (f/Subscript[f, 0])^2] Cosh[2 x/Sqrt[3/f]]]
  When 0 < f < Subscript[f, 0] for y large

    y[x] = Sqrt[3/(2 f)] Sqrt[1 - Cosh[2 x/Sqrt[3/f]] + Sqrt[f/Subscript[f, 0]]Sinh[2 x/Sqrt[3/f]]]   When 0 < f < Subscript[f, 0] for y small

    Where Sqrt[3/Subscript[f, 0]] = Sqrt[2] Subscript[y, 0] = 2 Sqrt[b]; all f, y and b are positive
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  • $\begingroup$ And your question is??? $\endgroup$ Nov 13, 2023 at 14:48
  • $\begingroup$ How to get those solution in mathematica!! because the solution on mathematica are different!!! $\endgroup$
    – Mathecis
    Nov 13, 2023 at 14:53
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    $\begingroup$ Please provide the definitions of Subscript[f, 0] and Subscript[y, 0] $\endgroup$ Nov 13, 2023 at 14:58
  • $\begingroup$ Sqrt[3/Subscript[f, 0]] = Sqrt[2] Subscript[y, 0] = 2 Sqrt[b] all f, y and b are positive $\endgroup$
    – Mathecis
    Nov 13, 2023 at 17:39

1 Answer 1

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To long for a comment!

You might check your expected solutions directly. Examplary for the first one

eqn /. y -> 
   Function[x, 
    Sqrt[3/(2 f)] Sqrt[
      1 - Cosh[2 x/Sqrt[3/f]] + 
       Sqrt[f/Subscript[f, 0]] Sinh[2 x/Sqrt[3/f]]] ] // 
 Simplify[#, f > Subscript[f, 0]] &

Result {(f (-3 + 4 b Subscript[f, 0]))/((-1 +Cosh[(2 x)/(Sqrt[3] Sqrt[1/f])] -Sinh[(2 x)/(Sqrt[3] Sqrt[1/f])] Sqrt[f/Subscript[f,0]]) Subscript[f, 0]) == 0} isn't True .

Your proposed solution doesn't solve eqn!

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