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I'm trying to DSolve these equations with each other:

q1[t_] := 3 (y'[t]/y[t])^2 - (1/(y[t]^3)) - (1/(y[t]^4)) - 1

q2[t_] := -2 y''[t]/y[t] - (y'[t]/y[t])^2 - (1/(3 y[t]^4)) + 1

q3[t_] := -3 (y''[t]/y[t] + (y'[t]/y[t])^2) + 1

When solving the first with the second by:

DSolve[{q1[t] == q2[t]}, y, t]

I get an output in the form: enter image description here

While when solving the second with the third

DSolve[{q2[t] == q3[t]}, y, t]

I get:

enter image description here

Any idea how to interpret or simplify these outputs to be for example like:

enter image description here

I tried many initial conditions- they are so arbitrary for y[t] or y'[t]- but always have in terms of Integral or InverseFunction .

Any help about this will be appreciated.

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  • 1
    $\begingroup$ Perhaps the solution of these ODEs just can't be expressed without InverseFunction or Integrate? Do you have any reason to believe the solution can be transformed to something free of them? $\endgroup$ – xzczd Sep 8 at 3:25
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With DSolve you get y[t] as an InverseFunction , diffucult to understand. Invert this function again with Solve to get t[y] in simple form you can plot with ParametricPlot .

f = q2[t] - q3[t] // Together // Numerator

dsol1 = DSolve[f == 0, y, t, 
   GeneratedParameters -> (ToExpression[
  StringJoin["c", ToString[#]]] &)]

(*   {{y -> Function[{t}, 
InverseFunction[-(1/2) Sqrt[
    3] (-3 c1 Log[#1 + Sqrt[3 c1 + #1^2]] + #1 Sqrt[
       3 c1 + #1^2]) &][c2 + t]]}, {y -> 
 Function[{t}, 
   InverseFunction[
  1/2 Sqrt[
    3] (-3 c1 Log[#1 + Sqrt[3 c1 + #1^2]] + #1 Sqrt[
       3 c1 + #1^2]) &][c2 + t]]}}   *)

t1[ys_, c1_, c2_] = t /. First@Solve[(y[t] /. dsol1[[1]]) == ys, t]

(*   -c2 - 1/2 Sqrt[3] (ys Sqrt[3 c1 + ys^2] - 3 c1 Log[ys + Sqrt[3 c1 + ys^2]])   *)

t2[ys_, c1_, c2_] = t /. First@Solve[(y[t] /. dsol1[[2]]) == ys, t]

(*   -c2 + 1/2 Sqrt[3] (ys Sqrt[3 c1 + ys^2] - 3 c1 Log[ys + Sqrt[3 c1 + ys^2]])   *)

Manipulate[
  ParametricPlot[{{t1[ys, c1, c2], ys}, {t2[ys, c1, c2], 
ys}}, {ys, -10, 10}, AspectRatio -> 1, 
     PlotRange -> All], {{c1, 1}, 0, 5}, {c2, -4, 4}]
| improve this answer | |
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FullSimplify[q2[t] - q3[t]]

(-1 + 6 y[t]^2 Derivative[1][y][t]^2 + 3 y[t]^3 (y^[Prime][Prime])[t])/(3 y[t]^4)

So the 3 y[t]^4 can be ignored for the solution.

The term

6 y[t]^2 y'[t]^2 + 3 y[t]^3 y''[t]

equals

y[t]D[y[t]^3, t, t].

Good practice is to entered the equation that are close to standards or standards.

So the homogenous equation can be solved by third root of a linear function.

The built-in InverseFunction works like this:

InverseFunction[c ArcTan][x]

InverseFunction[ArcTan c][x]

and

InverseFunction[ArcTan][x]

Tan[x]

It does chaining and works with numbers.

So the solution has to be dealt with by hand.

So the term in the InverseFunction of Your second DSolve problem does this:

Integrate[-3 ArcTanh[x/Sqrt[3 C[1] + x^2]] C[1] + 
  x Sqrt[3 C[1] + x^2], x]

-3 x ArcTanh[x/Sqrt[x^2 + 3 C1]] C1 + 1/3 Sqrt[x^2 + 3 C1] (x^2 + 12 C1)

The advantages of this strategy are not always visible to everybody.

The main problem in the q's are the inhomogeneities. The homogenous equation will give more closed form solutions.

Variation of Parameters Method Of Variation Of Parameters For Second Order Linear Differentia may lead to partly more closed form of solutions.

Since a lot power with ODE is stored in literature examples and equation collection these sources has to be made a requisite even for the word with the powerful Mathematica.

A look worth: equation world and many more.

This question shows, how to deal in Mathematica with InverseFunction: the-graph-of-inverse-function-f-1y. For C1==C2==1:

Plot[InverseFunction[
   Inactive[Integrate][-((Sqrt[3] K[1]^2)/Sqrt[
      3 1 + K[1]^2 + K[1]^3 + K[1]^6]), {K[1], 1, #1}] &][t + 1], {t, 
  0, 10}]

Plot

enter link description here

| improve this answer | |
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