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I have the following differential equation

DSolve[(Sin[c1]^2 Sin[c2] Sin[
    q[z]]^2 (3 Cos[q[z]] (1 + z^2 Derivative[1][q][z]^2) + 
     z Sin[q[z]] (3 Derivative[1][q][z] + 
        4 z^2 Derivative[1][q][z]^3 - z (q^\[Prime]\[Prime])[z])))/(
  z^5 (1 + z^2 Derivative[1][q][z]^2)^(3/2)) == 0, q[z], z]

And I would like Mathematica to solve it for me.

I know that the solution to the above is $ArcCos(m z)$.

To make it explicit.

In[102]:= (
   Sin[c1]^2 Sin[c2] Sin[
     q[z]]^2 (3 Cos[q[z]] (1 + z^2 Derivative[1][q][z]^2) + 
      z Sin[q[z]] (3 Derivative[1][q][z] + 
         4 z^2 Derivative[1][q][z]^3 - z (q^\[Prime]\[Prime])[z])))/(
   z^5 (1 + z^2 Derivative[1][q][z]^2)^(3/2)) == 0 /. 
  q -> (ArcCos[m #] &) // Factor

Out[102]= True

I understand that DSolve has difficulties when I input that particular D.E, so I was wondering if there is any way to go about it.

The message that I am getting is

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

But I don't understand how I could help Mathematica using the Reduce function. The DSolve command

DSolve[(Sin[c1]^2 Sin[c2] Sin[
    q[z]]^2 (3 Cos[q[z]] (1 + z^2 Derivative[1][q][z]^2) + 
     z Sin[q[z]] (3 Derivative[1][q][z] + 
        4 z^2 Derivative[1][q][z]^3 - z (q^\[Prime]\[Prime])[z])))/(
  z^5 (1 + z^2 Derivative[1][q][z]^2)^(3/2)) == 0, q[z], z]

and the result

During evaluation of In[104]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

During evaluation of In[104]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

Out[104]= DSolve[(
  Sin[c1]^2 Sin[c2] Sin[
    q[z]]^2 (3 Cos[q[z]] (1 + z^2 Derivative[1][q][z]^2) + 
     z Sin[q[z]] (3 Derivative[1][q][z] + 
        4 z^2 Derivative[1][q][z]^3 - z (q^\[Prime]\[Prime])[z])))/(
  z^5 (1 + z^2 Derivative[1][q][z]^2)^(3/2)) == 0, q[z], z]

Thanks in advance.

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  • 1
    $\begingroup$ Have you seen this? $\endgroup$ – J. M. is away Mar 14 '18 at 2:05
  • $\begingroup$ Thank you very much. No I hadn't seen that. There are some interesting approaches there. I will try these things, and if I get it to work I will update here with the solution. Thanks again. $\endgroup$ – Konstantinos Mar 14 '18 at 10:40
  • 1
    $\begingroup$ This also may be useful. (If it is, please up-vote the answer, which I think is extraordinary.) $\endgroup$ – bbgodfrey Mar 14 '18 at 15:38
  • $\begingroup$ The answer is indeed extraordinary and thanks for pointing this out, but I haven't made it to work for my example. I gave an upvote however as the answer is really good. $\endgroup$ – Konstantinos Mar 14 '18 at 16:53

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